Question 3

   Question: Let for any three distinct consecutive terms a, b, c of an A.P, the lines $ax + by + c = 0$ be concurrent at the point $P$ and $Q(\alpha, \beta)$ be a point such that the system of equations $x + y + z = 6$, $2x + 5y + \alpha z = \beta$ and $x + 2y + 3z = 4$ has infinitely many solutions. Then $(PQ)^2$ is equal to.

   Options: Numerical Answer Type.

   Correct Answer: 113.

   Year: 2024 (29-Jan-2024 Shift 2).

   Solution: ∵ a, b, c and in A.P $\Rightarrow 2b = a + c \Rightarrow a - 2b + c = 0$. $ax + by + c$ passes through fixed point $(1, -2)$. $\therefore P = (1, -2)$. For infinite solution, $D = D_1 = D_2 = D_3 = 0$. $D: \left| \begin{smallmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{smallmatrix} \right| = 0 \Rightarrow \alpha = 8$. $D_1: \left| \begin{smallmatrix} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{smallmatrix} \right| = 0 \Rightarrow \beta = 6$. $\therefore Q = (8, 6), \therefore PQ^2 = 113$.

   Step Solution:

    1.  Find Point P: In an A.P., the middle term relation $2b = a + c$ gives $a - 2b + c = 0$; comparing this to $ax + by + c = 0$ identifies the point of concurrency as $P(1, -2)$.

    2.  Determine $\alpha$ for Infinite Solutions: Set the coefficient determinant $\Delta = 0$, where $\left| \begin{smallmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{smallmatrix} \right| = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$.

    3.  Solve for $\alpha$: $15 - 2\alpha - 6 + \alpha - 1 = 0 \Rightarrow 8 - \alpha = 0 \Rightarrow \alpha = 8$.

    4.  Solve for $\beta$: Set $\Delta_x = 0$, where $\left| \begin{smallmatrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 4 & 2 & 3 \end{smallmatrix} \right| = 6(15-16) - 1(3\beta - 32) + 1(2\beta - 20) = 0$, resulting in $\beta = 6$.

    5.  Calculate $(PQ)^2$: Using $P(1, -2)$ and $Q(8, 6)$, $(PQ)^2 = (8-1)^2 + (6 - (-2))^2 = 49 + 64 = 113$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions) and Concurrency of Lines.

   Short cut solution: Observe the second equation is a linear combination of the first and third: $3 \times (x + 2y + 3z = 4) - (x + y + z = 6) = 2x + 5y + 8z = 6$. Thus, $\alpha = 8$ and $\beta = 6$.

 Question 5

   Question: Consider the system of linear equation $x + y + z = 4\mu, x + 2y + 2\lambda z = 10\mu, x + 3y + 4\lambda^2 z = \mu^2 + 15$ where $\lambda, \mu \in \mathbb{R}$. Which one of the following statements is NOT correct?

   Options: 

    A. The system has unique solution if $\lambda \neq 1 / 2$ and $\mu \neq 1, 15$.

    B. The system is inconsistent if $\lambda = 1 / 2$ and $\mu \neq 1$.

    C. The system has infinite number of solutions if $\lambda = 1 / 2$ and $\mu = 15$.

    D. The system is consistent if $\lambda \neq 1 / 2$.

   Correct Answer: B.

   Year: 2024 (30-Jan-2024 Shift 1).

   Solution: $\Delta = \left| \begin{smallmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{smallmatrix} \right| = (2\lambda - 1)^2$. For unique solution $\Delta \neq 0, 2\lambda - 1 \neq 0, (\lambda \neq 1/2)$. Let $\Delta = 0, \lambda = 1/2$. $\Delta_y = 0, \Delta_x = \Delta_z = \left| \begin{smallmatrix} 4\mu & 1 & 1 \\ 10\mu & 2 & 1 \\ \mu^2 + 15 & 3 & 1 \end{smallmatrix} \right| = (\mu - 15)(\mu - 1)$. For infinite solution $\lambda = 1/2$ and $\mu = 1$ or 15.

   Step Solution:

    1.  Find $\Delta$: The determinant of the coefficient matrix simplifies to $(2\lambda - 1)^2$.

    2.  Unique Solution Rule: A unique solution exists if $\Delta \neq 0$, implying $\lambda \neq 1/2$.

    3.  Evaluate $\lambda = 1/2$: If $\lambda = 1/2$, find conditions for consistency by setting $\Delta_x = 0$.

    4.  Solve for $\mu$: $\Delta_x = (\mu - 15)(\mu - 1) = 0$ implies the system has infinite solutions if $\mu = 1$ or $\mu = 15$.

    5.  Identify False Statement: Statement B is incorrect because if $\lambda = 1/2$ and $\mu = 15$, the system is consistent (infinite solutions), but B claims it is inconsistent for all $\mu \neq 1$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: For $\lambda = 1/2$, the equations become planes with parallel normals. Consistency then depends on the constants; solving the resulting two-variable system shows consistency only at $\mu = 1$ and $\mu = 15$.

 Question 6

   Question: Consider the system of linear equations $x + y + z = 5, x + 2y + \lambda^2 z = 9, x + 3y + \lambda z = \mu$, where $\lambda, \mu \in \mathbb{R}$. Then, which the following statement is NOT correct?

   Options: 

    A. System has infinite number of solution if $\lambda = 1$ and $\mu = 13$.

    B. System is inconsistent if $\lambda = 1$ and $\mu \neq 13$.

    C. System is consistent if $\lambda \neq 1$ and $\mu = 13$.

    D. System has unique solution if $\lambda \neq 1$ and $\mu \neq 13$.

   Correct Answer: D.

   Year: 2024 (30-Jan-2024 Shift 2).

   Solution: $\left| \begin{smallmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{smallmatrix} \right| = 0 \Rightarrow 2\lambda^2 - \lambda - 1 = 0 \Rightarrow \lambda = 1, -1/2$. $\left| \begin{smallmatrix} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{smallmatrix} \right| = 0 \Rightarrow \mu = 13$. Infinite solution $\lambda = 1$ & $\mu = 13$. For unique solution $\lambda \neq 1$. Considering the case when $\lambda = -1/2$ and $\mu \neq 13$ this will generate no solution case.

   Step Solution:

    1.  Find $\Delta$: The determinant of the coefficients is $-(2\lambda + 1)(\lambda - 1)$.

    2.  Locate $\Delta = 0$: The determinant is zero when $\lambda = 1$ or $\lambda = -1/2$.

    3.  Condition for Unique Solution: A unique solution requires $\Delta \neq 0$, meaning $\lambda$ cannot be $1$ and cannot be $-1/2$.

    4.  Evaluate Statement D: Statement D claims a unique solution exists as long as $\lambda \neq 1$. 

    5.  Verify Counter-example: If $\lambda = -1/2$, the system does not have a unique solution (it's inconsistent or infinite), proving Statement D incorrect.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Non-Homogeneous Systems).

   Short cut solution: Unique solution only exists if $\Delta \neq 0$. Since $\Delta$ has two roots ($1$ and $-1/2$), a statement claiming only one value of $\lambda$ must be excluded for uniqueness is logically incomplete and false.

 Question 7

   Question: If the system of linear equations $x - 2y + z = -4$, $2x + \alpha y + 3z = 5$, $3x - y + \beta z = 3$ has infinitely many solutions, then $12\alpha + 13\beta$ is equal to.

   Options: 

    A. 60

    B. 64

    C. 54

    D. 58.

   Correct Answer: D.

   Year: 2024 (31-Jan-2024 Shift 1).

   Solution: $\mathbb{D} = \left| \begin{smallmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{smallmatrix} \right| = \alpha\beta + 3 + 4\beta - 18 - 2 - 3\alpha$. For infinite solutions $\mathbb{D} = 0 \Rightarrow \alpha\beta - 3\alpha + 4\beta = 17$ (1). $D_2 = \left| \begin{smallmatrix} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{smallmatrix} \right| = 1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 0 \Rightarrow 13\beta = 54, \beta = \frac{54}{13}$. Put in (1): $\frac{54}{13}\alpha - 3\alpha + 4(\frac{54}{13}) = 17 \Rightarrow 15\alpha = 5, \alpha = \frac{1}{3}$. $12\alpha + 13\beta = 4 + 54 = 58$.

   Step Solution: 

    1.  Set $\Delta = 0$: Expand the coefficient determinant to get $\alpha\beta - 3\alpha + 4\beta = 17$.

    2.  Set $\Delta_2 = 0$: Solve the determinant formed by replacing the second column with constants.

    3.  Solve for $\beta$: $1(5\beta - 9) + 8\beta - 36 - 9 = 0$ leads to $\beta = \frac{54}{13}$.

    4.  Solve for $\alpha$: Substitute $\beta$ into the first equation to find $\alpha = \frac{1}{3}$.

    5.  Compute Value: $12(\frac{1}{3}) + 13(\frac{54}{13}) = 4 + 54 = 58$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: Use the consistency condition where $R_3 = m R_1 + n R_2$. By comparing constants and $x$-coefficients, $m+2n=3$ and $-4m+5n=3$; solving gives $m=\frac{9}{13}, n=\frac{15}{13}$. Then $\beta = m(1) + n(3) = \frac{54}{13}$ and $-1 = m(-2) + n(\alpha) \Rightarrow \alpha = \frac{1}{3}$.

 Question 10

   Question: If the system of equations $2x + 3y - z = 5$, $x + \alpha y + 3z = -4$, $3x - y + \beta z = 7$ has infinitely many solutions, then $13\alpha\beta$ is equal to.

   Options: 

    A. 1110

    B. 1120

    C. 1210

    D. (Not provided in source).

   Correct Answer: B.

   Year: 2024 (1-Feb-2024 Shift 1).

   Solution: Using family of planes: $2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$. $2 = k_1 + 3k_2$, $3 = k_1\alpha - k_2$, $-1 = 3k_1 + \beta k_2$, $-5 = 4k_1 - 7k_2$. Solving we get $k_2 = \frac{13}{19}, k_1 = \frac{-1}{19}, \alpha = -70, \beta = \frac{-16}{13}$. $13\alpha\beta = 13(-70)(\frac{-16}{13}) = 1120$.

   Step Solution: 

    1.  Define Plane Relation: Set Equation 1 as a linear combination of Equations 2 and 3 using constants $k_1$ and $k_2$.

    2.  Find $k_1, k_2$: Solve $k_1 + 3k_2 = 2$ and $4k_1 - 7k_2 = -5$ to find $k_1 = -\frac{1}{19}$ and $k_2 = \frac{13}{19}$.

    3.  Solve for $\alpha$: Use the $y$-coefficient equation $k_1\alpha - k_2 = 3$ to find $\alpha = -70$.

    4.  Solve for $\beta$: Use the $z$-coefficient equation $3k_1 + \beta k_2 = -1$ to find $\beta = -\frac{16}{13}$.

    5.  Compute Product: Multiply values: $13 \times (-70) \times (-\frac{16}{13}) = 1120$.

   Difficulty Level: Hard.

   The Concept Name: Family of Planes (Linear Dependence for Infinite Solutions).

   Short cut solution: Use determinants: $\Delta = 0$ and $\Delta_x = 0$ to create two equations in $\alpha$ and $\beta$. Expanding $\Delta_x$ directly gives $\beta$ because the constant column is independent of $\alpha$, then solve for $\alpha$ using $\Delta=0$.

 Question 11

   Question: Let the system of equations $x + 2y + 3z = 5$, $2x + 3y + z = 9$, $4x + 3y + \lambda z = \mu$ have infinite number of solutions. Then $\lambda + 2\mu$ is equal to :.

   Options: 

    A. 28

    B. 17

    C. 22

    D. 15.

   Correct Answer: B.

   Year: 2024 (1-Feb-2024 Shift 2).

   Solution: For infinite solutions $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. $\Delta = \left| \begin{smallmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{smallmatrix} \right| = 0 \Rightarrow \lambda = -13$. $\Delta_1 = \left| \begin{smallmatrix} 5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13 \end{smallmatrix} \right| = 0 \Rightarrow \mu = 15$. For $\lambda = -13, \mu = 15$, system has infinite solutions, hence $\lambda + 2\mu = 17$.

   Step Solution: 

    1.  Find $\lambda$: Set the coefficient determinant $\Delta = 0$ and expand: $1(3\lambda - 3) - 2(2\lambda - 4) + 3(6 - 12) = 0$.

    2.  Solve for $\lambda$: $3\lambda - 3 - 4\lambda + 8 - 18 = 0 \Rightarrow -\lambda - 13 = 0 \Rightarrow \lambda = -13$.

    3.  Find $\mu$: Set $\Delta_1 = 0$ by replacing the $x$-column with constants $[5, 9, \mu]$.

    4.  Solve for $\mu$: $5(-39 - 3) - 2(-117 - \mu) + 3(27 - 3\mu) = 0 \Rightarrow \mu = 15$.

    5.  Final Calculation: $\lambda + 2\mu = -13 + 2(15) = 17$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: Observe row transformations: $R_3$ can be expressed as $a R_1 + b R_2$. Solving $a+2b=4$ and $2a+3b=3$ gives $a=-6, b=5$. Then $\lambda = a(3) + b(1) = -18+5 = -13$ and $\mu = a(5) + b(9) = -30+45 = 15$. $\lambda + 2\mu = -13 + 30 = 17$.

 Question 12

   Question: If the system of equations $x + 2y + 3z = 3$, $4x + 3y - 4z = 4$, and $8x + 4y - \lambda z = 9 + \mu$ has infinitely many solutions, then the ordered pair $(\lambda, \mu)$ is equal to:

   Options: 

    A. $(\frac{72}{5}, \frac{21}{5})$

    B. $(-\frac{72}{5}, -\frac{21}{5})$

    C. $(\frac{72}{5}, -\frac{21}{5})$

    D. $(-\frac{72}{5}, \frac{21}{5})$

   Correct Answer: C

   Year: 2023 (24-Jan-2023 Shift 2)

   Solution: $x + 2y + 3z = 3$ (i); $4x + 3y - 4z = 4$ (ii); $8x + 4y - \lambda z = 9 + \mu$ (iii). (i) $\times 4 - (ii) \Rightarrow 5y + 16z = 8$ (iv). (ii) $\times 2 - (iii) \Rightarrow 2y + (\lambda - 8)z = -1 - \mu$ (v). (iv) $\times 2 - (v) \times 5 \Rightarrow (32 - 5(\lambda - 8))z = 16 - 5(-1 - \mu)$. For infinite solutions $\Rightarrow 72 - 5\lambda = 0 \Rightarrow \lambda = \frac{72}{5}$ and $21 + 5\mu = 0 \Rightarrow \mu = \frac{-21}{5}$.

   Step Solution:

    1.  Eliminate $x$ from first two equations: $(4x + 8y + 12z = 12) - (4x + 3y - 4z = 4)$ gives $5y + 16z = 8$.

    2.  Eliminate $x$ from last two equations: $(8x + 6y - 8z = 8) - (8x + 4y - \lambda z = 9 + \mu)$ gives $2y + (\lambda - 8)z = -1 - \mu$.

    3.  Create a single equation in $z$: Multiply the first result by 2 and the second by 5: $(10y + 32z = 16) - (10y + 5(\lambda - 8)z = -5 - 5\mu)$.

    4.  Solve for $\lambda$: For infinite solutions, the coefficient of $z$ must be zero: $32 - 5\lambda + 40 = 0 \Rightarrow 5\lambda = 72 \Rightarrow \lambda = \frac{72}{5}$.

    5.  Solve for $\mu$: The constant term must also be zero: $16 - (-5 - 5\mu) = 0 \Rightarrow 21 + 5\mu = 0 \Rightarrow \mu = -\frac{21}{5}$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use the determinant method. For infinite solutions, $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$. $\Delta = \left| \begin{smallmatrix} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{smallmatrix} \right| = 0$ leads to $1(-3\lambda + 16) - 2(-4\lambda + 32) + 3(16 - 24) = 0 \Rightarrow 5\lambda = 72$. Similarly, setting $\Delta_z = 0$ provides $\mu = -\frac{21}{5}$.

 Question 13

   Question: Let $S_1$ and $S_2$ be respectively the sets of all $a \in \mathbb{R} - \{0\}$ for which the system of linear equations $ax + 2ay - 3az = 1$, $(2a + 1)x + (2a + 3)y + (a + 1)z = 2$, and $(3a + 5)x + (a + 5)y + (a + 2)z = 3$ has unique solution and infinitely many solutions. Then:

   Options: 

    A. $n(S_1) = 2$ and $S_2$ is an infinite set

    B. $S_1$ is an infinite set and $n(S_2) = 2$

    C. $S_1 = \Phi$ and $S_2 = \mathbb{R} - \{0\}$

    D. $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$

   Correct Answer: D

   Year: 2023 (25-Jan-2023 Shift 1)

   Solution: $\Delta = \left| \begin{smallmatrix} a & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{smallmatrix} \right| = a(15a^2 + 31a + 36)$. $\Delta = 0 \Rightarrow a = 0$. $\Delta \neq 0$ for all $a \in \mathbb{R} - \{0\}$. Hence $S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$.

   Step Solution:

    1.  Set up the determinant $\Delta$: Use the coefficients of $x, y, z$ to form the matrix determinant.

    2.  Simplify the determinant: Factor out '$a$' from the first row: $a \times \left| \begin{smallmatrix} 1 & 2 & -3 \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{smallmatrix} \right|$.

    3.  Calculate the expression: Expanding the determinant results in the polynomial $a(15a^2 + 31a + 36)$.

    4.  Analyze the quadratic factor: The discriminant of $15a^2 + 31a + 36$ is $31^2 - 4(15)(36) = 961 - 2160 = -1199$. [Outside Source] Since the discriminant is negative and the leading coefficient is positive, the quadratic is always positive for all real $a$. [Outside Source]

    5.  Determine sets: $\Delta$ is only zero when $a=0$. Since the domain is $a \in \mathbb{R} - \{0\}$, $\Delta$ is never zero, meaning a unique solution always exists ($S_1 = \mathbb{R} - \{0\}$) and infinite solutions never exist ($S_2 = \Phi$).

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: In non-homogeneous systems, if the determinant of the coefficient matrix $\Delta$ is a polynomial that is never zero in the given domain, the system must have a unique solution for all values in that domain.

 Question 14

   Question: Consider the following system of equations: $\alpha x + 2y + z = 1$, $2\alpha x + 3y + \alpha z = 1$, $3x + \alpha y + 2z = \beta$ for some $\alpha, \beta \in \mathbb{R}$. Then which of the following is NOT correct?

   Options: 

    A. It has no solution if $\alpha = -1$ and $\beta \neq 2$

    B. It has no solution for $\alpha = -1$ and for all $\beta \in \mathbb{R}$

    C. It has no solution for $\alpha = 3$ and for all $\beta \neq 2$

    D. It has a solution for all $\alpha \neq -1$ and $\beta = 2$

   Correct Answer: B

   Year: 2023 (29-Jan-2023 Shift 1)

   Solution: $D = \left| \begin{smallmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{smallmatrix} \right| = 0 \Rightarrow \alpha = -1, 3$. $\Delta D_x = \dots = 0 \Rightarrow \beta = 2$. $D_y = 0, D_z = 0$.

   Step Solution:

    1.  Calculate $\Delta$: Expanding the coefficient determinant gives $\alpha(6 - \alpha) - 2(4\alpha - 3) + 1(2\alpha^2 - 9) = \alpha^2 - 2\alpha - 3$.

    2.  Solve for $\Delta = 0$: $(\alpha - 3)(\alpha + 1) = 0$ gives $\alpha = 3$ or $\alpha = -1$.

    3.  Determine condition for infinite solutions: For the system to have infinite solutions at these $\alpha$ values, $\Delta_x, \Delta_y, \Delta_z$ must also be zero; solving $\Delta_x = 0$ gives $\beta = 2$.

    4.  Analyze $\alpha = -1$: If $\alpha = -1$ and $\beta = 2$, the system has infinite solutions (consistent). If $\alpha = -1$ and $\beta \neq 2$, it has no solution.

    5.  Identify incorrect statement: Statement B claims the system has no solution for $\alpha = -1$ for all $\beta$, but if $\beta = 2$, a solution exists, making B false.

   Difficulty Level: Hard.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Check the boundaries of the statement. If a statement says "for all $\beta$" and you find even one $\beta$ (like $\beta=2$) where the system is consistent, that statement is immediately the "Not Correct" answer.

 Question 15

   Question: Let the system of linear equations $x + y + kz = 2$, $2x + 3y - z = 1$, $3x + 4y + 2z = k$ have infinitely many solutions. Then the system $(k + 1)x + (2k - 1)y = 7$ and $(2k + 1)x + (k + 5)y = 10$ has:

   Options: 

    A. infinitely many solutions

    B. unique solution satisfying $x - y = 1$

    C. no solution

    D. unique solution satisfying $x + y = 1$

   Correct Answer: D

   Year: 2023 (30-Jan-2023 Shift 1)

   Solution: $\begin{array} { r } { \left| \begin{array} { c c c } { 1 } & { 1 } & { \mathbf { k } } \\ { 2 } & { 3 } & { - 1 } \\ { 3 } & { 4 } & { 2 } \end{array} \right| \quad = 0 } \end{array}$ $\Rightarrow 1(10) - 1(7) + k(-1) = 0 \Rightarrow k = 3$. For $k = 3$, $2^{nd}$ system is $4x + 5y = 7 \dots (1)$ and $7x + 8y = 10 \dots (2)$. Clearly, they have a unique solution $(2) - (1) \Rightarrow 3x + 3y = 3 \Rightarrow x + y = 1$.

   Step Solution:

    1.  Find $k$ for Infinite Solutions: Set the determinant of the first system to zero: $1(6 - (-4)) - 1(4 - (-3)) + k(8 - 9) = 0$.

    2.  Solve for $k$: $10 - 7 - k = 0 \Rightarrow k = 3$.

    3.  Form the Second System: Substitute $k = 3$ into the second set of equations: $4x + 5y = 7$ and $7x + 8y = 10$.

    4.  Solve for Relation: Subtract the first equation from the second: $(7x - 4x) + (8y - 5y) = 10 - 7$, which gives $3x + 3y = 3$.

    5.  Identify Final Condition: Simplify the result to $x + y = 1$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: In the first system, observe that Row 1 + Row 2 = Row 3 would imply infinite solutions if the constants also match. Comparing $z$ coefficients: $k + (-1) = 2 \Rightarrow k=3$. Check constants: $2 + 1 = 3$ (which is $k$). Thus, $k=3$ is immediate.

 Question 16

   Question: For $\alpha$, $\beta \in \mathbb{R}$, suppose the system of linear equations $x - y + z = 5$, $2x + 2y + \alpha z = 8$, $3x - y + 4z = \beta$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of:

   Options: 

    A. $x^2 - 10x + 16 = 0$

    B. $x^2 + 18x + 56 = 0$

    C. $x^2 - 18x + 56 = 0$

    D. $x^2 + 14x + 24 = 0$

   Correct Answer: C

   Year: 2023 (30-Jan-2023 Shift 2)

   Solution: $\begin{array} { r l } { { \left| \begin{array} { l l l } { 1 } & { - 1 } & { 1 } \\ { 2 } & { 2 } & { \alpha } \\ { 3 } & { - 1 } & { 4 } \end{array} \right| } } & { { } = 0 ; 8 + \alpha - 2 ( - 4 + 1 ) + 3 ( - \alpha - 2 ) = 0 } \end{array}$ $\begin{array} { l } { 8 + \alpha + 6 - 3 \alpha - 6 = 0 } \\ { \alpha = 4 } \end{array}$. (Calculation for $\beta = 14$ is implied to form the quadratic $x^2 - 18x + 56 = 0$).

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant: $1(8 - (-\alpha)) - (-1)(8 - 3\alpha) + 1(-2 - 6) = 0$.

    2.  Solve for $\alpha$: $8 + \alpha + 8 - 3\alpha - 8 = 0 \Rightarrow 8 - 2\alpha = 0 \Rightarrow \alpha = 4$.

    3.  Solve for $\beta$: For infinite solutions, $\Delta_x$ (replacing the $x$-column with constants $[5, 8, \beta]$) must be 0: $5(8+\alpha) + 1(32 - 8\alpha) + 1(-8 - 2\beta) = 0$. Substituting $\alpha=4$ gives $60 + 0 - 8 - 2\beta = 0$. [Source calculation: $84 - 6\beta = 0 \Rightarrow \beta = 14$].

    4.  Find Equation Roots: The roots are 4 and 14.

    5.  Form Quadratic: Equation is $x^2 - (\text{sum})x + (\text{product}) = 0 \Rightarrow x^2 - 18x + 56 = 0$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: Use row operations to find $\beta$ faster. Row 3 - Row 1 = $2x + 3z = \beta - 5$. The second equation is $2x + 2y + \alpha z = 8$. For these to allow infinite solutions along with the first, they must be linearly dependent. Solving directly via determinants remains the most reliable method for finding both variables.

 Question 17

   Question: For the system of linear equations $x + y + z = 6$, $\alpha x + \beta y + 7z = 3$, $x + 2y + 3z = 14$, which of the following is NOT true?

   Options: 

    A. If $\alpha = \beta = 7$, then the system has no solution.

    B. If $\alpha = \beta$ and $\alpha \neq 7$ then the system has a unique solution.

    C. There is a unique point $(\alpha, \beta)$ on the line $x + 2y + 18 = 0$ for which the system has infinitely many solutions.

    D. For every point $(\alpha, \beta) \neq (7, 7)$ on the line $x - 2y + 7 = 0$, the system has infinitely many solutions.

   Correct Answer: D

   Year: 2023 (31-Jan-2023 Shift 1)

   Solution: By equation 1 and 3: $y + 2z = 8 \Rightarrow y = 8 - 2z$ and $x = -2 + z$. Now putting in equation 2: $\alpha ( z - 2 ) + \beta ( - 2 z + 8 ) + 7 z = 3 \Rightarrow ( \alpha - 2 \beta + 7 ) z = 2 \alpha - 8 \beta + 3$. Equations have unique solution if $\alpha - 2 \beta + 7 \neq 0$. Equations have no solution if $\alpha - 2 \beta + 7 = 0$ and $2 \alpha - 8 \beta + 3 \neq 0$. Equations have infinite solution if $\alpha - 2 \beta + 7 = 0$ and $2 \alpha - 8 \beta + 3 = 0$.

   Step Solution:

    1.  Reduce Variables: Use Eq 1 ($x+y+z=6$) and Eq 3 ($x+2y+3z=14$) to find $y=8-2z$ and $x=z-2$.

    2.  Substitute into Eq 2: Replace $x$ and $y$ in $\alpha x + \beta y + 7z = 3$ to get $\alpha(z-2) + \beta(8-2z) + 7z = 3$.

    3.  Group by $z$: Rearrange as $(\alpha - 2\beta + 7)z = 2\alpha - 8\beta + 3$.

    4.  Analyze Infinite solutions: This requires both the coefficient and constant to be zero: $\alpha - 2\beta + 7 = 0$ AND $2\alpha - 8\beta + 3 = 0$.

    5.  Evaluate D: Option D is false because infinite solutions only occur at the specific intersection point of those two lines, not for "every point" on $x - 2y + 7 = 0$.

   Difficulty Level: Hard.

   The Concept Name: Consistency of Linear Equations.

   Short cut solution: Infinite solutions require the second equation to be a specific linear combination of the first and third. This occurs at only one specific point $(\alpha, \beta)$. Any statement claiming infinite solutions for an entire line of points (like D) is mathematically impossible for this system.

 Question 18

   Question: Let $S$ denote the set of all real values of $\lambda$ such that the system of equations

    $\lambda x + y + z = 1$

    $x + \lambda y + z = 1$

    $x + y + \lambda z = 1$

    is inconsistent, then $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|)$ is equal to.

   Options: 

    A. 2 

    B. 12 

    C. 4 

    D. 6

   Correct Answer: D

   Year: 2023 (1-Feb-2023 Shift 1)

   Solution: $\begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0 \Rightarrow (\lambda + 2)(\lambda - 1)^2 = 0 \Rightarrow \lambda = -2, 1$. $\Sigma (|-2|^2 + |-2|) = 6$.

   Step Solution:

    1.  Set $\Delta = 0$: Find the values of $\lambda$ where the system is not unique by solving the determinant of coefficients: $\lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda) = 0$.

    2.  Factorize: This simplifies to $(\lambda + 2)(\lambda - 1)^2 = 0$, giving potential values $\lambda = 1$ and $\lambda = -2$.

    3.  Test $\lambda = 1$: When $\lambda = 1$, all equations become $x + y + z = 1$. This system has infinitely many solutions, so $\lambda = 1$ is not in set $S$. [Outside Source]

    4.  Test $\lambda = -2$: When $\lambda = -2$, adding the three equations gives $0 = 3$. This is a contradiction, meaning the system is inconsistent, so $S = \{-2\}$. [Outside Source]

    5.  Calculate Final Value: $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) = |-2|^2 + |-2| = 4 + 2 = 6$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: In a cyclic system where the right-hand side constants are equal ($1, 1, 1$), $\lambda = 1$ always results in identical planes (infinite solutions). For $\lambda = -(n-1) = -2$, the sum of the left sides is zero while the right side is $3$, immediately indicating inconsistency.

 Question 19

   Question: For the system of linear equations $ax + y + z = 1$, $x + ay + z = 1$, $x + y + az = \beta$, which one of the following statements is NOT correct?

   Options: 

    A. It has infinitely many solutions if $\alpha = 2$ and $\beta = -1$.

    B. It has no solution if $\alpha = -2$ and $\beta = 1$.

    C. $x + y + z = \frac{3}{4}$ if $\alpha = 2$ and $\beta = 1$.

    D. It has infinitely many solutions if $\alpha = 1$ and $\beta = 1$.

   Correct Answer: A

   Year: 2023 (1-Feb-2023 Shift 2)

   Solution: $\begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = (\alpha - 1)^2 (\alpha + 2)$. For $\alpha = 2$, $\Delta = (1)^2(4) = 4 \neq 0$, giving a unique solution.

   Step Solution:

    1.  Calculate Determinant: The coefficient determinant $\Delta = (\alpha - 1)^2(\alpha + 2)$.

    2.  Unique Solution Rule: If $\Delta \neq 0$, the system must have a unique solution. This occurs when $\alpha \neq 1$ and $\alpha \neq -2$.

    3.  Analyze $\alpha = 2$: Since $2 \neq 1$ and $2 \neq -2$, a unique solution exists for any $\beta$. Statement A claims "infinitely many solutions" for $\alpha = 2$, which is impossible.

    4.  Verify Statement D: If $\alpha = 1$ and $\beta = 1$, all equations are $x + y + z = 1$, which has infinite solutions. Statement D is correct.

    5.  Verify Statement C: If $\alpha = 2$ and $\beta = 1$, the equations are $2x+y+z=1, x+2y+z=1, x+y+2z=1$. Adding them gives $4(x+y+z) = 3 \Rightarrow x+y+z = 3/4$. Statement C is correct.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Non-Homogeneous Systems).

   Short cut solution: Identify values that make $\Delta = 0$ ($\alpha=1, -2$). Since $\alpha=2$ does not make the determinant zero, the system is guaranteed to have a unique solution, making any claim of "infinite solutions" for $\alpha=2$ immediately incorrect.

 Question 20

   Question: If the system of equations $x + y + az = b$, $2x + 5y + 2z = 6$, $x + 2y + 3z = 3$ has infinitely many solutions, then $2a + 3b$ is equal to :

   Options: 

    A. 28 

    B. 20 

    C. 25 

    D. 23

   Correct Answer: D

   Year: 2023 (6-Apr-2023 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 0 \Rightarrow 11 - 4 - a = 0 \Rightarrow a = 7$. $\Delta_z = \begin{vmatrix} 1 & 1 & b \\ 2 & 5 & 6 \\ 1 & 2 & 3 \end{vmatrix} = 0 \Rightarrow 3 - 0 - b = 0 \Rightarrow b = 3$. $2a + 3b = 23$.

   Step Solution:

    1.  Set $\Delta = 0$: Expand along the first row: $1(15-4) - 1(6-2) + a(4-5) = 0$.

    2.  Solve for $a$: $11 - 4 - a = 0 \Rightarrow a = 7$.

    3.  Set $\Delta_z = 0$: Use the determinant where the $z$-column is replaced by constants: $1(15-12) - 1(6-6) + b(4-5) = 0$.

    4.  Solve for $b$: $3 - 0 - b = 0 \Rightarrow b = 3$.

    5.  Final Calculation: $2a + 3b = 2(7) + 3(3) = 14 + 9 = 23$.

   Difficulty Level: Easy.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use row operations. $Eq_2 - 2 \times Eq_3$ gives $y - 4z = 0$. Also $Eq_3 - Eq_1$ gives $y + (3-a)z = 3-b$. For infinite solutions, these two equations in $y$ and $z$ must be identical: $3-a = -4 \Rightarrow \mathbf{a=7}$ and $3-b = 0 \Rightarrow \mathbf{b=3}$. $2(7)+3(3)=23$.

 Question 23

   Question: For the system of linear equations $2x - y + 3z = 5$, $3x + 2y - z = 7$, $4x + 5y + \alpha z = \beta$, which of the following is NOT correct?

   Options: 

    A. The system is inconsistent for $\alpha = -5$ and $\beta = 8$.

    B. The system has infinitely many solutions for $\alpha = -6$ and $\beta = 9$.

    C. The system has a unique solution for $\alpha \neq -5$ and $\beta = 8$.

    D. The system has infinitely many solutions for $\alpha = -5$ and $\beta = 9$.

   Correct Answer: B

   Year: 2023 (10-Apr-2023 Shift 1)

   Solution: $\Delta = \left| \begin{smallmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{smallmatrix} \right| = 7\alpha + 35 = 7(\alpha + 5)$. For unique solution $\Delta \neq 0 \Rightarrow \alpha \neq -5$. For $\alpha = -5$ and $\beta = 8$, $\Delta_1 = -5(\beta - 9) = 5 \neq 0$, so system is inconsistent. For infinite solution, $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0 \Rightarrow \alpha = -5, \beta = 9$.

   Step Solution:

    1.  Calculate $\Delta$: The coefficient determinant simplifies to $7\alpha + 35 = 7(\alpha + 5)$.

    2.  Unique Solution Rule: A unique solution exists if $\Delta \neq 0$, which means $\alpha \neq -5$.

    3.  Analyze Inconsistency: If $\alpha = -5$, the system is inconsistent if any of $\Delta_x, \Delta_y, \Delta_z \neq 0$. Calculating $\Delta_x$ gives $-5(\beta - 9)$.

    4.  Evaluate Infinite Solutions: Infinite solutions require $\Delta = 0$ (so $\alpha = -5$) and $\Delta_x = 0$ (so $\beta = 9$).

    5.  Identify False Statement: Option B is incorrect because if $\alpha = -6$, $\Delta \neq 0$, meaning the system must have a unique solution, not infinitely many.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: Focus on $\alpha$. The determinant $\Delta$ only depends on $\alpha$. If $\alpha \neq -5$, the solution is unique. Statement B claims infinite solutions for $\alpha = -6$, which is impossible because $\alpha = -6$ leads to a unique solution.

 Question 24

   Question: Let $S$ be the set of values of $\lambda$ for which the system of equations $2\lambda x - 3y + 3z = 4\lambda^2$, $2x + 6\lambda y + 4z = 1$, $3x + 2y + 3\lambda z = \lambda$ has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to.

   Options: Numerical Answer Type.

   Correct Answer: 24

   Year: 2023 (10-Apr-2023 Shift 2)

   Solution: $\Delta = \left| \begin{smallmatrix} 2\lambda & -3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda \end{smallmatrix} \right| = 0 \Rightarrow ( \lambda - 1 ) ( 3 \lambda + 1 ) ( 3 \lambda + 2 ) = 0$. Roots are $\lambda = 1, -1/3, -2/3$. For these values, $\Delta_1 \neq 0$. $12(1 + 1/3 + 2/3) = 24$.

   Step Solution:

    1.  Set up Determinant: Set the coefficient determinant $\Delta$ to zero to find values where the system is not unique.

    2.  Form the Equation: Expanding the determinant leads to the cubic equation $18\lambda^3 - 14\lambda - 4 = 0$.

    3.  Find the Roots: Factoring gives roots $\lambda = 1, -1/3, -2/3$.

    4.  Verify No Solution: Check if the constant-replaced determinant $\Delta_1$ is non-zero for these roots to ensure inconsistency.

    5.  Compute the Final Value: $12 \times (|1| + |-1/3| + |-2/3|) = 12 \times (1 + 1/3 + 2/3) = 12 \times 2 = 24$.

   Difficulty Level: Hard.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Use the rational root theorem to identify $\lambda = 1$ as a root quickly (sum of coefficients is zero). Perform synthetic division to find the other two fractional roots.

 Question 26

   Question: If the system of linear equations $7x + 11y + \alpha z = 13$, $5x + 4y + 7z = \beta$, $175x + 194y + 57z = 361$ has infinitely many solutions, then $\alpha + \beta + 2$ is equal to.

   Options: 

    A. 3

    B. 6

    C. 5

    D. 4

   Correct Answer: 4

   Year: 2023 (11-Apr-2023 Shift 2)

   Solution: Condition of Infinite Many solution $\Delta = 0 \& \Delta x, \Delta y, \Delta z = 0$. After solving we get $\alpha + \beta + 2 = 4$ [Source shows result of calculation directly].

   Step Solution:

    1.  Analyze Row Dependency: For infinite solutions, the third row must be a linear combination of the first two ($R_3 = m R_1 + n R_2$).

    2.  Solve for Multipliers: Solve $7m + 5n = 175$ and $11m + 4n = 194$ to find $m = 10$ and $n = 21$.

    3.  Find $\alpha$: Using the $z$-coefficients, $10(\alpha) + 21(7) = 57 \Rightarrow 10\alpha = -90$, so $\alpha = -9$.

    4.  Find $\beta$: Using the constant terms, $10(13) + 21(\beta) = 361 \Rightarrow 21\beta = 231$, so $\beta = 11$.

    5.  Final Calculation: $\alpha + \beta + 2 = -9 + 11 + 2 = 4$.

   Difficulty Level: Medium.

   The Concept Name: Linear Dependence of Equations (Infinite Solutions).

   Short cut solution: Use $R_3 = 10R_1 + 21R_2$. Once the multipliers $10$ and $21$ are found by comparing the $x$ and $y$ coefficients, the values for $\alpha$ and $\beta$ can be determined through simple arithmetic.

Question 29

   Question: For the system of linear equations $2x + 4y + 2az = b, x + 2y + 3z = 4, 2x - 5y + 2z = 8$, which of the following is NOT correct?

   Options: 

    A. It has infinitely many solutions if $a = 3, b = 8$.

    B. It has unique solution if $a = b = 8$.

    C. It has unique solution if $a = b = 6$.

    D. It has infinitely many solutions if $a = 3, b = 6$.

   Correct Answer: D

   Year: 2023 (13-Apr-2023 Shift 1)

   Solution: $\Delta = \left| \begin{smallmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{smallmatrix} \right| = 18(3 - a)$. $\Delta_x = (64 + 19b - 72a)$. For unique solution $\Delta \neq 0 \Rightarrow a \neq 3$ and $b \in R$. For infinitely many solution: $\Delta = \Delta_x = \Delta_y = \Delta_z = 0 \Rightarrow a = 3$.,

   Step Solution:

    1.  Calculate $\Delta$: Expanding the coefficient determinant: $2(4 - (-15)) - 4(2 - 6) + 2a(-5 - 4) = 38 + 16 - 18a = 54 - 18a = 18(3 - a)$.

    2.  Unique Solution Rule: A unique solution exists if $\Delta \neq 0$, which means $a \neq 3$.

    3.  Find $\Delta_x$: Substitute the constants into the first column: $\left| \begin{smallmatrix} b & 4 & 2a \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{smallmatrix} \right| = b(19) - 4(8-24) + 2a(-20-16) = 19b + 64 - 72a$.

    4.  Analyze Infinite Solutions: If $a=3$, $\Delta_x = 19b + 64 - 216 = 19b - 152$. For infinite solutions, $\Delta_x$ must be $0$, so $b = 152/19 = 8$.

    5.  Evaluate D: Option D claims infinite solutions for $a=3, b=6$. However, if $b=6$, then $\Delta_x = 114 - 152 = -38 \neq 0$, making the system inconsistent (no solution), not infinite.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: Check the $a$ value first. The system only fails to have a unique solution when $a=3$. Statements B and C are immediately correct because $a \neq 3$. For $a=3$, the equations are consistent only if the second constant $b$ follows the pattern; calculations show $b=8$ is required, so $b=6$ must be wrong.

 Question 30

   Question: If the system of equations $2x + y - z = 5$, $2x - 5y + \lambda z = \mu$, $x + 2y - 5z = 7$ has infinitely many solutions, then $(\lambda + \mu)^2 + (\lambda - \mu)^2$ is equal to.

   Options: 

    A. 904

    B. 916

    C. 912

    D. 920

   Correct Answer: B

   Year: 2023 (13-Apr-2023 Shift 2)

   Solution: $\Delta = 0 \Rightarrow \left| \begin{smallmatrix} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{smallmatrix} \right| = 0 \Rightarrow 2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5) = 0 \Rightarrow 51 - 3\lambda = 0 \Rightarrow \lambda = 17$. $\Delta_x = 0 \Rightarrow 5(25 - 34) - 1(-5\mu - 119) - 1(2\mu + 35) = 0 \Rightarrow 39 + 3\mu = 0 \Rightarrow \mu = -13$. Expression $= 4^2 + (30)^2 = 916$.

   Step Solution:

    1.  Find $\lambda$: Set the coefficient determinant $\Delta = 0$: $2(25-2\lambda) - 1(-10-\lambda) - 1(9) = 0$.

    2.  Solve for $\lambda$: $50 - 4\lambda + 10 + \lambda - 9 = 0 \Rightarrow 51 - 3\lambda = 0 \Rightarrow \lambda = 17$.

    3.  Find $\mu$: Set $\Delta_x = 0$ using $\lambda=17$: $\left| \begin{smallmatrix} 5 & 1 & -1 \\ \mu & -5 & 17 \\ 7 & 2 & -5 \end{smallmatrix} \right| = 5(-9) - 1(-5\mu - 119) - 1(2\mu + 35) = 0$.

    4.  Solve for $\mu$: $-45 + 5\mu + 119 - 2\mu - 35 = 0 \Rightarrow 3\mu + 39 = 0 \Rightarrow \mu = -13$.

    5.  Final Calculation: $(\lambda + \mu)^2 + (\lambda - \mu)^2 = (17 - 13)^2 + (17 + 13)^2 = 4^2 + 30^2 = 16 + 900 = 916$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use the identity $(\lambda + \mu)^2 + (\lambda - \mu)^2 = 2(\lambda^2 + \mu^2)$. After finding $\lambda = 17$ and $\mu = -13$, the calculation becomes $2(289 + 169) = 2(458) = 916$.

 Question 31

   Question: Let the system of linear equations $-x + 2y - 9z = 7, -x + 3y + 7z = 9, -2x + y + 5z = 8, -3x + y + 13z = \lambda$ has a unique solution $x = \alpha, y = \beta, z = \gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x - 2y + z = \lambda$ is.

   Options: 

    A. 7

    B. 9

    C. 13

    D. 11

   Correct Answer: A

   Year: 2023 (15-Apr-2023 Shift 1)

   Solution: Subtracting equations: $(2)-(1) \Rightarrow y + 16z = 2$. $(3)-2(1) \Rightarrow -3y + 23z = -6$. Solving gives $z=0, y=2, x=-3$. Point is $(-3, 2, 0)$. Put in fourth eq: $-3(-3)+2 = \lambda \Rightarrow \lambda = 11$. Distance $d = |(-6-4-11)/3| = 7$.

   Step Solution:

    1.  Eliminate $x$: From equations 1 and 2, $(-x + 3y + 7z) - (-x + 2y - 9z) = 9 - 7 \Rightarrow y + 16z = 2$.

    2.  Solve for variables: Using equation 3 and equation 1, $(-2x + y + 5z) - 2(-x + 2y - 9z) = 8 - 14 \Rightarrow -3y + 23z = -6$.

    3.  Coordinate Calculation: Multiplying the first result by 3 and adding to the second gives $71z = 0$, so $z=0$. This leads to $y=2$ and $x=-3$.

    4.  Find $\lambda$: Substitute $(-3, 2, 0)$ into the fourth equation: $-3(-3) + 2 + 13(0) = 11$. So, $\lambda = 11$.

    5.  Calculate Distance: Point $P(-3, 2, 0)$ to plane $2x - 2y + z - 11 = 0$: $d = \frac{|2(-3) - 2(2) + 0 - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-21|}{3} = 7$.

   Difficulty Level: Hard.

   The Concept Name: Solving Systems of Equations and Distance from Point to Plane.

   Short cut solution: For a unique solution to exist among four equations, the first three must be consistent. Solving them yields $z=0$ immediately. Once $z=0$, the system reduces to two variables, making the discovery of the point $(-3, 2, 0)$ very fast. Substitute and apply the standard distance formula.

 Question 33

   Question: The system of equations $-kx + 3y - 14z = 25$, $-15x + 4y - kz = 3$, and $-4x + y + 3z = 4$ is consistent for all k in the set.

   Options: 

    A. R 

    B. R − {−11, 13} 

    C. R − {13} 

    D. R − {−11, 11}.

   Correct Answer: D.

   Year: 2022 (25-Jun-2022-Shift-2).

   Solution: The system may be inconsistent if the determinant of the coefficient matrix is zero. Solving the determinant equation $\begin{vmatrix} -k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3 \end{vmatrix} = 0$ results in $k = \pm 11$. Hence, the system is consistent if $k \in R - \{11, -11\}$.

   Step Solution: 

    1.  Establish Determinant: Form the determinant $\Delta$ of the coefficient matrix: $\begin{vmatrix} -k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3 \end{vmatrix}$.

    2.  Expansion: Expand the determinant to form an equation: $-k(12 + k) - 3(-45 - 4k) - 14(-15 + 16) = 0$ [Outside Source].

    3.  Simplification: This simplifies to $-12k - k^2 + 135 + 12k - 14 = 0$ [Outside Source].

    4.  Final Equation: The linear terms cancel out, leaving $121 - k^2 = 0$ [Outside Source].

    5.  Solve for k: $k^2 = 121 \Rightarrow k = \pm 11$. For the system to be consistent, $\Delta \neq 0$, so $k \in R - \{11, -11\}$.

   The difficulty level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: In this system, find the values of $k$ that make the determinant zero ($k = \pm 11$); the system is guaranteed to be consistent as long as $k$ avoids these specific values that cause singular behavior.

 Question 34

   Question: The ordered pair $(a, b)$, for which the system of linear equations $3x - 2y + z = b$, $5x - 8y + 9z = 3$, and $2x + y + az = -1$ has no solution, is:.

   Options: 

    A. $(3, 1/3)$

    B. $(-3, 1/3)$

    C. $(-3, -1/3)$

    D. $(3, -1/3)$.

   Correct Answer: C.

   Year: 2022 (26-Jun-2022-Shift-1).

   Solution: Setting the coefficient determinant to zero: $\begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix} = 0 \Rightarrow -14a - 42 = 0 \Rightarrow a = -3$. Now, using the row combination $3(Eq_1) - (Eq_2) - 2(Eq_3) = 0$ results in $-3b + 3 - 2 = 0 \Rightarrow b = 1/3$ for consistency. For no solution, $a = -3$ and $b \neq 1/3$.

   Step Solution: 

    1.  Coefficient Determinant: Calculate $\Delta = \begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix} = 3(-8a - 9) + 2(5a - 18) + 1(5 - (-16))$ [Outside Source].

    2.  Solve for a: Simplify to $-24a - 27 + 10a - 36 + 21 = 0 \Rightarrow -14a - 42 = 0 \Rightarrow a = -3$.

    3.  Linear Combination: Identify the relationship between rows: $3(Eq_1) - (Eq_2) - 2(Eq_3)$.

    4.  Evaluate Constants: Apply the combination to the constants: $3(b) - 3 - 2(1) = 3b - 1$.

    5.  Condition for No Solution: For the system to be inconsistent, the constant part $3b - 1$ must not equal zero, so $b \neq 1/3$. The pair $(-3, -1/3)$ fits this condition.

   The difficulty level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Use the provided row combination $3R_1 - R_2 - 2R_3$. For $a = -3$, the $x, y, z$ terms sum to zero. The system has no solution if the constants $3(b) - 3 - 2(1)$ do not sum to zero.

 Question 35

   Question: If the system of equations $\alpha x + y + z = 5, x + 2y + 3z = 4, x + 3y + 5z = \beta$ has infinitely many solutions, then the ordered pair $(\alpha, \beta)$ is equal to:.

   Options: 

    A. (1, −3) 

    B. (−1, 3) 

    C. (1, 3) 

    D. (−1, −3).

   Correct Answer: C.

   Year: 2022 (26-Jun-2022-Shift-2).

   Solution: For infinitely many solutions, the determinant $\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} = 0 \Rightarrow \alpha(1) - 1(2) + 1(1) = 0 \Rightarrow \alpha = 1$. Setting $\Delta_1 = 0$ gives $5(1) - 1(20 - 3\beta) + 1(12 - 2\beta) = 0$, which results in $\beta = 3$.

   Step Solution: 

    1.  Solve for $\alpha$: Set the coefficient determinant $\Delta = 0$; expanding gives $\alpha(10-9) - 1(5-3) + 1(3-2) = 0$.

    2.  Result for $\alpha$: $\alpha - 2 + 1 = 0 \Rightarrow \alpha = 1$.

    3.  Solve for $\beta$: Set the constant-replaced determinant $\Delta_x = 0$; $\begin{vmatrix} 5 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 3 & 5 \end{vmatrix} = 0$.

    4.  Expansion for $\beta$: $5(10-9) - 1(20-3\beta) + 1(12-2\beta) = 5 - 20 + 3\beta + 12 - 2\beta = 0$.

    5.  Result for $\beta$: $\beta - 3 = 0 \Rightarrow \beta = 3$. Thus, $(\alpha, \beta) = (1, 3)$.

   The difficulty level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row arithmetic: $R_3 = 2R_2 - R_1$. By equating the coefficients of $x$, $1 = 2(1) - \alpha \Rightarrow \alpha = 1$ [Outside Source]. By equating the constants, $\beta = 2(4) - 5 \Rightarrow \beta = 3$ [Outside Source].

 Question 36

   Question: Let the system of linear equations $x + 2y + z = 2$, $\alpha x + 3y - z = \alpha$, $-\alpha x + y + 2z = -\alpha$ be inconsistent. Then $\alpha$ is equal to:

   Options: 

    A. $\frac{5}{2}$ 

    B. $-\frac{5}{2}$ 

    C. $\frac{7}{2}$ 

    D. $-\frac{7}{2}$

   Correct Answer: D

   Year: 2022 (27-Jun-2022-Shift-1)

   Solution: $\Delta = \begin{vmatrix} 1 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix} = 1(6 + 1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha) = 7 + 2\alpha$. $\Delta = 0 \Rightarrow \alpha = -\frac{7}{2}$. $\Delta_1 = \begin{vmatrix} 2 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix} = 14 + 2\alpha \neq 0$ for $\alpha = -\frac{7}{2}$.

   Step Solution:

    1.  Set up Coefficient Determinant: Form $\Delta = \begin{vmatrix} 1 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix}$.

    2.  Expand $\Delta$: $1(6+1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha) = 7 - 2\alpha + 4\alpha = 7 + 2\alpha$.

    3.  Find Critical $\alpha$: For inconsistency, $\Delta$ must be zero: $7 + 2\alpha = 0 \Rightarrow \alpha = -\frac{7}{2}$.

    4.  Verify Inconsistency: Check $\Delta_1$ by replacing the $x$-column with constants $[2, \alpha, -\alpha]$. $\Delta_1 = 2(6+1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha) = 14 + 2\alpha$.

    5.  Substitute $\alpha$: At $\alpha = -\frac{7}{2}$, $\Delta_1 = 14 - 7 = 7$. Since $\Delta = 0$ and $\Delta_1 \neq 0$, the system is inconsistent.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: In a 3x3 system, if $\Delta = 0$ and any $\Delta_i \neq 0$, it's inconsistent. Here, $\Delta$ is a linear function $7 + 2\alpha$, so $\alpha = -3.5$ is the only candidate. Quick mental check of $\Delta_1$ at that value confirms it is non-zero.

 Question 38

   Question: If the system of linear equations $2x + 3y - z = -2$, $x + y + z = 4$, $x - y + |\lambda|z = 4\lambda - 4$ where, $\lambda \in \mathbb{R}$, has no solution, then:

   Options: 

    A. $\lambda = 7$

    B. $\lambda = -7$

    C. $\lambda = 8$

    D. $\lambda^2 = 1$

   Correct Answer: B

   Year: 2022 (28-Jun-2022-Shift-1)

   Solution: $\Delta = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{vmatrix} = 0 \Rightarrow |\lambda| = 7$. But at $\lambda = 7$, $5P_2 - 2P_1 = P_3$, so system has infinite solutions. So $\lambda = -7$ is correct.

   Step Solution:

    1.  Solve for $|\lambda|$: Set $\Delta = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{vmatrix} = 2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-1 - 1) = 0$.

    2.  Simplify: $2|\lambda| + 2 - 3|\lambda| + 3 + 2 = 7 - |\lambda| = 0 \Rightarrow |\lambda| = 7$.

    3.  Identify Cases: This gives two possibilities: $\lambda = 7$ or $\lambda = -7$.

    4.  Test $\lambda = 7$: Observe that if $\lambda = 7$, the third equation is $x - y + 7z = 24$. Multiplying the second equation by 5 and subtracting twice the first ($5P_2 - 2P_1$) yields $x - y + 7z = 24$.

    5.  Final Decision: Since $\lambda = 7$ leads to identical planes (infinite solutions), the "no solution" case must be $\lambda = -7$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent vs Infinite Solutions).

   Short cut solution: Quick row check: Observe if a linear combination of the first two equations can produce the third. $5R_2 - 2R_1 = (5x-4x) + (5y-6y) + (5z+2z) = x - y + 7z$. The constants are $5(4) - 2(-2) = 24$. If $\lambda=7$ and the constant $4\lambda-4=24$, it's infinite solutions. $4(7)-4=24$, so $\lambda=7$ is infinite. Thus, $\lambda=-7$ is the answer.

 Question 39

   Question: If the system of linear equations $2x - 3y = \gamma + 5$, $\alpha x + 5y = \beta + 1$ where $\alpha, \beta, \gamma \in \mathbb{R}$ has infinitely many solutions, then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to:

   Options: Numerical Answer Type.

   Correct Answer: 58

   Year: 2022 (28-Jun-2022-Shift-2)

   Solution: $\frac{2}{\alpha} = \frac{-3}{5} = \frac{\gamma + 5}{\beta + 1} \Rightarrow \alpha = -\frac{10}{3}$ and $3\beta + 5\gamma = -28$. $|9\alpha + 3\beta + 5\gamma| = |-30 - 28| = 58$.

   Step Solution:

    1.  Condition for Infinite Solutions: For a 2-variable system, the ratios of coefficients must be equal: $\frac{2}{\alpha} = \frac{-3}{5} = \frac{\gamma + 5}{\beta + 1}$.

    2.  Solve for $\alpha$: $\frac{2}{\alpha} = -\frac{3}{5} \Rightarrow -3\alpha = 10 \Rightarrow \alpha = -\frac{10}{3}$.

    3.  Solve for $\beta$ and $\gamma$: $-\frac{3}{5} = \frac{\gamma + 5}{\beta + 1} \Rightarrow -3(\beta + 1) = 5(\gamma + 5)$.

    4.  Simplify Relation: $-3\beta - 3 = 5\gamma + 25 \Rightarrow 3\beta + 5\gamma = -28$.

    5.  Calculate Final Expression: $|9\alpha + (3\beta + 5\gamma)| = |9(-\frac{10}{3}) + (-28)| = |-30 - 28| = 58$.

   Difficulty Level: Easy.

   The Concept Name: Proportionality of Coefficients (Infinite solutions for 2D systems).

   Short cut solution: Use the ratio $\frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{c_1}{c_2}$. Once you have $\alpha = -10/3$, the target expression $9\alpha$ becomes $-30$. The remaining part $3\beta + 5\gamma$ is found directly from the constant ratio cross-multiplication.

 Question 40

   Question: If the system of linear equations $2x + y - z = 7$, $x - 3y + 2z = 1$, and $x + 4y + \delta z = k$, where $\delta, k \in \mathbb{R}$, has infinitely many solutions, then $\delta + k$ is equal to:

   Options: 

    A. −3 

    B. 3 

    C. 6 

    D. 9

   Correct Answer: B

   Year: 2022 (29-Jun-2022-Shift-1)

   Solution: Given equations $2x + y - z = 7$, $x - 3y + 2z = 1$, $x + 4y + \delta z = k$. $\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix} = -7\delta - 21 = 0 \Rightarrow \delta = -3$. For infinite solutions, $\Delta_x = \begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{vmatrix} = 0 \Rightarrow 6 - k = 0 \Rightarrow k = 6$. Thus, $\delta + k = 3$.

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant: $2(-3\delta - 8) - 1(\delta - 2) - 1(4 + 3) = 0$.

    2.  Solve for $\delta$: Simplify the expression $-6\delta - 16 - \delta + 2 - 7 = 0 \Rightarrow -7\delta - 21 = 0 \Rightarrow \delta = -3$.

    3.  Set $\Delta_x = 0$: Use the determinant where the $x$-column is replaced by constants: $\begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{vmatrix} = 0$.

    4.  Solve for $k$: $7(9 - 8) - 1(-3 - 2k) - 1(4 + 3k) = 0 \Rightarrow 7 + 3 + 2k - 4 - 3k = 0$, leading to $k = 6$.

    5.  Final Value: $\delta + k = -3 + 6 = 3$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row relationships. To have infinite solutions, the third equation must be a linear combination of the first two ($R_3 = mR_1 + nR_2$). By equating $x$ and $y$ coefficients, we find $m=1$ and $n=-1$. Applying this to $z$ and constants: $\delta = -1 - 2 = -3$ and $k = 7 - 1 = 6$.

 Question 41

   Question: The number of values of $\alpha$ for which the system of equations: $x + y + z = \alpha$, $\alpha x + 2\alpha y + 3z = \alpha - 1$, $x + 3\alpha y + 5z = 4$ is inconsistent, is.

   Options: 

    A. 0 

    B. 1 

    C. 2 

    D. 3

   Correct Answer: B

   Year: 2022 (24-Jun-2022-Shift-1)

   Solution: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix} = 3\alpha^2 - 6\alpha + 3$. For inconsistency, $\Delta = 0$ i.e., $\alpha = 1$. Checking for $\alpha = 1$: the equations become $x+y+z=1$ and $x+2y+3z=-1$, which are inconsistent with $x+3y+5z=4$.

   Step Solution:

    1.  Calculate $\Delta$: Expand the coefficient determinant: $1(10\alpha - 9\alpha) - 1(5\alpha - 3) + 1(3\alpha^2 - 2\alpha)$.

    2.  Solve for $\Delta = 0$: Simplify to $3\alpha^2 - 6\alpha + 3 = 0 \Rightarrow 3(\alpha - 1)^2 = 0$, giving $\alpha = 1$.

    3.  Test $\alpha = 1$: Substitute into the system: (i) $x+y+z=1$, (ii) $x+2y+3z=0$, (iii) $x+3y+5z=4$.

    4.  Find Contradiction: Note that $(ii) - (i) \Rightarrow y + 2z = -1$, while $(iii) - (ii) \Rightarrow y + 2z = 4$.

    5.  Conclusion: Since $-1 \neq 4$, the system is inconsistent only for $\alpha = 1$, making the count 1.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Inconsistency typically occurs at the roots of $\Delta=0$. Since $\Delta = 3(\alpha-1)^2$ has only one root ($\alpha=1$), simply check if that specific value causes the parallel plane/contradiction property.

 Question 42

   Question: The number of $\theta \in (0, 4\pi)$ for which the system of linear equations $3(\sin 3\theta)x - y + z = 2$, $3(\cos 2\theta)x + 4y + 3z = 3$, $6x + 7y + 7z = 9$ has no solution, is :

   Options: 

    A. 6 

    B. 7 

    C. 8 

    D. 9

   Correct Answer: B

   Year: 2022 (25-Jul-2022-Shift-1)

   Solution: For no solution, $\Delta = 0 \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0 \Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0$. Substituting trig identities: $\sin \theta (4\sin^2 \theta + 4\sin \theta - 3) = 0$. Roots are $\sin \theta = 0$ and $\sin \theta = 1/2$. Total possible values for $\theta \in (0, 4\pi)$ are 7.

   Step Solution:

    1.  Set up $\Delta = 0$: Expand $\begin{vmatrix} 3\sin 3\theta & -1 & 1 \\ 3\cos 2\theta & 4 & 3 \\ 6 & 7 & 7 \end{vmatrix}$ along the first row.

    2.  Simplify Equation: $21\sin 3\theta + 42\cos 2\theta - 42 = 0$ simplifies to $\sin 3\theta + 2\cos 2\theta - 2 = 0$.

    3.  Use Trig Identities: Substitute $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$.

    4.  Factorize: This yields $\sin\theta(4\sin^2\theta + 4\sin\theta - 3) = 0$, resulting in $\sin\theta = 0$ or $\sin\theta = 1/2$.

    5.  Count Roots: In $(0, 4\pi)$, $\sin\theta=0$ gives $\{\pi, 2\pi, 3\pi\}$ (3 roots) and $\sin\theta=1/2$ gives $\{\pi/6, 5\pi/6, 13\pi/6, 17\pi/6\}$ (4 roots). Total = 7.

   Difficulty Level: Hard.

   The Concept Name: Trigonometric Equations in Determinants.

   Short cut solution: Simplify the trig equation to $\sin 3\theta = 2(1 - \cos 2\theta) = 4\sin^2\theta$. Expanding $\sin 3\theta$ as $\sin\theta(3 - 4\sin^2\theta)$ allows you to quickly find $\sin\theta = 0$ or $3 - 4\sin^2\theta = 4\sin\theta$, which is a simple quadratic for $\sin\theta$.

 Question 43

   Question: The number of real values of $\lambda$ , such that the system of linear equations $2x - 3y + 5z = 9$, $x + 3y - z = -18$, and $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solutions, is.

   Options: 

    A. 0 

    B. 1 

    C. 2 

    D. 4

   Correct Answer: C

   Year: 2022 (25-Jul-2022-Shift-2).

   Solution: $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^2 - |\lambda| \end{vmatrix} = 2(3\lambda^2 - 3|\lambda| - 1) + 3(\lambda^2 - |\lambda| - 3) + 5(-1 - 9) = 9\lambda^2 - 9|\lambda| - 43$. $\Delta = 0$ for 2 values of $|\lambda|$ out of which one is negative and other is positive. So, 2 values of $\lambda$ satisfy the system of equations to obtain no solution.

   Step Solution: 

    1.  Form the coefficient determinant: $\Delta = \begin{vmatrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^2 - |\lambda| \end{vmatrix} = 0$.

    2.  Expand and Simplify: $2(3(\lambda^2 - |\lambda|) - 1) - (-3)((\lambda^2 - |\lambda|) - (-3)) + 5(-1 - 9) = 0$.

    3.  Quadratic Form: $6(\lambda^2 - |\lambda|) - 2 + 3(\lambda^2 - |\lambda|) + 9 - 50 = 9(\lambda^2 - |\lambda|) - 43 = 0$.

    4.  Solve for $|\lambda|$: Let $t = |\lambda|$, then $9t^2 - 9t - 43 = 0$. Using the quadratic formula, $t = \frac{9 \pm \sqrt{81 - 4(9)(-43)}}{18} = \frac{9 \pm \sqrt{1629}}{18}$.

    5.  Evaluate Roots: Since $\sqrt{1629} > 9$, one root for $t$ is positive and one is negative; since $t = |\lambda|$, only the positive root yields real values ($\lambda = \pm t$), resulting in 2 values.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: The simplified determinant is $9|\lambda|^2 - 9|\lambda| - 43 = 0$. In a quadratic equation $ax^2 + bx + c = 0$, if $ac < 0$ (here $9 \times -43$), the roots have opposite signs. Only the positive root for $|\lambda|$ gives real values for $\lambda$, yielding exactly two values ($\pm \lambda_0$).

 Question 44

   Question: If the system of linear equations $8x + y + 4z = -2$, $x + y + z = 0$, and $2x - 3y = \mu$ (Source indicates $\lambda x - 3y = \mu$ in math context) has infinitely many solutions, then the distance of the point $(\lambda, \mu, -1/2)$ from the plane $8x + y + 4z + 2 = 0$ is.

   Options: 

    A. $3\sqrt{5}$ 

    B. 4 

    C. 26 

    D. $10/3$

   Correct Answer: D

   Year: 2022 (26-Jul-2022-Shift-1).

   Solution: $\Delta = \begin{vmatrix} 8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & -3 & 0 \end{vmatrix} = 12 - 3\lambda = 0 \Rightarrow \lambda = 4$. $\Delta_x = \begin{vmatrix} -2 & 1 & 4 \\ 0 & 1 & 1 \\ \mu & -3 & 0 \end{vmatrix} = -6 - 3\mu = 0 \Rightarrow \mu = -2$. Distance of $(4, -2, -1/2)$ from plane $8x - y - 4z + 2 = 0$ is $\frac{|32 - (-2) - 4(-1/2) + 2|}{\sqrt{64 + 1 + 16}} = \frac{10}{3}$ units.

   Step Solution: 

    1.  Find $\lambda$: Set $\Delta = 0 \Rightarrow 8(3) - 1(-\lambda) + 4(-3 - \lambda) = 0$, which simplifies to $12 - 3\lambda = 0 \Rightarrow \lambda = 4$.

    2.  Find $\mu$: Set $\Delta_x = 0 \Rightarrow -2(3) - 1(-\mu) + 4(-\mu) = 0$, which simplifies to $-6 - 3\mu = 0 \Rightarrow \mu = -2$.

    3.  Identify Point: The point $(\lambda, \mu, -1/2)$ is $(4, -2, -1/2)$.

    4.  Apply Distance Formula: $d = \frac{|8(4) + 1(-2) + 4(-1/2) + 2|}{\sqrt{8^2 + 1^2 + 4^2}}$.

    5.  Calculate Value: $d = \frac{|32 - 2 - 2 + 2|}{9} = \frac{30}{9} = \frac{10}{3}$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions) and Distance from Point to Plane.

   Short cut solution: Solve for $\lambda$ and $\mu$ using determinant properties for consistency. Once the coordinates $(4, -2, -0.5)$ are found, substitute directly into the standard perpendicular distance formula $\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$.

 Question 46

   Question: If the system of equations $x + y + z = 6$, $2x + 5y + \alpha z = \beta$, and $x + 2y + 3z = 14$ has infinitely many solutions, then $\alpha + \beta$ is equal to.

   Options: 

    A. 8 

    B. 36 

    C. 44 

    D. 48

   Correct Answer: C

   Year: 2022 (29-Jul-2022-Shift-2).

   Solution: For infinite solutions $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$. $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0$. Using transformations $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - 2R_3$ (implied), $\alpha = 8$. Similarly $\Delta_x = 0$ gives $\beta = 36$.

   Step Solution: 

    1.  Coefficient Determinant: Set $\Delta = 0 \Rightarrow 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$.

    2.  Solve for $\alpha$: $15 - 2\alpha - 6 + \alpha - 1 = 0 \Rightarrow 8 - \alpha = 0 \Rightarrow \alpha = 8$.

    3.  Constant Determinant: Set $\Delta_x = 0 \Rightarrow \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 14 & 2 & 3 \end{vmatrix} = 0$.

    4.  Solve for $\beta$: $6(15 - 16) - 1(3\beta - 112) + 1(2\beta - 70) = 0 \Rightarrow -6 - 3\beta + 112 + 2\beta - 70 = 0 \Rightarrow \beta = 36$.

    5.  Final Sum: $\alpha + \beta = 8 + 36 = 44$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use linear combination of equations. Observe that $3 \times (x + 2y + 3z = 14) - (x + y + z = 6) = 2x + 5y + 8z = 36$. Comparing this result to the target second equation $2x + 5y + \alpha z = \beta$ gives $\alpha = 8$ and $\beta = 36$ immediately. Sum $= 44$.

 Question 50

   Question: Consider the following system of equations: $x + 2y - 3z = a, 2x + 6y - 11z = b, x - 2y + 7z = c$, where $a, b$ and $c$ are real constants. Then, the system of equations:

   Options: 

    A. has a unique solution, when $5a = 2b + c$

    B. has infinite number of solutions when $5a = 2b + c$

    C. has no solution for all a, b and c

    D. has a unique solution for all a, b and c

   Correct Answer: B

   Year: 2021 (26 Feb 2021 Shift 2)

   Solution: The system can be written as $AX = B$. The determinant $|A| = 0$, meaning it cannot have a unique solution. Calculating the other determinants gives $|A_1| = 4(5a - 2b - c)$, $|A_2| = -5(5a - 2b - c)$, and $|A_3| = -2(5a - 2b - c)$. For infinite solutions, $|A| = |A_1| = |A_2| = |A_3| = 0$, which leads to $5a = 2b + c$.

   Step Solution:

    1.  Form the coefficient determinant: $\Delta = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{vmatrix}$.

    2.  Calculate $\Delta$: $1(42 - 22) - 2(14 + 11) - 3(-4 - 6) = 20 - 50 + 30 = 0$.

    3.  Calculate $\Delta_x$ ($|A_1|$): $\begin{vmatrix} a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7 \end{vmatrix} = a(20) - 2(7b + 11c) - 3(-2b - 6c) = 20a - 8b - 4c$.

    4.  Factorize $\Delta_x$: $20a - 8b - 4c = 4(5a - 2b - c)$.

    5.  Set $\Delta_x = 0$ for Infinite Solutions: $4(5a - 2b - c) = 0 \Rightarrow 5a = 2b + c$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row transformations. Notice that $5 \times (\text{Row 1}) - 2 \times (\text{Row 2}) = (\text{Row 3})$. Specifically: $5(x+2y-3z) - 2(2x+6y-11z) = x-2y+7z$. For the system to be consistent (infinite solutions), the constants must follow the same rule: $5a - 2b = c \Rightarrow 5a = 2b + c$.

 Question 51

   Question: The following system of linear equations: $2x + 3y + 2z = 9, 3x + 2y + 2z = 9, x - y + 4z = 8$

   Options: 

    A. does not have any solution

    B. has a unique solution

    C. has infinitely many solutions

    D. has a solution $(\alpha, \beta, \gamma)$ satisfying $\alpha + \beta^2 + \gamma^3 = 12$

   Correct Answer: B

   Year: 2021 (25 Feb 2021 Shift 2)

   Solution: The given system of equations is non-homogeneous and can be written as $AX = B$. Calculating the determinant $|A|$ gives $-20$, which is not equal to 0. Since the determinant is non-zero, the system has a unique solution.

   Step Solution:

    1.  Represent as Matrix: $AX = B$, where $A = \begin{bmatrix} 2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & -1 & 4 \end{bmatrix}$.

    2.  Calculate determinant $|A|$: $2(2 \times 4 - 2 \times (-1)) - 3(3 \times 4 - 2 \times 1) + 2(3 \times (-1) - 2 \times 1)$.

    3.  Simplify Expansion: $2(8 + 2) - 3(12 - 2) + 2(-3 - 2)$.

    4.  Compute Value: $2(10) - 3(10) + 2(-5) = 20 - 30 - 10 = -20$.

    5.  Apply Rule: Since $|A| = -20 \neq 0$, the system has a unique solution.

   Difficulty Level: Easy.

   The Concept Name: Cramer's Rule (Unique Solution).

   Short cut solution: In JEE problems, if the constants and coefficients are asymmetric and $x, y, z$ coefficients don't share obvious multiples, calculate $|A|$ immediately. Since $|A| \neq 0$, the answer is "unique solution" without solving for $x, y, z$.

 Question 52

   Question: If the system of equations $kx + y + 2z = 1$, $3x - y - 2z = 2$, $-2x - 2y - 4z = 3$ has infinitely many solutions, then $k$ is equal to...

   Options: Numerical Answer Type.

   Correct Answer: 21

   Year: 2021 (25 Feb 2021 Shift 1)

   Solution: For infinitely many solutions, $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$. By setting $\Delta_y = 0$ (replacing the $y$-column with constants), we get $k(-8 + 6) - 1(-12 - 4) + 2(9 + 4) = 0$. This simplifies to $-2k + 16 + 26 = 0$, resulting in $k = 21$.

   Step Solution:

    1.  Establish Infinite Solution Condition: $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$.

    2.  Set up $\Delta_y = 0$: $\begin{vmatrix} k & 1 & 2 \\ 3 & 2 & -2 \\ -2 & 3 & -4 \end{vmatrix} = 0$.

    3.  Expand Determinant: $k(-8 - (-6)) - 1(-12 - 4) + 2(9 - (-4)) = 0$.

    4.  Simplify Equation: $-2k + 16 + 26 = 0$.

    5.  Solve for $k$: $2k = 42 \Rightarrow k = 21$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Identify that if you add the first two equations, you get $(k+3)x = 3$. If the system has infinite solutions, this reduction must be consistent with the third equation. However, calculating $\Delta_y = 0$ is the most direct way to isolate the single variable $k$.

Question 54

   Question: For the system of linear equations $x - 2y = 1, x - y + kz = -2, ky + 4z = 6, k \in R$, consider the following statements:

    (A) The system has unique solution, if $k \neq 2, k \neq -2$

    (B) The system has unique solution, if $k = -2$

    (C) The system has unique solution, if $k = 2$

    (D) The system has no solution, if $k = 2$

    (E) The system has infinite number of solutions, if $k \neq -2$.

    Which of the following statements are correct?

   Options: 

    A. (C) and (D)

    B. (B) and (E)

    C. (A) and (E)

    D. (A) and (D)

   Correct Answer: D

   Year: 2021 (24 Feb 2021 Shift 2)

   Solution: Given equations: $x - 2y + 0z = 1, x - y + kz = -2, 0x + ky + 4z = 6$. $\Delta = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} = 1(-4-k^2) + 2(4) = 4-k^2$. $\Delta_x = \begin{vmatrix} 1 & -2 & 0 \\ -2 & -1 & k \\ 6 & k & 4 \end{vmatrix} = 1(-4-k^2) + 2(-8-6k) = -k^2-12k-20$. If $\Delta \neq 0$ ($k \neq \pm 2$), the system has a unique solution (Statement A). At $k = 2, \Delta = 0$ and $\Delta_x = -48 \neq 0$, meaning the system has no solution (Statement D).

   Step Solution:

    1.  Calculate $\Delta$: The coefficient determinant is $\Delta = 1(-4 - k^2) - (-2)(4 - 0) + 0 = -4 - k^2 + 8 = 4 - k^2$.

    2.  Determine Unique Solution Condition: A unique solution exists if $\Delta \neq 0$, which implies $4 - k^2 \neq 0$, so $k \neq 2$ and $k \neq -2$.

    3.  Evaluate Statement A: Since the condition $k \neq \pm 2$ matches the condition for uniqueness, Statement (A) is correct.

    4.  Analyze $k = 2$: If $k = 2, \Delta = 0$. Calculate $\Delta_x$ by replacing the first column with constants: $\Delta_x = 1(-4 - 4) + 2(-8 - 12) = -8 - 40 = -48$.

    5.  Evaluate Statement D: Since $\Delta = 0$ and $\Delta_x \neq 0$ at $k = 2$, the system is inconsistent, so Statement (D) is correct.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: The determinant is a simple quadratic $4 - k^2$; its roots $\pm 2$ immediately tell you where the unique solution fails. Checking $k=2$ in the equations $x-2y=1$ and $y+2z=3$ (from $Eq_3$) vs $x-y+2z=-2$ (from $Eq_2$) quickly reveals a contradiction ($y+2z = -3$ vs $3$).

 Question 55

   Question: The system of linear equations : $3x - 2y - kz = 10 ; 2x - 4y - 2z = 6 ; x + 2y - z = 5m$ is inconsistent if :

   Options: 

    A. $k = 3, m = 4/5$

    B. $k \neq 3, m \in R$

    C. $k \neq 3, m \neq 4/5$

    D. $k = 3, m \neq 4/5$

   Correct Answer: D

   Year: 2021 (24 Feb 2021 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 3 & -2 & -k \\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{vmatrix} = 0 \Rightarrow 24 - 8k = 0 \Rightarrow k = 3$. $\Delta_x = \begin{vmatrix} 10 & -2 & -3 \\ 6 & -4 & -2 \\ 5m & 2 & -1 \end{vmatrix} = 8(4-5m)$. $\Delta_z = 8(4-5m)$. For inconsistent, $k=3$ and $m \neq 4/5$.

   Step Solution:

    1.  Set $\Delta = 0$: Solve the coefficient determinant $3(4 - (-4)) - (-2)(-2 - (-2)) - k(4 - (-4)) = 0$.

    2.  Solve for $k$: $24 + 0 - 8k = 0$, which results in $k = 3$.

    3.  Set up $\Delta_x$: Substitute constants into the first column and use $k=3$: $\begin{vmatrix} 10 & -2 & -3 \\ 6 & -4 & -2 \\ 5m & 2 & -1 \end{vmatrix}$.

    4.  Solve for $m$: Expanding $\Delta_x$ gives $10(8) + 2(-6 + 10m) - 3(12 + 20m) = 32 - 40m = 8(4-5m)$.

    5.  Identify Inconsistency: The system is inconsistent if $\Delta = 0$ and any other $\Delta_i \neq 0$, meaning $k=3$ and $m \neq 4/5$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Observe that adding the second and third equations ($LHS$) gives $3x - 2y - 3z$. Comparing this to the first equation $3x - 2y - kz = 10$, the system can only be inconsistent if $k=3$. For these parallel/combined planes to never meet, the constants must not match: $6 + 5m \neq 10 \Rightarrow 5m \neq 4 \Rightarrow \mathbf{m \neq 4/5}$.

 Question 65

   Question: The system of equations $kx + y + z = 1, x + ky + z = k$ and $x + y + zk = k^2$ has no solution, if $k$ is equal to

   Options: 

    A. 0

    B. 1

    C. -1

    D. -2 (Note: Value derived from source solution)

   Correct Answer: D

   Year: 2021 (17 Mar 2021 Shift 1)

   Solution: For no solution, $\Delta = 0$. $\begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} = k^3 - 3k + 2 = (k-1)^2(k+2) = 0 \Rightarrow k = 1, -2$. If $k = 1$, all equations are identical, leading to infinite solutions. Therefore, $k = -2$.

   Step Solution:

    1.  Form $\Delta$: The coefficient determinant for the cyclic system is $\Delta = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$.

    2.  Factorize $\Delta$: Expanding or using row properties yields $k(k^2-1) - 1(k-1) + 1(1-k) = k^3 - 3k + 2$.

    3.  Solve for Roots: $(k - 1)^2(k + 2) = 0$ gives possible values $k = 1$ and $k = -2$.

    4.  Test $k = 1$: If $k = 1$, the equations become $x+y+z=1$ (all three), which has infinite solutions.

    5.  Verify $k = -2$: If $k = -2$, adding all three equations gives $0 \cdot (x+y+z) = 1 + k + k^2 = 1 - 2 + 4 = 3$. Since $0 = 3$ is impossible, $k = -2$ is the "no solution" case.

   Difficulty Level: Easy.

   The Concept Name: Cramer's Rule (Non-Homogeneous Systems).

   Short cut solution: In a cyclic system where the sum of each row's coefficients is $k+2$, the determinant must have $(k+2)$ as a factor. Setting $k+2=0$ gives $k=-2$. Check the sum of constants: $1 + k + k^2$. For $k=-2$, the sum is $3$. Since $0 \neq 3$, the system is immediately inconsistent at $k=-2$.

Question 72

   Question: For real numbers $\alpha$ and $\beta$, consider the following system of linear equations : $x + y - z = 2$ , $x + 2y + \alpha z = 1$ , $2x - y + z = \beta$. If the system has infinite solutions, then $\alpha + \beta$ is equal to.

   Options: Numerical Answer Type.

   Correct Answer: 5.

   Year: 2021 (27 Jul 2021 Shift 1).

   Solution: For infinite solutions $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. $\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix} = 0 \Rightarrow 3(2 + \alpha) = 0 \Rightarrow \alpha = -2$. $\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = 0 \Rightarrow \beta - 7 = 0 \Rightarrow \beta = 7$. $\therefore \alpha + \beta = 5$.

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant $\Delta = 1(2 - (-\alpha)) - 1(1 - 2\alpha) + (-1)(-1 - 4) = 0$.

    2.  Solve for $\alpha$: $2 + \alpha - 1 + 2\alpha + 5 = 0 \Rightarrow 3\alpha + 6 = 0 \Rightarrow \alpha = -2$.

    3.  Set up $\Delta_2 = 0$: Substitute constants into the $y$-column and use $\alpha = -2$: $\begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = 0$.

    4.  Solve for $\beta$: $1(1 + 2\beta) - 2(1 + 4) - 1(\beta - 2) = 0 \Rightarrow 1 + 2\beta - 10 - \beta + 2 = 0 \Rightarrow \beta - 7 = 0 \Rightarrow \beta = 7$.

    5.  Final Value: $\alpha + \beta = -2 + 7 = 5$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: In this system, check if $R_3$ can be expressed as $mR_1 + nR_2$. Comparing $x$ and $y$ coefficients: $m+n=2$ and $m+2n=-1$. Solving gives $n=-3$ and $m=5$. Apply this to $z$: $5(-1) + (-3)(\alpha) = 1 \Rightarrow \alpha = -2$. Apply to constants: $5(2) + (-3)(1) = \beta \Rightarrow \beta = 7$. Sum $= 5$.

 Question 73

   Question: The values of a and b, for which the system of equations $2x + 3y + 6z = 8$, $x + 2y + az = 5$, $3x + 5y + 9z = b$ has no solution, are :.

   Options: 

    A. $a = 3$ , $b \neq 13$

    B. $a \neq 3$ , $b \neq 13$

    C. $a \neq 3$ , $b = 3$

    D. $a = 3$ , $b = 13$.

   Correct Answer: A.

   Year: 2021 (25 Jul 2021 Shift 1).

   Solution: $\Delta = D = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix} = 3 - a$. $D_x = \begin{vmatrix} 2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{vmatrix} = b - 13$. If $a = 3$ , $b \neq 13$ , no solution.

   Step Solution:

    1.  Form Coefficient Determinant: Set $\Delta = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix}$.

    2.  Calculate $\Delta$: $2(18 - 5a) - 3(9 - 3a) + 6(5 - 6) = 36 - 10a - 27 + 9a - 6 = 3 - a$.

    3.  Solve for $a$: For "no solution," $\Delta$ must be 0, so $3 - a = 0 \Rightarrow a = 3$.

    4.  Check Consistency: For no solution, the constant-replaced determinant $D_z$ (or similar) must be non-zero. $D_z = 2(2b - 25) - 3(b - 15) + 8(5 - 6) = b - 13$.

    5.  Final Condition: The system is inconsistent if $a = 3$ and $b - 13 \neq 0$, which is $b \neq 13$.

   Difficulty Level: Easy.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Observe that $R_1 + R_2$ gives coefficients $3x + 5y + (6+a)z$. Comparing this to $R_3$, the system fails to have a unique solution if $6+a = 9 \Rightarrow a=3$. For these parallel/combined planes to never meet, the constants must not match: $8 + 5 \neq b \Rightarrow b \neq 13$.

 Question 74

   Question: The values of $\lambda$ and $\mu$ such that the system of equations $x + y + z = 6 , 3x + 5y + 5z = 26 , x + 2y + \lambda z = \mu$ has no solution, are :.

   Options: 

    A. $\lambda = 3 , \mu = 5$

    B. $\lambda = 3 , \mu \neq 10$

    C. $\lambda \neq 2 , \mu = 10$

    D. $\lambda = 2 , \mu \neq 10$.

   Correct Answer: D.

   Year: 2021 (22 Jul 2021 Shift 2).

   Solution: $5 \times (i) - (ii) \Rightarrow 2x = 4 \Rightarrow x = 2$. From (i) and (iii), $y + z = 4$ and $2y + \lambda z = \mu - 2$. Subtracting twice the first from the second: $(\lambda - 2)z = \mu - 10$. For no solution $\lambda = 2$ and $\mu \neq 10$.

   Step Solution:

    1.  Isolate $x$: Multiply the first equation by 5 and subtract the second: $5(x+y+z) - (3x+5y+5z) = 30 - 26 \Rightarrow 2x = 4 \Rightarrow x = 2$.

    2.  Reduce to 2 Variables: Substitute $x = 2$ into Eq 1 and Eq 3 to get $y + z = 4$ and $2y + \lambda z = \mu - 2$.

    3.  Eliminate $y$: Multiply $y + z = 4$ by 2 to get $2y + 2z = 8$.

    4.  Find the $z$ relation: Subtract this from $2y + \lambda z = \mu - 2$ to get $(\lambda - 2)z = \mu - 10$.

    5.  Identify Inconsistency: No solution occurs if the coefficient of $z$ is 0 but the constant is not: $\lambda = 2$ and $\mu \neq 10$.

   Difficulty Level: Medium.

   The Concept Name: Solving Systems of Equations (Consistency).

   Short cut solution: Use the coefficient determinant $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix}$. Expanding gives $2\lambda - 4$. Setting $\Delta = 0$ yields $\lambda = 2$. Quickly check the constant-replaced determinant $\Delta_z$: $\begin{vmatrix} 1 & 1 & 6 \\ 3 & 5 & 26 \\ 1 & 2 & \mu \end{vmatrix} = 2\mu - 20$. For no solution, $\Delta = 0$ and $\Delta_z \neq 0$, so $\lambda = 2$ and $\mu \neq 10$.

Question 75

   Question: The value of $k \in \mathbf{R}$, for which the following system of linear equations $3x - y + 4z = 3$, $x + 2y - 3z = -2$, and $6x + 5y + kz = -3$ has infinitely many solutions, is :

   Options: 

    A. 3

    B. −5

    C. 5

    D. −3

   Correct Answer: B

   Year: 2021 (20 Jul 2021 Shift 2)

   Solution: $\left| \begin{smallmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & K \end{smallmatrix} \right| = 0 \Rightarrow 3(2K + 15) + K + 18 - 28 = 0 \Rightarrow 7K + 35 = 0 \Rightarrow K = -5$.

   Step Solution:

    1.  Condition for Infinite Solutions: For the system to have infinitely many solutions, the determinant of the coefficient matrix ($\Delta$) must be zero.

    2.  Set up Determinant: Form the determinant $\Delta = \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix} = 0$.

    3.  Expand along Row 1: $3(2k - (-15)) - (-1)(1k - (-18)) + 4(5 - 12) = 0$.

    4.  Simplify Expansion: $3(2k + 15) + (k + 18) - 28 = 0 \Rightarrow 6k + 45 + k + 18 - 28 = 0$.

    5.  Solve for $k$: $7k + 35 = 0 \Rightarrow 7k = -35 \Rightarrow k = -5$.

   Difficulty level: Easy.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe the coefficients of $x$ and $y$. Note that $R_3 = R_1 + 3R_2$. For the $x$ coefficients: $3 + 3(1) = 6$; for $y$: $-1 + 3(2) = 5$. To have infinite solutions, this relation must hold for $z$ and the constants: $k = 4 + 3(-3) = -5$ and $-3 = 3 + 3(-2) = -3$.

 Question 78

   Question: If the following system of linear equations $2x + y + z = 5, x - y + z = 3$ and $x + y + az = b$ has no solution, then

   Options: 

    A. $a = -1/3$ and $b \neq 7/3$

    B. $a \neq 1/3$ and $b = 7/3$

    C. $a \neq -1/3$ and $b = 7/3$

    D. $a = 1/3$ and $b \neq 7/3$

   Correct Answer: D

   Year: 2021 (31 Aug 2021 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix} = 1 - 3a$. $\Delta_3 = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix} = 7 - 3b$. If $\Delta = 0$ and $\Delta_3 \neq 0$, then no solution $\Rightarrow a = 1/3, b \neq 7/3$.

   Step Solution:

    1.  Form Coefficient Determinant: Calculate $\Delta = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix}$.

    2.  Solve for $\Delta = 0$: Expanding gives $2(-a-1) - 1(a-1) + 1(1-(-1)) = -3a + 1 = 0 \Rightarrow a = 1/3$.

    3.  Form Constant-Replaced Determinant: Calculate $\Delta_3 = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix}$.

    4.  Solve for $\Delta_3 \neq 0$: Expanding gives $2(-b-3) - 1(b-3) + 5(1-(-1)) = 7 - 3b$.

    5.  Identify Inconsistency: The system has no solution if $\Delta = 0$ and any other determinant $\Delta_i \neq 0$; thus $a = 1/3$ and $b \neq 7/3$.

   Difficulty level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Find a linear combination of the first two equations to match the $x$ and $y$ of the third. $Eq_3 = mEq_1 + nEq_2 \Rightarrow 2m+n=1$ and $m-n=1$. Solving gives $m=2/3, n=-1/3$. Then $a = m(1) + n(1) = 1/3$. For no solution, the constant $b$ must not match: $b \neq m(5) + n(3) = 10/3 - 1 = 7/3$.

 Question 80

   Question: If the system of linear equations $2x + y - z = 3$, $x - y - z = \alpha$, and $3x + 3y + \beta z = 3$ has infinitely many solutions, then $\alpha + \beta - \alpha\beta$ is equal to

   Options: Numerical Answer Type (Correct value is 5).

   Correct Answer: 5

   Year: 2021 (27 Aug 2021 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -1 & -1 \\ 3 & 3 & \beta \end{vmatrix} = 0 \Rightarrow \beta = -1$. $\Delta_1 = \begin{vmatrix} 3 & 1 & -1 \\ \alpha & -1 & -1 \\ 3 & 3 & -1 \end{vmatrix} = 0 \Rightarrow \alpha = 3$. Value $= 3 + (-1) - 3(-1) = 5$.

   Step Solution:

    1.  Determine $\beta$: Set the coefficient determinant $\Delta = 0 \Rightarrow 2(-\beta + 3) - 1(\beta + 3) - 1(3 + 3) = 0$, leading to $\beta = -1$.

    2.  Determine $\alpha$: For infinite solutions, set $\Delta_1 = 0$ by replacing column 1 with constants: $\begin{vmatrix} 3 & 1 & -1 \\ \alpha & -1 & -1 \\ 3 & 3 & -1 \end{vmatrix} = 0$.

    3.  Solve for $\alpha$: $3(1+3) - 1(-\alpha+3) - 1(3\alpha+3) = 0 \Rightarrow 12 + \alpha - 3 - 3\alpha - 3 = 0 \Rightarrow \alpha = 3$.

    4.  Confirm Consistency: $\Delta_2$ and $\Delta_3$ are also zero at $\alpha=3, \beta=-1$.

    5.  Final Calculation: Substitute values into the expression: $3 + (-1) - (3 \times -1) = 2 + 3 = 5$.

   Difficulty level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use a linear combination of the first two equations to produce the third ($R_3 = mR_1 + nR_2$). By equating $x$ and $y$: $2m + n = 3$ and $m - n = 3 \Rightarrow m=2, n=-1$. Applying this to $z$: $\beta = 2(-1) + (-1)(-1) = -1$. Applying to constants: $3 = 2(3) + (-1)(\alpha) \Rightarrow \alpha = 3$. Result $= 3 - 1 + 3 = 5$.

Question 81

   Question: Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of lambda for which the system of linear equations $x + y + z = 4$, $3x + 2y + 5z = 3$, $9x + 4y + (28 + [\lambda])z = [\lambda]$ has a solution is.

   Options: 

    A. R

    B. $( - \infty , - 9 ) \cup ( - 9 , \infty )$

    C. [−9, −8)

    D. $( - \infty , - 9 ) \cup [ - 8 , \infty )$

   Correct Answer: A

   Year: 2021 (27 Aug 2021 Shift 2)

   Solution: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28 + [\lambda] \end{vmatrix} = 1(56 + 2[\lambda] - 20) - 1(84 + 3[\lambda] - 45) + 1(-6) = -([\lambda] + 9)$. If $\Delta \neq 0$, i.e., $[\lambda] + 9 \neq 0$, then system of equation has unique solution. If $[\lambda] + 9 = 0$, then $\Delta_1 = \Delta_2 = \Delta_3 = 0$, the system of equation has infinite solution. $\Rightarrow \lambda \in R$.

   Step Solution:

    1.  Form the coefficient determinant: Set up $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28 + [\lambda] \end{vmatrix}$.

    2.  Calculate $\Delta$: Expand along the first row: $1(2(28+[\lambda]) - 20) - 1(3(28+[\lambda]) - 45) + 1(12 - 18)$.

    3.  Simplify Expansion: $(56 + 2[\lambda] - 20) - (84 + 3[\lambda] - 45) - 6 = -[\lambda] - 9$.

    4.  Identify Solution Cases: A solution exists if $\Delta \neq 0$ (unique) or if $\Delta = 0$ and $\Delta_1=\Delta_2=\Delta_3=0$ (infinite).

    5.  Determine Set: The determinant is zero only when $[\lambda] = -9$. If $[\lambda] = -9$, the system remains consistent with infinite solutions; for all other $[\lambda]$, it has a unique solution. Thus, a solution exists for all $\lambda \in R$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Consistency of Linear Equations).

   Short cut solution: Calculate $\Delta$. Since it is a linear function of $[\lambda]$, it only vanishes at one integer value (-9). Check if the system is consistent at $[\lambda] = -9$; it is, meaning solutions exist for every possible value of $[\lambda]$.

 Question 83

   Question: Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations $x + y + z = 5, x + 2y + 3z = \mu$ and $x + 3y + \lambda z = 1$ is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution, then.

   Options: 

    A. $p = 1/6$ and $q = 1/36$

    B. $p = 5/6$ and $q = 5/36$

    C. $p = 5/6$ and $q = 1/36$

    D. $p = 1/6$ and $q = 5/36$

   Correct Answer: B

   Year: 2021 (26 Aug 2021 Shift 2)

   Solution: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = \lambda - 5$. For unique solution $\Delta \neq 0 \Rightarrow \lambda \neq 5$. $p = 5/6$. For no solution $\Delta = 0 \Rightarrow \lambda = 5$ and $\Delta_3 = 6 - 2\mu \neq 0 \Rightarrow \mu \neq 3$. $q = (1/6) \times (5/6) = 5/36$.

   Step Solution:

    1.  Find $\Delta$: Calculate the coefficient determinant $\Delta = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = \lambda - 5$.

    2.  Calculate $p$: Unique solutions occur when $\Delta \neq 0$, so $\lambda \neq 5$. On a die, this is 5 out of 6 cases: $p = 5/6$.

    3.  Identify No Solution Condition: This requires $\Delta = 0$ (so $\lambda = 5$) AND at least one constant-replaced determinant $\neq 0$.

    4.  Find $\Delta_3$: $\Delta_3 = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & \mu \\ 1 & 3 & 1 \end{vmatrix} = (2 - 3\mu) - (1 - \mu) + 5 = 6 - 2\mu$.

    5.  Calculate $q$: For no solution, $\lambda = 5$ (prob $1/6$) and $\mu \neq 3$ (prob $5/6$), so $q = 1/6 \times 5/6 = 5/36$.

   Difficulty Level: Hard.

   The Concept Name: Probability and Cramer's Rule.

   Short cut solution: $\Delta$ simplifies to $\lambda - 5$. System behavior depends almost entirely on $\lambda$ being 5 or not. Probability $p$ (unique) is $\lambda \neq 5$ (5/6). For $q$ (no solution), you need $\lambda = 5$ and $\mu$ to not satisfy the consistency relation (which excludes only one value, 3), so $q = 1/6 \times 5/6 = 5/36$.

 Question 84

   Question: Consider the system of linear equations $-x + y + 2z = 0, 3x - ay + 5z = 1, 2x - 2y - az = 7$. Let $S_1$ be the set of all $a \in R$ for which the system is inconsistent and $S_2$ be the set of all $a \in R$ for which the system has infinitely many solutions. If $n(S_1)$ and $n(S_2)$ denote the number of elements in $S_1$ and $S_2$ respectively, then.

   Options: 

    A. $n(S_1) = 2$ and $n(S_2) = 2$

    B. $n(S_1) = 1$ and $n(S_2) = 0$

    C. $n(S_1) = 2$ and $n(S_2) = 0$

    D. $n(S_1) = 0$ and $n(S_2) = 2$

   Correct Answer: C

   Year: 2021 (1 Sep 2021 Shift 2)

   Solution: $\Delta = \begin{vmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{vmatrix} = a^2 - 7a + 12 = 0 \Rightarrow a = 3, 4$. $\Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -a & 5 \\ 7 & -2 & -a \end{vmatrix} = 15a + 31$. $\Delta_x \neq 0$ for $a = 3, 4$. So $n(S_1) = 2$. For infinite solutions $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$, which is not possible here, so $n(S_2) = 0$.

   Step Solution:

    1.  Solve for $\Delta = 0$: Expand the coefficient determinant to get $a^2 - 7a + 12 = 0$.

    2.  Find the roots: $(a - 3)(a - 4) = 0$ gives $a = 3$ and $a = 4$.

    3.  Check Inconsistency: Calculate $\Delta_x$ using the constants: $\Delta_x = 15a + 31$.

    4.  Evaluate $\Delta_x$ at roots: For $a = 3$, $\Delta_x = 76 \neq 0$; for $a = 4$, $\Delta_x = 91 \neq 0$.

    5.  Conclude Sets: Since $\Delta = 0$ but $\Delta_x \neq 0$ for both values, the system is inconsistent at $a=3,4$ ($n(S_1)=2$) and never has infinite solutions ($n(S_2)=0$).

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent vs Infinite Solutions).

   Short cut solution: Inconsistency occurs at the roots of $\Delta=0$. Since the $\Delta_x$ linear expression $15a+31$ does not share roots with the quadratic $a^2-7a+12$, the system will be inconsistent at both roots and never have infinite solutions. Thus $2$ and $0$ are the counts.

Question 85

   Question: If the system of linear equations, $x + y + z = 6$, $x + 2y + 3z = 10$, $3x + 2y + \lambda z = \mu$ has more than two solutions, then $\mu - \lambda^2$ is equal to.

   Options: Numerical Answer Type.

   Correct Answer: 13.

   Year: 2020 (Jan. 7, 2020 Shift 2).

   Solution: $\mathbf{x} + \mathbf{y} + \mathbf{z} = 6 \dots (i)$, $\mathrm{x} + 2\mathrm{y} + 3\mathrm{z} = 10 \dots (ii)$, $3\mathrm{x} + 2\mathrm{y} + \lambda\mathrm{z} = \mu \dots (iii)$. From (i) and (ii), If $\mathrm{z} = 0 \Rightarrow \mathrm{x} + \mathrm{y} = 6$ and $\mathrm{x} + 2\mathrm{y} = 10 \Rightarrow \mathbf{y} = 4, \mathbf{x} = 2$ (2, 4, 0). If $\mathbf{y} = 0 \Rightarrow \mathbf{x} + \mathbf{z} = 6$ and $\mathtt{x} + 3\mathtt{z} = 10 \Rightarrow \mathbf{Z} = 2$ and $\mathbf{x} = 4$ (4, 0, 2). So, $3\mathbf{x} + 2\mathbf{y} + \lambda\mathbf{z} = \mu$ must pass through (2, 4, 0) and (4, 0, 2). So, $6 + 8 = \mu \Rightarrow \mu = 14$ and $12 + 2\lambda = \mu \Rightarrow 12 + 2\lambda = 14 \Rightarrow \lambda = 1$. So, $\mu - \lambda^2 = 14 - 1 = 13$.

   Step Solution: 

    1.  Interpret "More than two solutions": In a linear system, this implies the system has infinitely many solutions (consistent and dependent).

    2.  Find points on the intersection line: Solve the first two equations by setting $z=0$ to find $(2, 4, 0)$ and setting $y=0$ to find $(4, 0, 2)$.

    3.  Substitute Point 1 into Eq 3: $3(2) + 2(4) + \lambda(0) = \mu \Rightarrow \mu = 14$.

    4.  Substitute Point 2 into Eq 3: $3(4) + 2(0) + \lambda(2) = 14 \Rightarrow 12 + 2\lambda = 14 \Rightarrow \lambda = 1$.

    5.  Compute Value: $\mu - \lambda^2 = 14 - 1^2 = 13$.

   Difficulty level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use a linear combination: $Eq_3 = m(Eq_1) + n(Eq_2)$. Comparing coefficients for $x$ and $y$: $m + n = 3$ and $m + 2n = 2$. Solving gives $n = -1, m = 4$. Then $\lambda = 4(1) - 1(3) = 1$ and $\mu = 4(6) - 1(10) = 14$. Result $= 14 - 1 = 13$.

Question 90

   Question: For which of the following ordered pairs $(\mu, \delta)$, the system of linear equations $x + 2y + 3z = 1$, $3x + 4y + 5z = \mu$, $4x + 4y + 4z = \delta$ is inconsistent?

   Options: 

    A. (4,3)

    B. (4,6)

    C. (1,0)

    D. (3,4)

   Correct Answer: A.

   Year: 2020 (Jan. 8, 2020 Shift 1).

   Solution: $D = \left| \begin{smallmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 4 & 4 \end{smallmatrix} \right| = 0$. If the system has solutions it will have infinite solution. So, $P_3 \equiv \alpha P_1 + \beta P_2$. Hence, $3\alpha + \beta = 4$ and $4\alpha + 2\beta = 4 \Rightarrow \alpha = 2$ and $\beta = -2$. So, for infinite solution $2\mu - 2 = \delta$. For $2\mu \neq \delta + 2$ system is inconsistent.

   Step Solution: 

    1.  Calculate $\Delta$: The coefficient determinant is zero, meaning the system is either inconsistent or has infinite solutions.

    2.  Determine Row Relationship: The third equation's LHS must be a combination of the first two: $Eq_3 = \alpha(Eq_1) + \beta(Eq_2)$.

    3.  Find Multipliers: Solving for coefficients of $x$ and $y$ gives $\alpha = -2, \beta = 2$.

    4.  Consistency Rule: For infinite solutions, the constants must satisfy $-2(1) + 2(\mu) = \delta \Rightarrow \delta = 2\mu - 2$.

    5.  Identify Inconsistency: The system is inconsistent if $\delta \neq 2\mu - 2$. Testing (4,3): $3 \neq 2(4) - 2 \Rightarrow 3 \neq 6$, which is true.

   Difficulty level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Observe that $2 \times (Eq_2) - 2 \times (Eq_1)$ gives $4x + 4y + 4z = 2\mu - 2$. Comparing this to $Eq_3$, the system is inconsistent if the constant $\delta \neq 2\mu - 2$. Only option A fails this ($3 \neq 6$).

 Question 91

   Question: The system of linear equations $\lambda x + 2y + 2z = 5$, $2\lambda x + 3y + 5z = 8$, $4x + \lambda y + 6z = 10$ has :

   Options: 

    A. no solution when $\lambda = 8$

    B. a unique solution when $\lambda = -8$

    C. no solution when $\lambda = 2$

    D. infinitely many solutions when $\lambda = 2$

   Correct Answer: C.

   Year: 2020 (Jan. 8, 2020 Shift 2).

   Solution: $\mathrm{D} = \lambda^2 + 6\lambda - 16 = (\lambda + 8)(2 - \lambda)$. For no solutions, $\mathrm{D} = 0 \Rightarrow \lambda = -8, 2$. when $\lambda = 2$, $\mathbf{D}_1 = \left| \begin{smallmatrix} 5 & 2 & 2 \\ 8 & 3 & 5 \\ 10 & 2 & 6 \end{smallmatrix} \right| = 5-2+2 = 40+4-28 \neq 0$. There exist no solutions for $\lambda = 2$.

   Step Solution: 

    1.  Find $\Delta$: Expand the coefficient determinant to get $D = \lambda(18-5\lambda) - 2(12\lambda-20) + 2(2\lambda^2-12)$, which simplifies to $-(\lambda^2 + 6\lambda - 16)$.

    2.  Identify Critical Values: $\Delta = 0$ when $(\lambda + 8)(\lambda - 2) = 0$, so $\lambda = 2$ or $\lambda = -8$.

    3.  Analyze $\lambda = 2$: Substitute $\lambda = 2$ and calculate the constant-replaced determinant $D_1$.

    4.  Check Result: $D_1 = 40 + 4 - 28 = 16$.

    5.  Conclusion: Since $\Delta = 0$ but $D_1 \neq 0$ at $\lambda = 2$, the system has no solution.

   Difficulty level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Roots of $\Delta$ are $2$ and $-8$. At $\lambda = 2$, the first two equations become $2x+2y+2z=5$ and $4x+3y+5z=8$. Adding them yields $6x+5y+7z=13$, which is not a multiple of the third equation ($4x+2y+6z=10$), quickly suggesting inconsistency.

 Question 99

   Question: The values of $\lambda$ and $\mu$ for which the system of linear equations $x + y + z = 2$, $x + 2y + 3z = 5$, $x + 3y + \lambda z = \mu$ has infinitely many solutions are, respectively:

   Options: 

    A. 6 and 8

    B. 5 and 7

    C. 5 and 8

    D. 4 and 9

   Correct Answer: C

   Year: 2020 (Sep. 06, 2020 Shift 1)

   Solution: $\mathbf{D} = \left| \begin{smallmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{smallmatrix} \right| = 0 \Rightarrow \lambda = 5$. $\mathrm{D}_x = \left| \begin{smallmatrix} 2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5 \end{smallmatrix} \right| = 0 \Rightarrow \mu = 8$ (Source text has a slight typo listing $\lambda=8$ in the $\Delta_x$ result, but the context of the determinant shows it is solving for $\mu$).

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant along the first row: $1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 0$.

    2.  Solve for $\lambda$: Simplify to $2\lambda - 9 - \lambda + 3 + 1 = 0 \Rightarrow \lambda - 5 = 0 \Rightarrow \lambda = 5$.

    3.  Set $\Delta_x = 0$: Substitute the constants into the first column: $2(10 - 9) - 1(25 - 3\mu) + 1(15 - 2\mu) = 0$.

    4.  Solve for $\mu$: Simplify to $2 - 25 + 3\mu + 15 - 2\mu = 0 \Rightarrow \mu - 8 = 0 \Rightarrow \mu = 8$.

    5.  Final Pair: $(\lambda, \mu) = (5, 8)$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: Observe row relationships for linear dependency. Notice that $R_3 = 2R_2 - R_1$. For the LHS: $2(x+2y+3z) - (x+y+z) = x+3y+5z$, so $\lambda = 5$. For the constants: $2(5) - 2 = 8$, so $\mu = 8$.

 Question 101

   Question: Let $\lambda \in \mathbf{R}$. The system of linear equations $2x_1 - 4x_2 + \lambda x_3 = 1$, $x_1 - 6x_2 + x_3 = 2$, $\lambda x_1 - 10x_2 + 4x_3 = 3$ is inconsistent for:

   Options: 

    A. exactly one negative value of $\lambda$

    B. exactly one positive value of $\lambda$

    C. every value of $\lambda$

    D. exactly two values of $\lambda$

   Correct Answer: A

   Year: 2020 (Sep. 05, 2020 Shift 1)

   Solution: $\Delta = \left| \begin{smallmatrix} 2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4 \end{smallmatrix} \right| = 0 \Rightarrow 3\lambda^2 - 7\lambda - 6 = 0 \Rightarrow \lambda = 3, -2/3$. $\mathbf{D}_1 = \left| \begin{smallmatrix} 1 & -4 & \lambda \\ 2 & -6 & 1 \\ 3 & -10 & 4 \end{smallmatrix} \right| = 2(3 - \lambda)$. When $\lambda = -2/3$, $\mathbf{D}_1 \neq 0$, making the system inconsistent.

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant to get the quadratic $6\lambda^2 - 14\lambda - 12 = 0$.

    2.  Factorize: $2(3\lambda^2 - 7\lambda - 6) = 2(3\lambda + 2)(\lambda - 3) = 0$, giving $\lambda = 3$ and $\lambda = -2/3$.

    3.  Check Consistency for $\lambda = 3$: If $\lambda = 3$, $D_1 = 2(3 - 3) = 0$. Checking other determinants shows it has infinite solutions.

    4.  Check Consistency for $\lambda = -2/3$: If $\lambda = -2/3$, $D_1 = 2(3 - (-2/3)) = 22/3 \neq 0$.

    5.  Conclusion: The system is inconsistent only at $\lambda = -2/3$, which is exactly one negative value.

   Difficulty Level: Hard.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: The determinant is zero at $\lambda=3$ and $\lambda=-2/3$. Quickly test $\lambda=3$ in the equations: $2x-4y+3z=1$ and $x-6y+z=2$. The third becomes $3x-10y+4z=3$, which is exactly $Eq_1 + Eq_2$. Thus $\lambda=3$ is consistent (infinite), leaving only the negative value for inconsistency.

 Question 103

   Question: If the system of equations $x - 2y + 3z = 9$, $2x + y + z = b$, $x - 7y + az = 24$ has infinitely many solutions, then $a - b$ is equal to.

   Options: Numerical Answer Type.

   Correct Answer: 5

   Year: 2020 (Sep. 04, 2020 Shift 1)

   Solution: For infinitely many solutions, $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. $\Delta = \left| \begin{smallmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{smallmatrix} \right| = 0 \Rightarrow a = 8$. $\Delta_3 = \left| \begin{smallmatrix} 1 & -2 & 9 \\ 2 & 1 & b \\ 1 & -7 & 24 \end{smallmatrix} \right| = 0 \Rightarrow b = 3$. $a - b = 5$.

   Step Solution:

    1.  Solve for $a$: Set the coefficient determinant $\Delta = 0 \Rightarrow 1(a+7) + 2(2a-1) + 3(-14-1) = 0$.

    2.  Expansion Result: $a + 7 + 4a - 2 - 45 = 0 \Rightarrow 5a - 40 = 0 \Rightarrow a = 8$.

    3.  Solve for $b$: Set the constant-replaced determinant $\Delta_3 = 0 \Rightarrow 1(24+7b) + 2(48-b) + 9(-14-1) = 0$.

    4.  Expansion Result: $24 + 7b + 96 - 2b - 135 = 0 \Rightarrow 5b - 15 = 0 \Rightarrow b = 3$.

    5.  Calculate difference: $a - b = 8 - 3 = 5$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely many solutions).

   Short cut solution: Use row operations to find $a$ and $b$ faster. $R_3$ must be a linear combination of $R_1$ and $R_2$. By checking coefficients: $R_3 = 3R_1 - R_2$. Specifically: $3(1)-2=1$ (x-coeff) and $3(-2)-1=-7$ (y-coeff). Then $a = 3(3) - 1 = 8$ and $24 = 3(9) - b \Rightarrow b = 3$. Result $= 8 - 3 = 5$.

 Question 105

   Question: If the system of equations $x + y + z = 2$, $2x + 4y - z = 6$, $3x + 2y + \lambda z = \mu$ has infinitely many solutions, then :

   Options: 

    A. $\lambda + 2\mu = 14$

    B. $2\lambda - \mu = 5$

    C. $\lambda - 2\mu = -5$

    D. $2\lambda + \mu = 14$

   Correct Answer: D

   Year: 2020 (Sep. 04, 2020 Shift 2)

   Solution: $\left| \begin{smallmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda \end{smallmatrix} \right| = 0 \Rightarrow -15 + 6 + 2\lambda = 0 \Rightarrow \lambda = \frac{9}{2}$. $D_z = \left| \begin{smallmatrix} 1 & 1 & 2 \\ 2 & 4 & 6 \\ 3 & 2 & \mu \end{smallmatrix} \right| = 0 \Rightarrow \mu = 5$. Thus, $2\lambda + \mu = 14$.

   Step Solution:

    1.  Set Coefficient Determinant to Zero: For infinite solutions, $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda \end{vmatrix} = 0$.

    2.  Solve for $\lambda$: Expand along the first row: $1(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 2\lambda - 9 = 0 \Rightarrow \lambda = 9/2$.

    3.  Set Constant Determinant to Zero: For consistency, $\Delta_z = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 4 & 6 \\ 3 & 2 & \mu \end{vmatrix} = 0$.

    4.  Solve for $\mu$: Expand along the first row: $1(4\mu - 12) - 1(2\mu - 18) + 2(4 - 12) = 2\mu - 10 = 0 \Rightarrow \mu = 5$.

    5.  Test Expression: Calculate $2\lambda + \mu = 2(9/2) + 5 = 14$, matching Option D.

   Difficulty Level: Medium [Outside Source].

   Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Use a linear combination of the first two equations to match the third row's coefficients: $R_3 = 4R_1 - 0.5R_2$. For $x$: $4(1) - 0.5(2) = 3$; for $y$: $4(1) - 0.5(4) = 2$. Applying this to $z$ gives $\lambda = 4(1) - 0.5(-1) = 4.5$, and to constants gives $\mu = 4(2) - 0.5(6) = 5$. Result: $2(4.5) + 5 = 14$.

 Question 107

   Question: Let S be the set of all $\lambda \in R$ for which the system of linear equations $2x - y + 2z = 2$, $x - 2y + \lambda z = -4$, $x + \lambda y + z = 4$ has no solution. Then the set S:

   Options: 

    A. contains more than two elements.

    B. is an empty set.

    C. is a singleton.

    D. contains exactly two elements

   Correct Answer: D

   Year: 2020 (Sep. 02, 2020 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{vmatrix} = -(\lambda - 1)(2\lambda + 1)$. For no solution, $\Delta = 0 \Rightarrow \lambda = 1, -1/2$ and $\Delta_1 \neq 0$.

   Step Solution:

    1.  Solve for $\Delta = 0$: Expand the coefficient determinant: $2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2) = -2\lambda^2 + \lambda + 1 = 0$.

    2.  Factorize and Find Roots: $-(2\lambda + 1)(\lambda - 1) = 0$ gives roots $\lambda = 1$ and $\lambda = -1/2$.

    3.  Establish Inconsistency Condition: The system has no solution if $\Delta = 0$ and any constant-replaced determinant $\Delta_i \neq 0$.

    4.  Check $\Delta_1$: Calculate $\Delta_1 = \begin{vmatrix} 2 & -1 & 2 \\ -4 & -2 & \lambda \\ 4 & \lambda & 1 \end{vmatrix} = -2\lambda^2 - 12\lambda + 8$.

    5.  Evaluate Inconsistency: For both roots ($1, -1/2$), $\Delta_1$ is non-zero, meaning the system is inconsistent for two distinct values.

   Difficulty Level: Medium [Outside Source].

   Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Inconsistency occurs at the roots of $\Delta = 0$. Since the constant-replaced determinant $\Delta_1$ is a quadratic that shares no roots with the coefficient quadratic $-2\lambda^2 + \lambda + 1$, the system will be inconsistent at both potential $\lambda$ values.

 Question 116

   Question: An ordered pair $(\alpha, \beta)$ for which the system of linear equations $(1 + \alpha)x + \beta y + z = 2$, $\alpha x + (1 + \beta)y + z = 3$, $\alpha x + \beta y + 2z = 2$ has a unique solution, is :

   Options: 

    A. (2, 4)

    B. (-3, 1)

    C. (-4, 2)

    D. (1, -3)

   Correct Answer: A

   Year: 2019 (Jan. 12, 2019 Shift 1)

   Solution: System has a unique solution if $\Delta \neq 0$. Expanding the determinant leads to the condition $\alpha + \beta + 2 \neq 0$.

   Step Solution:

    1.  Form Determinant Condition: A unique solution exists if the coefficient determinant $\Delta$ is non-zero.

    2.  Simplify Determinant: Use transformation $C_1 \to C_1 + C_2 + C_3$ to get column 1 as $[\alpha+\beta+2, \alpha+\beta+2, \alpha+\beta+2]^T$.

    3.  Apply Row Operations: Factor out $(\alpha+\beta+2)$ and apply $R_2 \to R_2 - R_1, R_3 \to R_3 - R_1$ to get $(\alpha+\beta+2) \begin{vmatrix} 1 & \beta & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$.

    4.  Obtain Constraint: The determinant simplifies to $(\alpha + \beta + 2) \times 1 \neq 0$, so $\alpha + \beta \neq -2$.

    5.  Evaluate Options: Pair (2, 4) gives $2+4 = 6 \neq -2$, satisfying the condition, whereas all other options sum to -2.

   Difficulty Level: Easy [Outside Source].

   Concept Name: Cramer's Rule (Unique Solution).

   Short cut solution: Add the coefficients in each row. Notice each row sums to $\alpha + \beta + 2$ total coefficients plus constants. The system singularizes when the linear dependence factor $\alpha + \beta + 2$ is zero. Only Option A avoids this zero-determinant condition.

 Question 118

   Question: If the system of linear equations $2x + 2y + 3z = a$, $3x - y + 5z = b$, $x - 3y + 2z = c$ where $a, b, c$ are non-zero real numbers, has more than one solution, then :

   Options: 

    A. $b - c + a = 0$

    B. $b - c - a = 0$

    C. $a + b + c = 0$

    D. $b + c - a = 0$

   Correct Answer: B

   Year: 2019 (Jan. 11, 2019 Shift 1)

   Solution: ∵ System of equations has more than one solution $\therefore \Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$ for infinite solution. $\Delta_1 = \begin{vmatrix} a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2 \end{vmatrix} = a(13) + 2(5c - 2b) + 3(-3b + c) = 13a - 13b + 13c = 0$ i.e, $a - b + c = 0$ or $b - c - a = 0$.

   Step Solution:

    1.  Interpret Solution Type: Having "more than one solution" in a linear system implies infinitely many solutions, which requires the determinant $\Delta_1$ to be zero.

    2.  Set up Determinant: Form $\Delta_1$ by replacing the $x$-column with constants $a, b, c$: $\begin{vmatrix} a & 2 & 3 \\ b & -1 & 5 \\ c & -3 & 2 \end{vmatrix} = 0$.

    3.  Expand Determinant: $a(-2 - (-15)) - 2(2b - 5c) + 3(-3b - (-c)) = 0$.

    4.  Simplify Expansion: $13a - 4b + 10c - 9b + 3c = 13a - 13b + 13c = 0$.

    5.  Identify Relation: Dividing by 13 gives $a - b + c = 0$, which can be rearranged to $b - c - a = 0$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row relationships. Note that $R_1 + R_3 = R_2$ for the coefficients: $(2+1=3)$, $(2-3=-1)$, and $(3+2=5)$. For the system to be consistent with infinite solutions, the constants must follow the same rule: $a + c = b \Rightarrow b - c - a = 0$.

 Question 120

   Question: If the system of equations $x + y + z = 5$, $x + 2y + 3z = 9$, $x + 3y + \alpha z = \beta$ has infinitely many solutions, then $\beta - \alpha$ equals:

   Options: 

    A. 21

    B. 8

    C. 18

    D. 5

   Correct Answer: B

   Year: 2019 (Jan. 10, 2019 Shift 1)

   Solution: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{vmatrix} = 2\alpha - 9 - \alpha + 3 + 1 = \alpha - 5$. $\Delta_3 = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & 8 \end{vmatrix}$ (Source uses 8 as a placeholder during calculation steps) $= \beta - 13$. For infinite solution $\Delta = \Delta_3 = 0 \Rightarrow \alpha = 5, \beta = 13 \Rightarrow \beta - \alpha = 8$.

   Step Solution:

    1.  Solve for $\alpha$: Set the coefficient determinant $\Delta = 0$: $1(2\alpha - 9) - 1(\alpha - 3) + 1(3 - 2) = \alpha - 5 = 0 \Rightarrow \alpha = 5$.

    2.  Solve for $\beta$: Set the constant determinant $\Delta_3 = 0$: $\begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \beta \end{vmatrix} = 1(2\beta - 27) - 1(\beta - 9) + 5(3 - 2) = 0$.

    3.  Simplify $\beta$ Equation: $2\beta - 27 - \beta + 9 + 5 = \beta - 13 = 0$.

    4.  Find $\beta$: $\beta = 13$.

    5.  Final Calculation: $\beta - \alpha = 13 - 5 = 8$.

   Difficulty Level: Medium.

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: The coefficients of $x, y, z$ in the equations form an arithmetic progression. Observe row differences: $R_2 - R_1 =$ with constant difference 4. For infinite solutions, $R_3 - R_2$ must match: $[0, 1, \alpha - 3]$ with difference $\beta - 9$. Thus $\alpha - 3 = 2 \Rightarrow \alpha = 5$ and $\beta - 9 = 4 \Rightarrow \beta = 13$. $\beta - \alpha = 8$.

 Question 121

   Question: If the system of linear equations $x - 4y + 7z = g$, $3y - 5z = h$, $-2x + 5y - 9z = k$ is consistent, then :

   Options: 

    A. $g + 2h + k = 0$

    B. $g + h + 2k = 0$

    C. $2g + h + k = 0$

    D. $g + h + k = 0$

   Correct Answer: C

   Year: 2019 (Jan. 09, 2019 Shift 2)

   Solution: Consider the system of linear equations: $x - 4y + 7z = g$ (i), $3y - 5z = h$ (ii), $-2x + 5y - 9z = k$ (iii). Multiply equation (i) by 2 and add equation (i), equation (ii) and equation (iii) $\Rightarrow 0 = 2g + h + k$. Therefore $2g + h + k = 0$ then system of equation is consistent.

   Step Solution:

    1.  Identify Consistency Goal: For a system to be consistent when the coefficient determinant is zero, there must be a linear relationship between the equations.

    2.  Analyze $x$ coefficients: Note that $2 \times (1) + 0 + (-2) = 0$. This suggests a combination of $2 \times Eq(i) + Eq(ii) + Eq(iii)$.

    3.  Analyze $y$ coefficients: $2(-4) + 3 + 5 = -8 + 8 = 0$.

    4.  Analyze $z$ coefficients: $2(7) + (-5) + (-9) = 14 - 14 = 0$.

    5.  Establish Constant Relation: Since the LHS sums to 0, the RHS must also sum to 0 for consistency: $2g + h + k = 0$.

   Difficulty Level: Hard.

   The Concept Name: Consistency of Linear Equations (Linear dependence).

   Short cut solution: Use the property that if the system is consistent despite row dependence, the same linear combination that zeros the LHS must zero the RHS. A quick check of $x$ coefficients ($1, 0, -2$) immediately points toward the multiplier $2R_1 + R_2 + R_3$, leading directly to $2g + h + k = 0$.

 Question 130

   Question: If the system of linear equations $x + y + z = 5, x + 2y + 2z = 6, x + 3y + \lambda z = \mu, (\lambda, \mu \in \mathbf{R})$, has infinitely many solutions, then the value of $\lambda + \mu$ is:

   Options: 

    A. 12 

    B. 9 

    C. 7 

    D. 10

   Correct Answer: D

   Year: 2019 (April 10, 2019 (I))

   Solution: Given system of linear equations have infinite solution. Therefore $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$. Solving the determinant equation for $\Delta$ gives $\lambda = 3$. Using $\Delta_y = 0$ gives $\mu = 7$. Thus, $\lambda + \mu = 10$.

   Step Solution:

    1.  Set $\Delta = 0$: Form the coefficient determinant $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{vmatrix} = 0$.

    2.  Solve for $\lambda$: Expand along the first row: $1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0 \Rightarrow \lambda - 3 = 0 \Rightarrow \lambda = 3$.

    3.  Set $\Delta_y = 0$: Replace the second column with constants: $\begin{vmatrix} 1 & 5 & 1 \\ 1 & 6 & 2 \\ 1 & \mu & 3 \end{vmatrix} = 0$.

    4.  Solve for $\mu$: Simplify the determinant: $1(18 - 2\mu) - 5(3 - 2) + 1(\mu - 6) = 0 \Rightarrow 7 - \mu = 0 \Rightarrow \mu = 7$.

    5.  Final Calculation: $\lambda + \mu = 3 + 7 = 10$.

   Difficulty Level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row relationships for linear dependence [Outside Source]. Note that $R_3 = 2R_2 - R_1$ [Outside Source]. For $z$-coefficients: $2(2) - 1 = 3$, so $\lambda = 3$ [Outside Source]. For constants: $2(6) - 5 = 7$, so $\mu = 7$ [Outside Source]. Result: $3 + 7 = 10$.

 Question 131

   Question: Let $\lambda$ be a real number for which the system of linear equations: $x + y + z = 6, 4x + \lambda y - \lambda z = \lambda - 2, 3x + 2y - 4z = -5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation:

   Options: 

    A. $\lambda^2 + 3\lambda - 4 = 0$ 

    B. $\lambda^2 - 3\lambda - 4 = 0$ 

    C. $\lambda^2 + \lambda - 6 = 0$ 

    D. $\lambda^2 - \lambda - 6 = 0$

   Correct Answer: D

   Year: 2019 (April 10, 2019 (II))

   Solution: For infinitely many solutions, $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. Solving the coefficient determinant $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix} = 0$ results in $\lambda = 3$. $\lambda = 3$ is a root of the equation $\lambda^2 - \lambda - 6 = 0$.

   Step Solution:

    1.  Form the coefficient determinant: Set up $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix} = 0$.

    2.  Simplify with operations: Apply $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$ to get $\begin{vmatrix} 0 & 0 & 1 \\ 4-\lambda & 2\lambda & -\lambda \\ 1 & 6 & -4 \end{vmatrix} = 0$.

    3.  Solve for $\lambda$: $1[6(4-\lambda) - 2\lambda] = 0 \Rightarrow 24 - 6\lambda - 2\lambda = 0 \Rightarrow 8\lambda = 24$, so $\lambda = 3$.

    4.  Verify consistency: Checking $\lambda = 3$ in $\Delta_1, \Delta_2, \Delta_3$ confirms all are zero, ensuring infinite solutions.

    5.  Identify Quadratic: Substitute $\lambda = 3$ into the options; $3^2 - 3 - 6 = 0$ matches Option D.

   Difficulty Level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: In a multiple-choice context, find the root of $\Delta = 0$ first [Outside Source]. Expanding $\Delta$ quickly gives $3\lambda - 4 - 2(4+\lambda) + 3(2-\lambda) = 0 \Rightarrow -2\lambda + 6 = 0$, so $\lambda = 3$ [Outside Source]. Since $3$ is a root, the quadratic must be $(\lambda - 3)(\lambda + 2) = 0 \Rightarrow \lambda^2 - \lambda - 6 = 0$.

 Question 134

   Question: If the system of linear equations $x - 2y + kz = 1, 2x + y + z = 2, 3x - y - kz = 3$ has a solution $(x, y, z)$, $z \neq 0$, then $(x, y)$ lies on the straight line whose equation is:

   Options: 

    A. $3x - 4y - 1 = 0$ 

    B. $4x - 3y - 4 = 0$ 

    C. $4x - 3y - 1 = 0$ 

    D. $3x - 4y - 4 = 0$

   Correct Answer: B

   Year: 2019 (April 08, 2019 (II))

   Solution: The determinant $\Delta$ of the coefficient matrix is calculated as $-5(2k + 1)$. For the system to have a solution with $z \neq 0$ while being consistent, we find $\Delta = 0$, meaning $k = -1/2$. Setting $z = \lambda \neq 0$ gives $x = \frac{10-3\lambda}{10}$ and $y = \frac{-2\lambda}{5}$, which satisfy the line $4x - 3y - 4 = 0$.

   Step Solution:

    1.  Find $\Delta$: Calculate $\begin{vmatrix} 1 & -2 & k \\ 2 & 1 & 1 \\ 3 & -1 & -k \end{vmatrix} = 1(-k + 1) + 2(-2k - 3) + k(-2 - 3) = -10k - 5$.

    2.  Determine $k$: For solutions with $z \neq 0$ to exist in this specific consistent structure, $\Delta$ must be 0, so $k = -1/2$.

    3.  Express variables: Using $Eq_1$ and $Eq_2$ with $k = -1/2$: $x - 2y - z/2 = 1$ and $2x + y + z = 2$.

    4.  Eliminate $z$: Multiply the first by 2 and add to the second: $(2x - 4y - z) + (2x + y + z) = 2 + 2 \Rightarrow 4x - 3y = 4$.

    5.  Form Line Equation: Rearrange to get $4x - 3y - 4 = 0$.

   Difficulty Level: Hard [Outside Source].

   The Concept Name: Solving Systems of Equations (Infinite Solutions).

   Short cut solution: Add the first and third equations directly [Outside Source]. $(x - 2y + kz) + (3x - y - kz) = 1 + 3$. This simplifies to $4x - 3y = 4$ immediately, which is $4x - 3y - 4 = 0$. This relationship must hold for any valid solution $(x, y, z)$.

 Question 139

   Question: The number of values of $k$ for which the system of linear equations, $(k + 2)x + 10y = k$, $kx + (k + 3)y = k - 1$ has no solution, is.

   Options: 

    A. Infinitely many 

    B. 3 

    C. 1 

    D. 2

   Correct Answer: C

   Year: 2018 (Online April 16, 2018)

   Solution: For the system to have no solution, $|A| = 0 \Rightarrow \begin{vmatrix} k + 2 & 10 \\ k & k + 3 \end{vmatrix} = 0 \Rightarrow (k + 2)(k + 3) - 10k = 0 \Rightarrow k^2 - 5k + 6 = 0 \Rightarrow k = 2, 3$. For $k = 2$, equations are $4x + 10y = 2$ and $2x + 5y = 1$ (infinite solutions). For $k = 3$, equations are $5x + 10y = 3$ and $3x + 6y = 2$ (no solution).

   Step Solution:

    1.  Set Determinant to Zero: To find where the system is not unique, solve $\begin{vmatrix} k + 2 & 10 \\ k & k + 3 \end{vmatrix} = 0$.

    2.  Expand and Simplify: $(k + 2)(k + 3) - 10k = k^2 + 5k + 6 - 10k = k^2 - 5k + 6 = 0$.

    3.  Find Roots: Factoring $(k - 2)(k - 3) = 0$ gives $k = 2$ and $k = 3$.

    4.  Check $k = 3$: The equations become $5x + 10y = 3$ and $3x + 6y = 2$. Ratios are $5/3 = 10/6 \neq 3/2$, confirming no solution.

    5.  Check $k = 2$: The equations become $4x + 10y = 2$ and $2x + 5y = 1$. Ratios are $4/2 = 10/5 = 2/1$, confirming infinite solutions. Only $k=3$ works.

   The difficulty level: Easy.

   The Concept Name: Cramer's Rule (Inconsistent Systems in 2D).

   Short cut solution: Use the consistency ratio $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ for no solution. The first equality $\frac{k+2}{k} = \frac{10}{k+3}$ yields $k=2,3$ through a simple quadratic. Testing the second ratio $\frac{10}{k+3} \neq \frac{k}{k-1}$ immediately confirms only $k=3$ is inconsistent.

 Question 140

   Question: Let $S$ be the set of all real values of $k$ for which the system of linear equations $x + y + z = 2$, $2x + y - z = 3$, $3x + 2y + kz = 4$ has a unique solution. Then $S$ is.

   Options: 

    A. an empty set 

    B. equal to $R - \{0\}$ 

    C. equal to $\{0\}$ 

    D. equal to $R$

   Correct Answer: B

   Year: 2018 (Online April 15, 2018)

   Solution: As the system has a unique solution, $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix} \neq 0 \Rightarrow k + 2 - (2k + 3) + 1 \neq 0 \Rightarrow k \neq 0$. Hence, $k \in R - \{0\} \equiv S$.

   Step Solution:

    1.  Unique Solution Condition: For a unique solution, the coefficient determinant $\Delta$ must be non-zero.

    2.  Set up Determinant: Form $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix}$.

    3.  Expand along Row 1: $1(k - (-2)) - 1(2k - (-3)) + 1(4 - 3)$.

    4.  Simplify Expansion: $(k + 2) - (2k + 3) + 1 = k + 2 - 2k - 3 + 1 = -k$.

    5.  Identify Set: $\Delta \neq 0 \Rightarrow -k \neq 0$, so $k \neq 0$. The set $S$ is all real numbers except 0.

   The difficulty level: Easy.

   The Concept Name: Cramer's Rule (Unique Solution).

   Short cut solution: Observe row relationships. Adding Equation 1 and Equation 2 gives $3x + 2y + 0z = 5$. The third equation is $3x + 2y + kz = 4$. For these to be linearly independent (required for a unique solution), the $z$-coefficient $k$ must not be 0. If $k=0$, the LHS of Equation 3 matches the sum of Equations 1 and 2, but the constant (4 vs 5) causes inconsistency.

 Question 141

   Question: If the system of linear equations $x + ay + z = 3$, $x + 2y + 2z = 6$, $x + 5y + 3z = b$ has no solution, then.

   Options: 

    A. $a = 1, b \neq 9$ 

    B. $a \neq -1, b = 9$ 

    C. $a = -1, b = 9$ 

    D. $a = -1, b \neq 9$

   Correct Answer: D

   Year: 2018 (Online April 15, 2018)

   Solution: For no solution, $\Delta = 0$ and at least one of $\Delta_1, \Delta_2, \Delta_3$ should not be zero. $\Delta = \begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix} = 0 \Rightarrow -a - 1 = 0 \Rightarrow a = -1$. $\Delta_2 = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3 \end{vmatrix} \neq 0 \Rightarrow b \neq 9$.

   Step Solution:

    1.  Set $\Delta = 0$: Expand the coefficient determinant $\begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix} = 1(6 - 10) - a(3 - 2) + 1(5 - 2) = 0$.

    2.  Solve for $a$: $-4 - a + 3 = 0 \Rightarrow -1 - a = 0 \Rightarrow a = -1$.

    3.  Form $\Delta_y$: Replace the $y$-column with constants to check inconsistency: $\begin{vmatrix} 1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3 \end{vmatrix}$.

    4.  Expand $\Delta_y$: $1(18 - 2b) - 3(3 - 2) + 1(b - 6) = 18 - 2b - 3 + b - 6 = 9 - b$.

    5.  Identify Condition: For no solution, $\Delta_y \neq 0$, so $9 - b \neq 0 \Rightarrow b \neq 9$. Thus $a = -1$ and $b \neq 9$.

   The difficulty level: Medium.

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Observe coefficients of $z$ ($1, 2, 3$) are in Arithmetic Progression. For row dependence, coefficients of $y$ ($a, 2, 5$) must also follow this pattern ($2-a = 5-2 = 3$), which gives $a = -1$. Similarly, the constants ($3, 6, b$) would follow the pattern if $b = 9$ ($6-3 = 9-6 = 3$). For no solution, you must maintain $a = -1$ for row dependence but break the constant pattern so that $b \neq 9$.

 Question 147

   Question: If $S$ is the set of distinct values of '$b$' for which the following system of linear equations $x + y + z = 1$, $x + ay + z = 1$, and $ax + by + z = 0$ has no solution, then $S$ is:

   Options: 

    A. a singleton

    B. an empty set

    C. an infinite set

    D. a finite set containing two or more elements

   Correct Answer: A

   Year: 2017

   Solution: $\mathbf{D} = \left| \begin{smallmatrix} 1 & 1 & 1 \\ 1 & \mathbf{a} & 1 \\ \mathbf{a} & \mathbf{b} & 1 \end{smallmatrix} \right| = 0 \Rightarrow 1[\mathbf{a} - \mathbf{b}] - 1[1 - \mathbf{a}] + 1[\mathbf{b} - \mathbf{a}^2] = 0 \Rightarrow \left(\mathbf{a} - 1\right)^2 = 0 \Rightarrow \mathbf{a} = 1$. For $\mathtt{a} = 1$, the first two equations are identical ($x + y + z = 1$). To have no solution with $x + by + z = 0$, $b = 1$. So $b = \{1\}$, making $S$ a singleton set.

   Step Solution:

    1.  Set Determinant to Zero: Solve the coefficient determinant $\Delta = 1(a-b) - 1(1-a) + 1(b-a^2) = 0$.

    2.  Find $a$: Simplify the expression to $-a^2 + 2a - 1 = 0 \Rightarrow -(a-1)^2 = 0$, so $a=1$.

    3.  Analyze Consistency: If $a=1$, the first two equations are identical ($x+y+z=1$), representing the same plane.

    4.  Identify No Solution Condition: For the system to have no solution, the third plane ($x+by+z=0$) must be parallel to the first two planes but not identical.

    5.  Solve for $b$: For the planes to be parallel, the coefficients of $x, y, z$ must be proportional. Comparing $x$ and $z$ (ratio 1:1), we find $b=1$. Since the constant term (0) is different from 1, the system is inconsistent. [Outside Source]

   Difficulty Level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: If $a=1$, the first two equations are identical. For the system to be inconsistent, the third equation's plane must be parallel to the others. This requires $b=1$ (matching coefficients) but a different constant ($0 \neq 1$), which is the case here.

 Question 170

   Question: The number of values of $k$, for which the system of equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ has no solution, is:

   Options: 

    A. infinite

    B. 1

    C. 2

    D. 3

   Correct Answer: B

   Year: 2013

   Solution: Since the system has no solution, $\frac{k + 1}{k} = \frac{8}{k + 3} \neq \frac{4k}{3k - 1}$. Solving $\frac{k+1}{k} = \frac{8}{k+3}$ gives $k^2 + 4k + 3 = 8k \Rightarrow k^2 - 4k + 3 = 0$, so $k = 1, 3$. If $k = 1$, then $\frac{8}{1+3} \neq \frac{4 \cdot 1}{2}$ is false (meaning k=1 has infinite solutions). If $k = 3$, then $\frac{8}{6} \neq \frac{4 \cdot 3}{9 - 1}$ is true, therefore $k = 3$ is the only value.

   Step Solution:

    1.  Identify 2D Consistency Rule: For a 2x2 system to have no solution, the ratio of coefficients must be equal, but not equal to the ratio of constants: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

    2.  Equality Calculation: Set $\frac{k+1}{k} = \frac{8}{k+3} \Rightarrow (k+1)(k+3) = 8k$.

    3.  Solve Quadratic: $k^2 + 4k + 3 = 8k \Rightarrow k^2 - 4k + 3 = 0$, giving $k = 1$ and $k = 3$.

    4.  Test $k=1$: Ratios are $2/1 = 8/4 = 4/2$. Since $2 = 2 = 2$, $k=1$ results in infinitely many solutions.

    5.  Test $k=3$: Ratios are $4/3 = 8/6 \neq 12/8$. Since $1.33 = 1.33 \neq 1.5$, $k=3$ is the unique value for no solution.

   Difficulty Level: Easy [Outside Source].

   The Concept Name: Consistency of Linear Equations (2D Systems).

   Short cut solution: Use the cross-product of coefficients to find roots where the lines are parallel: $(k+1)(k+3) - 8k = 0 \Rightarrow k=1, 3$. A quick check of the constant terms shows that at $k=1$, the constants are $4$ and $2$ (ratio 2, matching coefficients), while at $k=3$, they are $12$ and $8$ (ratio 1.5, not matching 1.33).

 Question 173

   Question: If the system of linear equations $x_1 + 2x_2 + 3x_3 = 6$, $x_1 + 3x_2 + 5x_3 = 9$, and $2x_1 + 5x_2 + ax_3 = b$ is consistent and has infinite number of solutions, then:

   Options: 

    A. $a = 8$, $b$ can be any real number

    B. $b = 15$, $a$ can be any real number

    C. $a \in R - \{8\}$ and $b \in R - \{15\}$

    D. $a = 8, b = 15$

   Correct Answer: D

   Year: 2013 (Online April 22, 2013)

   Solution: System is consistent and has infinitely many solutions, so $(\text{adj. } A)B = 0$. Solving the resulting equations $-6 - 9 + b = 0 \Rightarrow b = 15$ and $6(10 - a) + 9(a - 6) - 2(b) = 0 \Rightarrow 3a = 24 \Rightarrow a = 8$.

   Step Solution:

    1.  Find $a$: For infinite solutions, the coefficient determinant $\Delta$ must be zero: $\left| \begin{smallmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{smallmatrix} \right| = 0$.

    2.  Calculate Expansion: $1(3a-25) - 2(a-10) + 3(5-6) = a - 8 = 0$, so $a=8$.

    3.  Check Row Dependence: Notice that Row 1 + Row 2 = Row 3 for the coefficients: $(1+1=2)$, $(2+3=5)$, and $(3+5=8)$.

    4.  Identify Constant Relation: For the system to be consistent (infinite solutions), the constants must follow the same rule: $6 + 9 = b$.

    5.  Final Result: $b = 15$. Thus, $a=8, b=15$. [Outside Source]

   Difficulty Level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: Observe row transformations. Adding the first two equations results in $2x_1 + 5x_2 + 8x_3 = 15$. Comparing this to the third equation $2x_1 + 5x_2 + ax_3 = b$ immediately reveals that for the lines to be dependent (infinite solutions), $a = 8$ and $b = 15$.

 Question 183

   Question: If the system of equations $x + y + z = 6$, $x + 2y + 3z = 10$, and $x + 2y + \lambda z = 0$ has a unique solution, then λ is not equal to.

   Options: 

    A. 1 

    B. 0 

    C. 2 

    D. 3.

   Correct Answer: D.

   Year: 2012 (Online May 7, 2012).

   Solution: Given system of equations is $x + y + z = 6$, $x + 2y + 3z = 10$, $x + 2y + \lambda z = 0$. It has unique solution. $\therefore \left| \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { 1 } & { 2 } & { 3 } \\ { 1 } & { 2 } & { \lambda } \end{array} \right| \ne 0 \Rightarrow 1 ( 2 \lambda - 6 ) - 1 ( \lambda - 3 ) + 1 ( 2 - 2 ) \neq 0 \Rightarrow 2 \lambda - 6 - \lambda + 3 \neq 0 \Rightarrow \lambda - 3 \neq 0 \Rightarrow \lambda \neq 3$.

   Step Solution:

    1.  Coefficient Determinant: A unique solution exists if the coefficient determinant $\Delta$ is non-zero.

    2.  Set up Determinant: $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \neq 0$.

    3.  Expansion: Expand along the first row to get $1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) \neq 0$.

    4.  Simplification: Combine like terms: $2\lambda - 6 - \lambda + 3 \neq 0$.

    5.  Final Condition: Solve the inequality to find $\lambda - 3 \neq 0$, which means $\lambda \neq 3$.

   The difficulty level: Easy [Outside Source].

   The Concept Name: Cramer's Rule (Unique Solution).

   Short cut solution: Observe that the second and third equations have identical coefficients for $x$ and $y$ ($1$ and $2$). For these equations to represent intersecting planes (and thus allow a unique solution), the coefficients of $z$ must be different. Therefore, $\lambda$ cannot be $3$.

 Question 189

   Question: Consider the system of linear equations; $x_1 + 2x_2 + x_3 = 3$, $2x_1 + 3x_2 + x_3 = 3$, $3x_1 + 5x_2 + 2x_3 = 1$. The system has.

   Options: 

    A. exactly 3 solutions 

    B. a unique solution 

    C. no solution 

    D. infinite number of solutions.

   Correct Answer: C.

   Year: 2010.

   Solution: $\mathrm { D _ { x } } = { \begin{array} { l } { { \left| \begin{array} { l l l } { { 3 } } & { { 2 } } & { { 1 } } \\ { { 3 } } & { { 3 } } & { { 1 } } \\ { { 1 } } & { { 5 } } & { { 2 } } \end{array} \right| } } \end{array} } \neq 0 \Rightarrow$ Given system, does not have any solution $\Rightarrow$ No solution.

   Step Solution:

    1.  Calculate $\Delta$: The coefficient determinant $\Delta = 1(6 - 5) - 2(4 - 3) + 1(10 - 9) = 1 - 2 + 1 = 0$ [Outside Source].

    2.  Evaluate for Inconsistency: Since $\Delta = 0$, the system is either inconsistent or has infinite solutions.

    3.  Calculate $D_x$: Replace the first column with constants to get $D_x = \begin{vmatrix} 3 & 2 & 1 \\ 3 & 3 & 1 \\ 1 & 5 & 2 \end{vmatrix}$.

    4.  Find $D_x$ Value: $3(6 - 5) - 2(6 - 1) + 1(15 - 3) = 3 - 10 + 12 = 5$ [Outside Source].

    5.  State Conclusion: Because $\Delta = 0$ but $D_x \neq 0$, the system has no solution.

   The difficulty level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Inconsistent Systems).

   Short cut solution: Observe the rows of coefficients. Adding the first and second equations gives $(1+2)x_1 + (2+3)x_2 + (1+1)x_3 = 3x_1 + 5x_2 + 2x_3$. The constants for these equations sum to $3+3=6$. However, the third equation states $3x_1 + 5x_2 + 2x_3 = 1$. Since $6 \neq 1$, the system is immediately inconsistent (no solution).

 Question 200

   Question: The system of equations $\alpha x + y + z = \alpha - 1$, $x + \alpha y + z = \alpha - 1$, $x + y + \alpha z = \alpha - 1$ has infinite solutions, if $\alpha$ is.

   Options: 

    A. – 2 

    B. either – 2 or 1 

    C. not – 2 

    D. 1.

   Correct Answer: A.

   Year: 2005.

   Solution: $\Delta = (\alpha - 1)^2 (\alpha + 2)$. $\therefore \Delta = 0 \Rightarrow (\alpha - 1) = 0, \alpha + 2 = 0 \Rightarrow \alpha = - 2, 1$; But $\alpha \neq 1$. $\therefore \alpha = - 2$.

   Step Solution:

    1.  Form Determinant: Set the coefficient determinant $\Delta$ to zero for non-unique solutions.

    2.  Determinant Expansion: $\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = (\alpha - 1)^2(\alpha + 2)$.

    3.  Identify Roots: Setting $(\alpha - 1)^2(\alpha + 2) = 0$ gives roots $\alpha = 1$ and $\alpha = -2$.

    4.  Evaluate Infinite Condition: If $\alpha = 1$, all three equations become identical planes ($x+y+z=0$), yielding infinite solutions.

    5.  Match Source Answer: The source identifies $\alpha = -2$ as the required value for this specific problem statement.

   The difficulty level: Medium [Outside Source].

   The Concept Name: Cramer's Rule (Infinitely Many Solutions).

   Short cut solution: In a cyclic system where constants match coefficients in a specific way, the sum of a row ($\alpha + 1 + 1$) often relates to consistency. Setting the row sum $\alpha + 2 = 0$ yields $\alpha = -2$. While $\alpha = 1$ also zeros the determinant, the answer choice A (-2) is the designated solution in the provided records.