Matrices and Determinants

We begin with the system:

$ \begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned} $

Step 1. Solve the first equation for $x$:

$ x = 6 - y - 2z. $

Step 2. Substitute $x = 6-y-2z$ into the second equation:

$ 2(6-y-2z) + 3y + az = a+1. $

Expanding and simplifying:

$ 12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11. $

Call this Equation (I).

Step 3. Substitute $x = 6-y-2z$ into the third equation:

$ -(6-y-2z) - 3y + bz = 2b. $

Expanding and simplifying:

$ -6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6. $

Call this Equation (II).

Step 4. For the system to have infinitely many solutions, the two equations in $y$ and $z$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $k$ such that

$ -2 = k \cdot 1 \quad \Longrightarrow \quad k = -2. $

Apply this to the coefficient of $z$ and the constant term.

For the $z$-coefficient in Equations (I) and (II):

$ b+2 = k(a-4) = -2(a-4) = -2a+8. $

Thus,

$ b = -2a+6. $

For the constant term:

$ 2b+6 = k(a-11) = -2(a-11) = -2a + 22. $

Substitute $b = -2a+6$ into this equation:

$ 2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22. $

Simplify:

$ -4a + 18 = -2a + 22. $

Solve for $a$:

$ -4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22, $

$ 2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2. $

Substitute $a = -2$ into $b = -2a+6$:

$ b = -2(-2) + 6 = 4 + 6 = 10. $

Step 5. We now compute

$ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. $

Thus, the value of $7a+3b$ is

$ \boxed{16}. $

$ B = \operatorname{adj}(\operatorname{adj}(2A)) = \det(2A) \cdot (2A) $

Since $ A $ is a $ 3 \times 3 $ matrix with

$ \det(A) = \frac{1}{2}, $

the determinant of $ 2A $ is computed as

$ \det(2A) = 2^3 \det(A) = 8 \cdot \frac{1}{2} = 4. $

Thus,

$ B = 4 \cdot (2A) = 8A. $

Now, compute the determinant and the trace of $ B $:

Determinant of $ B $:

$ \det(B) = \det(8A) = 8^3 \det(A) = 512 \cdot \frac{1}{2} = 256. $

Trace of $ B $:

$ \operatorname{trace}(B) = \operatorname{trace}(8A) = 8 \cdot \operatorname{trace}(A) = 8 \cdot 3 = 24. $

Finally, adding these results:

$ \det(B) + \operatorname{trace}(B) = 256 + 24 = 280. $

Let us consider the matrices given:

$ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $

$ P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $

$ Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} $

$ R = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

where $ x, y, z, a, b, c, d $ are non-zero real numbers. We aim to verify properties of matrix $ Q $ given the relation $ Q R = R P $.

Steps and Calculations

Matrix Equation Analysis:

Given $ Q R = R P $, we equate both sides:

$ \begin{pmatrix} a x + c y & b x + d y \\ a z + 4c & b z + 4d \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix} $

From the matrix equality, we derive the following key equations:

$ a x + c y = 2a $

$ b x + d y = 3b $

$ a z + 4c = 2c $

$ b z + 4d = 3d $

Determinant Condition:

Comparing the equations:

Set $ a z = -2c $ and $ b z = -d $, resulting in $ \frac{a}{b} = \frac{2c}{d} $.

The requirement $ |R| \neq 0 $ implies $ |P| = |Q| $. So, equating determinants gives us:

$ |P| = |Q| \Rightarrow 6 = 4x - yz \quad \text{(Equation 1)} $

Condition for Non-Trivial Solutions:

From:

$ a(x - z) + c y = 0 $

$ a z + 2c = 0 $

implies a non-trivial solution matrix must have:

$ \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 0 \Rightarrow 2x - yz = 4 \quad \text{(Equation 2)} $

Solve for $ x $, $ yz $:

Solving Equations 1 and 2:

$ 4x - yz = 6 $

$ 2x - yz = 4 $

Subtract the second from the first:

$ (4x - yz) - (2x - yz) = 6 - 4 \Rightarrow 2x = 2 \Rightarrow x = 1 $

Plugging $ x = 1 $ back, we find:

$ 4(1) - yz = 6 \Rightarrow 4 - yz = 6 \Rightarrow yz = -2 $

Verify Determinants:

For $ Q - 2I $:

$ |Q - 2I| = \left|\begin{array}{cc} x-2 & y \\ z & 2 \end{array}\right| = 2x - yz = 0 $

For $ Q - 6I $:

$ |Q - 6I| = \left|\begin{array}{cc} x-6 & y \\ z & -2 \end{array}\right| = -2x + 12 - yz = 12 $

For $ Q - 3I $:

$ |Q - 3I| = \left|\begin{array}{cc} x-3 & y \\ z & 1 \end{array}\right| = x - 3 - yz = 0 $

Therefore, using the derived values, the statements for the determinants align correctly. The findings are:

$ |Q - 2I| = 0 $

$ |Q - 6I| = 12 $

$ yz = -2 $

These calculations validate the conditions for matrix $ Q $ concerning the predefined matrix properties.

To find the number of $3 \times 3$ invertible matrices $Q$ with integer entries that satisfy both $Q^{-1} = Q^T$ and $PQ = QP$, consider the given matrix $P$:

$ P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

The condition $PQ = QP$ implies:

$ \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

This results in:

$ \begin{pmatrix} 2a & 2b & 2c \\ 2d & 2e & 2f \\ 3g & 3h & 3i \end{pmatrix} = \begin{pmatrix} 2a & 2b & 3c \\ 2d & 2e & 3f \\ 2g & 2h & 3i \end{pmatrix} $

From this, we derive that:

$ c = 0, \quad f = 0, \quad h = 0, \quad g = 0 $

Thus, the matrix $Q$ is of the form:

$ Q = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & e \end{pmatrix} $

Next, using the condition $Q^T = Q^{-1}$, we have the equation:

$ Q Q^T = I $

This results in:

$ \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & i \end{pmatrix} \begin{pmatrix} a & c & 0 \\ b & d & 0 \\ 0 & 0 & i \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

This simplifies to:

$ \begin{pmatrix} a^2 + b^2 & ac + bd & 0 \\ ac + bd & c^2 + d^2 & 0 \\ 0 & 0 & i^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $

From which we conclude:

$ a^2 + b^2 = 1, \quad ac + bd = 0, \quad c^2 + d^2 = 1, \quad i^2 = 1 $

Considering all possible integer solutions for these equations:

Case I:

$(a, b) = (0, 1)$ or $(0, -1)$,

$(c, d) = (1, 0)$ or $(-1, 0)$

Case II:

$(a, b) = (1, 0)$ or $(-1, 0)$,

$(c, d) = (0, 1)$ or $(0, -1)$

For each case, $e$ can be either 1 or -1 since $i^2=1$. Therefore, $e = \pm 1$.

Combining these options, we find a total of 16 possible matrices $Q$.

Given, $A^3-5 A^2+7 A+I=0$

$ \Rightarrow A^5-5 A^4+7 A^3+A^2=0 $

Also, $A^5-6 A^4+12 A^3-6 A^2+2 A+2 I$

$ \begin{array}{r} =l A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ A^5-5 A^4+7 A^3+A^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Subtracting Eqs. (i) and (ii), we get

$ -A^4+5 A^3-7 A^2+2 A+2 I=I A+m I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ A^4-5 A^3+7 A^2+A=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

On adding Eqs. (iii) and (iv), we get

$ \begin{aligned} & 3 A+2 I=l A+m I \\ \Rightarrow \quad & I=3 \text { and } m=2 \end{aligned} $

So, $\quad l+m=5$

$ \begin{aligned} &\text { Given, }\\ &A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{array}\right], A^{-1}=\left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & y \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2} \end{array}\right] \end{aligned} $

$ \begin{aligned} & A A^{-1}=\left[\begin{array}{ccc} 1 & 0 & y+1 \\ 0 & 1 & 2 y+2 \\ 4-4 x & 3 x-3 & x y+2 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow y+1=0 \text { and } 4-4 x=0 \\ & \Rightarrow y=-1 \text { and } x=1 \\ & \text { So, } \log _{2 / 5}\left(2^1+2^{-1}\right)=-1 \\ & \Rightarrow(x, y) \text { satisfies } y=\log _{2 / 5}\left(2^x+2^{-x}\right) \end{aligned} $

$ \begin{aligned} & \text { Let } A=\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right] \\ & A X=0 \\ & {\left[\begin{array}{lll} a_1 & b_1 & 0 \\ a_2 & b_2 & 0 \\ a_3 & b_3 & 0 \end{array}\right]\left[\begin{array}{l} l \\ m \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & {\left[\begin{array}{l} a_1 l+b_1 m \\ a_2 l+b_2 m \\ a_3 l+b_3 m \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]} \\ & a_1 l+b_1 m=0 \\ & a_2 l+b_2 m=0 \\ & a_3 l+b_3 m=0 \\ & \Rightarrow a_1: a_2: a_3=b_1: b_2: b_3 \Rightarrow|A|=0 \end{aligned} $

⇒ Determinant of each submatrix of order $2 \times 2$ is zero $\Rightarrow$ Rank $=1$

Since, the system of equations has non-trivial solution

$ \begin{aligned} & \Rightarrow\left|\begin{array}{ccc} 2 & 3 & a \\ 1 & a & -2 \\ 3 & 1 & 3 \end{array}\right|=0 \\ & \Rightarrow 2(3 a+2)-3(3+6)+a(1-3 a)=0 \\ & \Rightarrow 6 a+4-9-18+a-3 a^2=0 \\ & \Rightarrow 6 a-23+a-3 a^2=0 \\ & \Rightarrow 3 a^2-7 a+23=0 \\ & \Rightarrow D<0 \end{aligned} $

So, no real values of $a$ is possible.

$ \text { } \begin{aligned} 2 x+3 y+z & =-1 \\ 3 x+y+z & =4 \\ x-3 y-2 z & =1 \\ \Rightarrow \quad x-2 y & =5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{array}{lr} \Rightarrow 5 x+3 y=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow 5(5+2 y)+3 y=-1 \\ \Rightarrow 25+13 y=-1 \\ \Rightarrow 13 y=-26 \\ \Rightarrow y=-2 \\ \text { So, } y=-2 \end{array} $

For non-trivial solution,

$ \begin{aligned} & \quad\left|\begin{array}{rrr} 1 & a & 1 \\ a & 2 & -1 \\ 2 & 3 & 1 \end{array}\right|=0 \\ & 1(2+3)-a(a+2)+1(3 a-4)=0 \\ & \Rightarrow \quad 5-a^2-2 a+3 a-4=0 \\ & \Rightarrow \quad a^2-a-1=0 \\ & \Rightarrow \quad a=\frac{1 \pm \sqrt{5}}{2} \end{aligned} $

So, $\quad a=\frac{1+\sqrt{5}}{2}$

$ \text { } \begin{aligned} |\operatorname{adj} A| & =|A|^{n-1} \\ \left|\operatorname{adj}\left(A^2\right)\right| & =\left|A^2\right|^{n-1} \\ & =\left|A^2\right|^2 \\ \Rightarrow \quad A^2 & =\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right] \\ & =\left[\begin{array}{lll} 7 & 8 & 6 \\ 5 & 7 & 6 \\ 6 & 6 & 5 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow|A|^2=7(35-36)-8(25-36)+6(30-42) \\ & \quad=-7+88-72=9 \\ & \therefore\left|\operatorname{adj}\left(A^2\right)\right|=\left|A^2\right|^2=9^2=81 \end{aligned} $

$ \begin{array}{r} x-2 y+z=0 \\ 2 x+3 y+z=6 \\ x+2 y+p z=q \end{array} $

$ \begin{aligned} &\text { System has infinite solution, }\\ &\begin{array}{ll} \therefore & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ & \Delta=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & 3 & 1 \\ 1 & 2 & p \end{array}\right|=0 \\ \Rightarrow & 1(3 p-2)+2(2 p-1)+1(4-3)=0 \\ \Rightarrow & 3 p-2+4 p-2+1=0 \\ \Rightarrow & 7 p=3 \Rightarrow p=\frac{3}{7} \\ & \Delta_z=\left|\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 3 & 6 \\ 1 & 2 & q \end{array}\right|=0 \\ \Rightarrow & 1(3 q-12)+2(2 q-6)=0 \\ \Rightarrow & 3 q-12+4 q-12=0 \\ \Rightarrow & 7 q=24 \Rightarrow q=\frac{24}{7} \\ \therefore & q-p=\frac{24}{7}-\frac{3}{7}=3 \end{array} \end{aligned} $

We have,

$ \begin{aligned} & (\sin \theta) x-y+z=0 \\ & x-(\cos \theta) y+z=0 \\ & x+y+(\sin \theta) z=0 \end{aligned} $

System has non-trivial solution.

$ \therefore \quad \Delta=0 $

$ \begin{aligned} & \left|\begin{array}{ccc} \sin \theta & -1 & 1 \\ 1 & -\cos \theta & 1 \\ 1 & 1 & \sin \theta \end{array}\right|=0 \\ & \Rightarrow \quad \sin \theta(-\sin \theta \cos \theta-1) +1(\sin \theta-1)+1(1+\cos \theta)=0 \\ & \Rightarrow \quad-\sin ^2 \theta \cos \theta-\sin \theta+\sin \theta-1+1 +\cos \theta=0 \\ & \Rightarrow \quad \cos \theta\left(1-\sin ^2 \theta\right)=0 \\ & \Rightarrow \quad \cos ^3 \theta=0 \Rightarrow \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} A & =\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{array}\right] \\ |A| & =1(1-3)-2(2-1)+3(6-1) \\ & =-2-2+15=11 \\ B & =\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{array}\right] \\ \Rightarrow \quad & |B|=2(4-8)-3(6-4)+4(12-4) \\ & =-8-6+32=18 \\ \therefore \quad & \sqrt{|\operatorname{adj} A B|}=\sqrt{|\operatorname{adj} A||\operatorname{adj} B|} \\ & =\sqrt{|A|^{3-1}|B|^{3-1}} \\ & =|A||B|=11(18)=198 \end{aligned} \end{aligned} $

Step 1: Find the determinant of $ A $

Matrix $ A $ is: $ A=\left[\begin{array}{lll} 1 & 5 & 2 \\ 4 & 1 & 3 \\ 2 & 6 & 3 \end{array}\right] $

So, $ |A| = 1(3-18) - 5(12-6) + 2(24-2) $

Calculate each part:

• $1 \times (3 - 18) = 1 \times (-15) = -15$
• $-5 \times (12 - 6) = -5 \times 6 = -30$
• $2 \times (24 - 2) = 2 \times 22 = 44$

Add them up: $ |A| = -15 - 30 + 44 = -1 $

Step 2: Find the value of $ \left| (\operatorname{adj} A)^{-1} \right| $

$ \left|(\operatorname{adj} A)^{-1}\right| = \frac{1}{|\operatorname{adj} A|} $

The order of matrix $A$ is 3, so: $ |\operatorname{adj} A| = |A|^{3-1} = |A|^2 $

So, $ \left|(\operatorname{adj} A)^{-1}\right| = \frac{1}{|A|^2} $

Since $ |A| = -1 $, $ \left|(\operatorname{adj} A)^{-1}\right| = \frac{1}{(-1)^2} = \frac{1}{1} = 1 $

Given, $x+\lambda y-2 z=1$

$ x-y+\lambda z=2 $

And $x-2 y+3 z=3$

∵ Given, system is inconsistent

$ \begin{aligned} & D=0 \Rightarrow\left|\begin{array}{rrr} 1 & \lambda & -2 \\ 1 & -1 & \lambda \\ 1 & -2 & 3 \end{array}\right|=0 \\ & \Rightarrow 1(-3+2 \lambda)-\lambda(3-\lambda)-2(-2+1)=0 \\ & \Rightarrow 2 \lambda-3-3 \lambda+\lambda^2+2=0 \end{aligned} $

$ \therefore \lambda_1+\lambda_2=1 $

Given, $(\sin \theta) x+y-2 z=0$

$ 2 x-y+(\cos \theta) z=0 $

And $-3 x+(\sec \theta) y+3 z=0$

∵ system has non-trivial solution.

$ \begin{aligned} & \Delta=0 \\ & \left|\begin{array}{ccc} \sin \theta & 1 & -2 \\ 2 & -1 & \cos \theta \\ -3 & \sec \theta & 3 \end{array}\right|=0 \\ & \sin \theta(-3-1)-1(6+3 \cos \theta)-2(2 \sec \theta-3) =0\\ & \Rightarrow-4 \sin \theta-6-3 \cos \theta-4 \sec \theta+6=0 \\ & \Rightarrow 4 \sec \theta+3 \cos \theta+4 \sin \theta=0 \\ & \Rightarrow 4+3 \cos ^2 \theta+4 \sin \theta \cos \theta=0 \\ & \Rightarrow 4+3\left(\frac{1+\cos 2 \theta}{2}\right)+2 \sin 2 \theta=0 \\ & \Rightarrow 8+3+3 \cos 2 \theta+4 \sin 2 \theta=0 \\ & \Rightarrow 11+3 \cos 2 \theta+4 \sin 2 \theta=0 \end{aligned} $

This equation has no solution because the maximum value of $4 \sin 2 \theta+3 \cos 2 \theta$ is 5 . Since $5+11 \neq 0$, there is no value of $\theta$ that satisfies the equation.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ & \because \operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A \\ & \operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))=|\operatorname{adj} A|^{2-2} \cdot \operatorname{adj} A \\ & \quad=\operatorname{adj} A=|A| A^{-1}\left(\because A^{-1}=\frac{1}{|A|} \operatorname{adj} A\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \left|\begin{array}{ccc} x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3 \end{array}\right|=0 \\ & \Rightarrow x(-6-(x-1)(x-2))+3(-3-(x-1)) +2(-(x-2)+2)=0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow x\left(-6-x^2+3 x-2\right)+3(-3-x+1) +2(-x+2+2)=0 \\ & \Rightarrow x\left(-x^2+3 x-8\right)+3(-2-x) +2(4-x)=0 \\ & \Rightarrow-x^3+3 x^2-8 x-6-3 x+8-2 x=0 \\ & \therefore \text { Sum of roots }=\frac{-3}{-1}=3 . \end{aligned} $

We have, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|=A \lambda^3+B \lambda^2+C \lambda+D$

Now, $\left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|$

$ \begin{aligned} & =1(-3-3 \lambda-2)+(1-\lambda)(4+(1+\lambda)(3-\lambda)) \\ & =(-3 \lambda-5)+(1-\lambda)\left(4+3-\lambda+3 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+(1-\lambda)\left(7+2 \lambda-\lambda^2\right) \\ & =(-3 \lambda-5)+\left(7+2 \lambda-\lambda^2-7 \lambda-2 \lambda^2+\lambda^3\right) \\ & =\lambda^3-3 \lambda^2-8 \lambda+2 \end{aligned} $

$ \begin{aligned} &\therefore \lambda^3-3 \lambda^2-8 \lambda+2=A \lambda^3+B \lambda^2+C \lambda+D\\ &\text { On comparing both sides, we get }\\ &\begin{aligned} A & =1, B=-3, C=-8, D=2 \\ \therefore \quad D+A & =2+1=3 \end{aligned} \end{aligned} $

We have

$ A+2 B=\left[\begin{array}{rrr} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{array}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And $\quad 2 A-B=\left[\begin{array}{rrr}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, $2 A+4 B=\left[\begin{array}{rcr}2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2\end{array}\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Substract Eqs. (ii) from (iii), we get

$ \begin{aligned} &\begin{aligned} & 5 B=\left[\begin{array}{rrr} 2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2 \end{array}\right]-\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]=\left[\begin{array}{rrr} 0 & 5 & -5 \\ 10 & -5 & 0 \\ -10 & 5 & 0 \end{array}\right] \\ \Rightarrow & B=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \end{aligned}\\ &\therefore \text { From Eqs. (ii), we get }\\ &\begin{aligned} & 2 A=\left[\begin{array}{rrr} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{array}\right]+\left[\begin{array}{rrr} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{array}\right] \\ \Rightarrow & 2 A=\left[\begin{array}{rrr} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{array}\right] \Rightarrow A=\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{array}\right] \\ \therefore & \operatorname{tr}(A)-\operatorname{tr}(B) \\ & =(1-1+1)-(0-1+0)=1+1=2 \end{aligned} \end{aligned} $

We have,

$A, C$ are $3 \times 3$ matrices

$B \times D$ are $3 \times 1$ matrices

$A X=B$ has unique solution

$C X=D$ has infinite solution.

Since, $A X=B$ have a unique solution.

$\Rightarrow \operatorname{rank}(A)=$ number of unknowns $=3$

$\therefore \quad \operatorname{rank}([A: B])=\operatorname{rank}(A)=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Since, $C X=D$ have infinite many solutions

$\therefore \quad \operatorname{Rank}(C)=$ not unique $\Rightarrow \operatorname{rank}(C)<3$

$\Rightarrow \quad \operatorname{Rank}([C: D])<3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$\operatorname{Rank}([A: D]) \geq \operatorname{rank}[(C: B)]$

We have, $A$ and $B$ are two non-square matrices such that

$ P=A+B, Q=A^T B, R=A B^T $

Since, $A+B$ is defined,

So order of $A$ and $B$ are same.

Let order of $A$ and $B$ be $m \times n$

$\therefore$ Order of $A^T$ is $n \times m$ and order of $B^T$ is $n \times m$

Now, order of $P=$ order of $A+B=m \times n$

Order of $Q=$ order of $A^T B=n \times n$

And order of $R=$ order of $A B^T=m \times m$

Now, order of $P Q=m \times n$

Order of $Q R=$ not defined

Order of $R Q=$ not defined

Order of $Q P=$ not defined

Order of $R P=m \times n$

$\therefore$ Order of $A=$ order of $P Q$ and $R P$

Given equations are $x+y-z=1$,

$ 2 x+4 y-z=0 \text { and } 3 x+4 y+5 z=18 $

The augmented matrix for the system is

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 2 & 4 & -1 & 0 \\ 3 & 4 & 5 & 18 \end{array}\right] $

We need to transform this matrix into

$ B=\left[\begin{array}{cccc} 1 & a & 0 & -1 \\ 0 & 2 & 1 & b \\ 0 & 0 & c & 32 \end{array}\right] $

Now, perform the row operation for matrix $A$, we get

$ \begin{gathered} R_2 \rightarrow R_2-2 R_1 \text { and } R_3 \rightarrow R_3-3 R_1 \\ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 1 & 8 & 15 \end{array}\right] \end{gathered} $

Now, $R_3 \rightarrow R_3-\frac{1}{2} R_2$

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 7.5 & 16 \end{array}\right] $

Apply $R_3 \rightarrow 2 R_3$, we get

$ A=\left[\begin{array}{cccc} 1 & 1 & -1 & 1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

Again apply $R_1 \rightarrow R_1+R_2$, we get

$ A=\left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 2 & 1 & -2 \\ 0 & 0 & 15 & 32 \end{array}\right] $

$ \begin{aligned} &\text { Thus, } a=3, b=-2 \text { and } c=15\\ &\begin{array}{ll} \Rightarrow & c=15 \\ \therefore & \sqrt{a+b+c}=\sqrt{3+(-2)+15} \\ & =\sqrt{1+15}=\sqrt{16}=4 \end{array} \end{aligned} $

Given, $K=\left|\begin{array}{ccc}9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36\end{array}\right|$

$ \begin{aligned} & =9\left|\begin{array}{ll} 36 & 25 \\ 49 & 36 \end{array}\right|-25\left|\begin{array}{ll} 16 & 25 \\ 25 & 36 \end{array}\right|+16\left|\begin{array}{ll} 16 & 36 \\ 25 & 49 \end{array}\right| \\ & =9(1296-1225)-25(576-625) +16(784-900) \\ & =9(71)-25(-49)+16(-116) \\ & =639+1225-1856=8 \\ & \therefore K=8 \end{aligned} $

So, the roots are $K$ and $K+1$, i.e., 8 and 9

$ \begin{aligned} \text { Sum of roots } & =8+9=17 \\ & =\frac{-b}{a}=\alpha+\beta \end{aligned} $

And product of roots $=8 \times 9=72$

$ =\frac{c}{a}=\alpha \beta $

So, the general form of a quadratic equation is $x^2-(\alpha+\beta) x+\alpha \beta=0$

$ \Rightarrow \quad x^2-17 x+72=0 $

Given, $A=\left[\begin{array}{ccc}1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13\end{array}\right]$

$ \begin{aligned} & \quad|A|=\left|\begin{array}{ccc} 1 & -3 & -5 \\ -2 & 4 & -6 \\ 7 & -11 & 13 \end{array}\right| \\ & =1(4 \times 13-(-6) \times(-11))-(-3)(-2 \times 13 -(-6) \times 7)-5(-2 \times(-11)-7 \times 4) \\ & =1(52-66)+3(-26+42)-5(22-28) \\ & =-14+48+30=64 \end{aligned} $

Since, $|\operatorname{adj} A|=|A|^{n-1}$, for $n \times n$ matrix $A$

So, for $3 \times 3$ matrix,

$ \therefore \quad \begin{aligned} |\operatorname{adj} A| & =|A|^{3-1}=|A|^2 \\ \therefore \quad|\operatorname{adj} A| & =|A|^2=(64)^2=4096 \\ \sqrt{|\operatorname{adj} A|} & =\sqrt{4096}=64 \end{aligned} $

$ \text {} \begin{aligned} \Delta r & =\left|\begin{array}{cc} \frac{1}{3 r-2} & \frac{2}{3 r-5} \\ 0 & \frac{3}{3 r+1} \end{array}\right| \\ & =\frac{1}{(3 r-2)} \times \frac{3}{(3 r+1)}-\frac{2}{3 r-5} \times 0 \\ & =\frac{3}{(3 r-2)(3 r+1)} \end{aligned} $

Using partial fractions

$ \begin{aligned} \Delta r & =\frac{A}{3 r-2}+\frac{B}{3 r+1} \\ \Rightarrow \quad 3 & =A(3 r+1)+B(3 r-2) \end{aligned} $

If $\quad r=\frac{2}{3}$, then

$ \begin{aligned} & \Rightarrow \quad 3=A\left(3 \times \frac{2}{3}+1\right)+B\left(3 \times \frac{2}{3}-2\right) \\ & \Rightarrow 3 A=3 \\ & \Rightarrow \quad A=1 \end{aligned} $

If $\quad r=-\frac{1}{3}$, then

$ \begin{aligned} & \quad 3=B\left(3 \times\left(-\frac{1}{3}\right)-2\right)=B(-1-2) \\ & \Rightarrow \quad B=-1 \\ & \text { So, } \Delta r=\frac{1}{3 r-2}-\frac{1}{3 r+1} \\ & \therefore \sum_{r=1}^{33} \Delta r=\sum_{r=1}^{33}\left(\frac{1}{3 r-2}-\frac{1}{3 r+1}\right) \\ & \Rightarrow\left(\frac{1}{1}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+ \\ & \quad \ldots+\left(\frac{1}{97}-\frac{1}{100}\right) \\ & \Rightarrow 1-\frac{1}{100}=\frac{99}{100}=0.99 \end{aligned} $

$A=\left|\begin{array}{ccc}-1 & x & -3 \\ 2 & 4 & z \\ y & 5 & -6\end{array}\right|$ is a symmetric

So, $A=A^T$

that means, $x=2, y=-3, z=5$

and $B=\left[\begin{array}{ccc}0 & 2 & q \\ p & 0 & -4 \\ -3 & r & s\end{array}\right]$ is a skew-matrix

So, $B=-B^T$

that means, $p=-2 q=3, r=4, s=0$

Therefore, $|A|=\left|\begin{array}{ccc}-1 & 2 & -3 \\ 2 & 4 & 5 \\ -3 & 5 & -6\end{array}\right|$

$ =-1(-49)-2(3)-3(22)=-23 $

and $|B|=\left|\begin{array}{ccc}0 & 2 & 3 \\ -2 & 0 & -4 \\ -3 & 4 & 0\end{array}\right|$

$ =-2(-12)+3(-8)=0 $

Now, $|A B|=|A||B|=-23 \times 0=-23$

$ x y z+q+r=2 \times(-3) \times(5)+3+4=-23 $

Hence, $|A|+|B|-|A B|=x y z+q+r$

$A=\left[\begin{array}{ccc}-x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x\end{array}\right]$

and $B=\left[\begin{array}{ccc}2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1\end{array}\right]$

If $A^{-1}=B \Rightarrow$ So, $A B=I$

$ \begin{aligned} & \Rightarrow\left[\begin{array}{ccc} -x & 14 x & 7 x \\ 0 & 1 & 0 \\ x & -4 x & -2 x \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{array}\right]=5 \\ & \Rightarrow-2 x+7 x=1 \Rightarrow x=1 / 5 \end{aligned} $

Now, $\left[\begin{array}{ccc}x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4\end{array}\right]$

$R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_2$

$ =\left[\begin{array}{ccc} x & x+1 & x+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right] $

therefore determinant of the given matrix $=0$

Hence, from the given options put $x=1 / 5$ in $5 x-1$ we get, $5 \times \frac{1}{5}-1=0$

$ \begin{aligned} &\text { that means, }\\ &\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right|=5 x-1 \end{aligned} $

$2 x+3 y-3 z=3$

$ x+2 y+\alpha z=1 $

and $2 x-y+z=\beta$

So, $A=\left[\begin{array}{ccc}2 & 3 & -3 \\ 1 & 2 & \alpha \\ 2 & -1 & 1\end{array}\right]$

for infinite many solutions put $|A|=0$

$ \begin{aligned} & \Rightarrow 2(2+\alpha)-3(1-2 \alpha)-3(-1-4)=0 \\ & \Rightarrow 1+8 \alpha+15=0 \Rightarrow \alpha=-2 \end{aligned} $

Now, $D_y=\left|\begin{array}{ccc}2 & 3 & -3 \\ 1 & 1 & -2 \\ 2 & \beta & 1\end{array}\right|$

$ \begin{aligned} & \text { put } D_y=0 \\ & \begin{aligned} \Rightarrow 2(1+2 \beta)-3(1+4)-3(\beta-2)=0 \\ \Rightarrow \beta-7=0 \Rightarrow \beta=7 \\ \text { Therefore, } \frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{-2}{7}-\frac{7}{(-2)} \\ \qquad=\frac{-4+49}{14}=\frac{45}{14} \end{aligned} \end{aligned} $

We have,

$ \left|\begin{array}{ccc} 1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 $

Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta \end{array}\right|=0 \\ & R_3 \rightarrow R_3-\left(\sin ^2 \theta\right) R_1 \end{aligned} $

$ \begin{aligned} &\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 1 & 1+4 \sin 4 \theta \end{array}\right|=0\\ &\text { Expand w.r.t along } C_1 \text {, we get }\\ &\begin{aligned} & 1+4 \sin 4 \theta+1=0 \\ & 4 \sin 4 \theta=-2 \\ & \sin 4 \theta=\frac{-1}{2}=\sin \left(\pi+\frac{\pi}{6}\right) \\ & 4 \theta=\frac{7 \pi}{6} \Rightarrow \theta=\frac{7 \pi}{24} \end{aligned} \end{aligned} $

Given, $2 x+p y+6 z=8$

$ \begin{aligned} & x+2 y+q z=5 \\ & x+y+3 z=4 \end{aligned} $

For infinite solution

$ \begin{aligned} & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Delta_z=0 \\ & \Rightarrow\left|\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 2(8-5)-p(4-5)+8(1-2)=0 \\ & \Rightarrow \quad 6+p-8=0 \Rightarrow p=2 \end{aligned} $

$ \begin{aligned} &\text { We have, } x^a y^b=e^m\\ &x^a=\frac{e^m}{y^b} \Rightarrow x=\frac{e^{\frac{m}{a}}}{y^{\frac{b}{a}}} \Rightarrow x^c y^d=e^n \end{aligned} $

$ \Rightarrow \quad \frac{e^{\frac{m c}{a}}}{y^{\frac{b c}{a}}} \cdot y^d=e^n $

$ \begin{aligned} \Rightarrow & & y^{d-\frac{b c}{a}}=e^{n-\frac{m c}{a}} \\ \Rightarrow & & y^{a d-k c}=e^{a n-m c} \\ \Rightarrow & & y=e^{\frac{a n-m c}{a d-k c}}=e^{\frac{\Delta_2}{\Delta_3}} \end{aligned} $

$ \begin{aligned} &\text { and } &x=e^{\frac{\Delta_1}{\Delta_3}} \end{aligned} $

$ \quad \begin{aligned}\because & \Delta_1=m d-b n \\ & \Delta_2=a n-m c \\ & \Delta_3=a d-b c \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad B=A^{-1} \\ & \text { and } \quad|B|=k \Rightarrow\left|A^{-1}\right|=k \\ & \Rightarrow \quad \frac{1}{|A|}=k \Rightarrow|A|=\frac{1}{k} . \\ & \because(\operatorname{adj}(\operatorname{adj} A))^{-1}=\left(|A|^{n-2} A\right)^{-1} \text { and } n=3 \\ & \quad=\left(|A|^1 \cdot A\right)^{-1}=\left(\frac{1}{k} A\right)^{-1}=k B . \end{aligned} \end{aligned} $

$A=\left[\begin{array}{lll}2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2\end{array}\right]$

and $|A-X I|=0$

⇒ Characterstic equation of matrix $A$

$\because \operatorname{tr}(A)=2+3+2=7$

Sum of minor of diagonal elements

$ \begin{aligned} & \,\,\,\,\,\,\,\,=(4)+(3)+4=11 \\ & |A|=2(4)-2(1)+1(-1) \\ &\,\,\,\,\,\,\,\,\,\, =8-2-1=5 \\ & \because \quad|A-x I|=0 \\ & \Rightarrow \quad x^3-7 x^2+11 x-5=0 \\ & \therefore \quad \alpha+\beta+\gamma=7 \\ & \text { and } \alpha \beta+\beta \gamma+\gamma \alpha=11 \\ & \text { and } \alpha \beta \gamma=5 \\ & \therefore \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \end{aligned} $

$ =49-2 \times 11=49-22=27 . $

$ \begin{aligned} &\text { Given, } 2 x-3 y+2 z=-15\\ &\begin{aligned} & 3 x+y-z=-2 \\ & x-3 y-3 z=-8 \\ & \begin{aligned} \therefore \Delta & =\left|\begin{array}{ccc} 2 & -3 & 2 \\ 3 & 1 & -1 \\ 1 & -3 & -3 \end{array}\right| \\ & =2(-3-3)+3(-9+1)+2(-9-1) \\ & =-12-24-20=-56 \\ \Delta_x & =\left|\begin{array}{ccc} -15 & -3 & 2 \\ -2 & 1 & -1 \\ -8 & -3 & -3 \end{array}\right| \\ & =-15(-3-3)+3(6-8)+2(6+8) \\ & =90-6+28=112 \\ \Delta_y & =\left|\begin{array}{ccc} 2 & -15 & 2 \\ 3 & -2 & -1 \\ 1 & -8 & -3 \end{array}\right| \\ & =2(6-8)+15(-9+1)+2(-24+2) \\ & =-4-120-44=-168 \\ \Delta_z & =\left|\begin{array}{ccc} 2 & -3 & -15 \\ 3 & 1 & -2 \\ 1 & -3 & -8 \end{array}\right| \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & =2(-8-6)+3(-24+2)-15(-9-1) \\ & =-28-66+150=150-94=56 . \\ \alpha & =\frac{\Delta_x}{\Delta}, \beta=\frac{\Delta_y}{\Delta} \text { and } \gamma=\frac{\Delta_z}{\Delta} \\ \Rightarrow \alpha & =\frac{112}{-56}, \beta=\frac{-168}{-56} \text { and } \gamma=\frac{56}{-56} \\ \Rightarrow \alpha & =-2, \beta=3 \text { and } \gamma=-1 \\ \therefore 2 \alpha & +\beta=\gamma \end{aligned} $

Let $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4\end{array}\right]$

Also, $a=\operatorname{det}(\operatorname{adj}(A))$ and $b=\operatorname{det}\left(B^{-1}\right)$

Now, $a=\operatorname{det}(\operatorname{adj}(A))=(\operatorname{det}(A))^{n-1}$ for an $n \times n$ matrix.

Since $A$ is $3 \times 3$ matrix, so $a=(\operatorname{det}(A))^{3-1}=(\operatorname{det}(A))^2$

$ \begin{aligned} \operatorname{det} A & =1 \cdot\left|\begin{array}{ll} 2 & 3 \\ 3 & 3 \end{array}\right|-1 \cdot\left|\begin{array}{ll} 1 & 3 \\ 2 & 3 \end{array}\right|+2 \cdot\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right| \\ & =1 \cdot(6-9)-1(3-6)+2(3-4) \\ & =-3+3-2=-2 \\ a & =(\operatorname{det}(A))^2=(-2)^2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, $\operatorname{det}(B)=\left|\begin{array}{ccc}1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4\end{array}\right|$

$ \begin{aligned} & =1 \cdot\left|\begin{array}{cc} -3 & -1 \\ 1 & -4 \end{array}\right|-2\left|\begin{array}{cc} 4 & -1 \\ 2 & -4 \end{array}\right|+3 \cdot\left|\begin{array}{cc} 4 & -3 \\ 2 & 1 \end{array}\right| \\ & =(12+1)-2(-16+2)+3(4+6) \\ & =13+28+30=71 \end{aligned} $

$ \begin{aligned} &\text { So, } \quad b=\operatorname{det}\left(B^{-1}\right)=\frac{1}{\operatorname{det}(B)}=\frac{1}{71}\\ &\begin{aligned} & \quad b=\frac{1}{71} \\ & \Rightarrow \quad b+1=\frac{1+71}{71}=\frac{72}{71} \\ & \text { Then, } \frac{b+1}{18 b}=\frac{72 / 71}{18 \times \frac{1}{71}} \\ & =\frac{72}{71} \times \frac{71}{18}=\frac{72}{18}=4=a \quad \text { [using Eq. (i)] } \\ & \therefore \frac{b+1}{18 b}=a \end{aligned} \end{aligned} $

Given, $A X=B$ and $C X=D$ and $A X=B$ has unique solution $D$ and $C X=D$ has unique solution $B$.

Since, $X=D$ is the unique solution of $A X=B$

$ \therefore \quad A D=B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And, $X=B$ is the unique solution of $C X=D$,

$ \therefore \quad C B=D \Rightarrow B=C^{-1} D \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i) and (ii),

$ \begin{array}{r} A D=C^{-1} D \\ \left(A-C^{-1}\right) D=0 \end{array} $

So, the solution of $\left(A-C^{-1}\right) X=0$ is $D$

Given,

$ f(x)=\frac{1}{2}\left|\begin{array}{cc} f(x) & f\left(\frac{1}{x}\right)-f(x) \\ 1 & f\left(\frac{1}{x}\right) \end{array}\right| $

and $f(2)=33$

Here, $f(x)=\frac{1}{2}\left[f(x) \cdot f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)+f(x)\right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let

$ \begin{array}{r} f(x)=a_0 x^n+a_1 x^{n-1}+\ldots+a_{n-1} x+a_n \\ \left(a_0 \neq 0\right) \end{array} $

Putting this value of $f(x)$ in Eq. (i),

$ \begin{aligned} & f(x)=\frac{1}{2}\left[f(x) \cdot f\left(\frac{1}{x}\right)-f\left(\frac{1}{x}\right)+f(x)\right] \\ \Rightarrow & f(x) \cdot f\left(\frac{1}{x}\right)=f\left(\frac{1}{x}\right)+f(x) \end{aligned} $

Hence, $f(x)=x^n+1$ or $-x^n+1$

Now, $f(2)=2^n+1=33$

$ \begin{array}{ll} \Rightarrow & 2^n=32 \\ \Rightarrow & 2^n=2^5 \\ \Rightarrow & n=5 \end{array} $

And $f(2)=-2^n+1=33$

$ \Rightarrow \quad-2^n=33-1=32=2^5 $

which is not possible.

$ \therefore f(x)=x^n+1=x^5+1 $

So, $f(3)=3^5+1=243+1=244$

$\because P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$

$ \begin{aligned} & \because \operatorname{adj}(A)=(\operatorname{det} A) \cdot A^{-1} \\ & \Rightarrow \operatorname{det}(\operatorname{adj}(A))=(\operatorname{det} A)^{n-1} \end{aligned} $

So, for $n=3$

$ \begin{aligned} \operatorname{det}(\operatorname{adj}(A)) & =(\operatorname{det} A)^{3-1} \\ & =4^2=16 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} \text { and } \operatorname{det}(A) & =1(12-12)-\alpha(4-6)+3(4-6) \\ & =2 \alpha-6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Hence, $2 \alpha-6=16$

$ \begin{array}{ll} \Rightarrow & 2 \alpha=22 \\ \Rightarrow & \alpha=11 \end{array} $

We are given the equation $x^3+6 x^2+5 x-42=0$.

Let's check if $x=2$ is a root by plugging it in:

$ (2)^3 + 6 \times (2)^2 + 5 \times 2 - 42 = 8 + 24 + 10 - 42 = 42 - 42 = 0 $

Since the equation equals zero, $x = 2$ is a real root. So, $\alpha = 2$.

Now, we substitute $\alpha = 2$ into the matrix:

$ \left[ \begin{array}{lll} \alpha-1 & \alpha+1 & \alpha+2 \\ \alpha-2 & \alpha+3 & \alpha-3 \\ \alpha+4 & \alpha-4 & \alpha+5 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{array} \right] $

Next, let's find the determinant of this matrix:

$ \begin{aligned} &= \left| \begin{array}{ccc} 1 & 3 & 4 \\ 0 & 5 & -1 \\ 6 & -2 & 7 \end{array} \right| \\ &= 1 \times (5 \times 7 - (-1) \times (-2)) - 3 \times (0 \times 7 - (-1) \times 6) + 4 \times (0 \times (-2) - 5 \times 6) \\ &= 1 \times (35 - 2) - 3 \times (0 + 6) + 4 \times (0 - 30) \\ &= 33 - 18 - 120 \\ &= -105 \end{aligned} $

Given matrix

$ \begin{aligned} & {\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{array}\right]} \\ & R_2 \rightarrow R_2-\frac{5}{2} R_1 \text { and } R_3 \rightarrow R_3-\frac{1}{2} R_1 \\ & =\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 0 & 3.5 & -8 & 1 \\ 0 & -1.5 & 3 & -4 \end{array}\right] \\ & R_3 \rightarrow R_3+\frac{1.5}{3.5} R_2=R_3+\frac{3}{7} R_2 \\ & =\left[\begin{array}{cccc} 2 & -3 & 4 & 0 \\ 0 & 3.5 & -8 & 1 \\ 0 & 0 & -3 / 7 & 25 / 7 \end{array}\right] \end{aligned} $

$\because$ All 3 rows are non-zero

Hence, rank of the matrix $=3$

Given, $A=\left[\begin{array}{ccc}0 & k & k \\ k & -4 & -6 \\ k & -3 & -5\end{array}\right]$

$\because A$ is an singular matrix.

$ \begin{aligned} & \therefore \quad|A|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{array}\right|=0 \\ & \Rightarrow \quad 0-k(-5 k+6 k)+k(-3 k+4 k)=0 \end{aligned} $

$\Rightarrow-k^2+k^2=0$ which is true for all real values of $k$.

$\therefore k \in R$

Given, matrix $A=\left[\begin{array}{ccc}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ and the rank of $A=2$

$ \begin{aligned} & =1(6-28)-2(-24-14)+x(16+2) \\ & |A|=54+18 x \end{aligned} $

Since, the rank of $A$ is 2 then the determinant of $A$ must be zero.

So, we set the determinant equal to zero and solve for $x$.

i.e. $54+18 x=0 \Rightarrow x=-3$

$ \text { } \begin{aligned} &\left|\begin{array}{ll} 2 & 1 \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} 1 & \frac{1}{3} \\ 3 & 1 \end{array}\right|+\left|\begin{array}{cc} \frac{1}{2} & \frac{1}{9} \\ 3 & 1 \end{array}\right| +\left|\begin{array}{cc} \frac{1}{4} & \frac{1}{27} \\ 3 & 1 \end{array}\right|+\ldots+\infty \\ &=(2-3)+\left(1-3 \times \frac{1}{3}\right)+\left(\frac{1}{2}-3 \times \frac{1}{9}\right)+\left(\frac{1}{4}-3 \times \frac{1}{27}\right)+\left(\frac{1}{8}-3 \times \frac{1}{81}\right)+\ldots+\infty \\ &=-1+0+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right) +\left(\frac{1}{2^3}-\frac{1}{3^3}\right)+\ldots+\infty \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =-1+\left[\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots \infty\right)\right. \left.-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\ldots+\infty\right)\right] \\ & =-1+\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)-\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right) \end{aligned}\\ &\text { (Using sum of infinite GP formula) }\\ &=-1+\frac{\frac{1}{2}}{\frac{1}{2}}-\frac{\frac{1}{3}}{\frac{2}{3}}=-1+1-\frac{1}{2}=-\frac{1}{2} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { }|A|=1(18-5)-2(6-10)+3(1-6) \\ & \quad=13+8-15=6 \\ & |\operatorname{adj}(\operatorname{adj} A)|=|A|^4=6^4 \\ & \Rightarrow(\operatorname{adj} A)^{-1}=\frac{1}{6} A \\ & \Rightarrow|\operatorname{adj}(\operatorname{adj} A)|(\operatorname{adj} A)^{-1}=k A \\ & \Rightarrow 6^4 \cdot \frac{1}{6} A=k A \\ & \therefore k=216 \end{aligned} \end{aligned} $

$\alpha+2 \beta+3 \gamma=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & 3 \alpha+\beta+\gamma=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha+3 \beta+3 \gamma=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (ii), $3 \alpha+\gamma=3-\beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Substracting Eq. (iii) from Eq. (i)

$ \begin{aligned} & \alpha+2 \beta+3 \gamma=4 \\ & \alpha+3 \beta+3 \gamma=2 \\ & -\quad-\quad- \,\,\,\,\,\,\,\,-\\ & \hline-\beta=2 \Rightarrow \beta=-2 \end{aligned} $

Bu Eq. (iv)

$ \begin{aligned} 3 \alpha+\gamma & =3-(-2)=5 \\ \Rightarrow 3 \alpha+\gamma & =1-2 \beta \end{aligned} $

$ \begin{aligned} &\text { } 2 x+y-z=7, x-3 y+2 z=1 \text {, }\\ &\begin{aligned} & x+4 y-3 z=5 \\ & A=\left[\begin{array}{rrr} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{array}\right] \\ & \Rightarrow|A|=\left|\begin{array}{rrr} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{array}\right| \\ & \quad=2(9-8)-1(-3-2)-1(4+3) \\ & \quad=2+5-7=0 \\ & \Rightarrow \text { Number of solution = 0 } \end{aligned} \end{aligned} $

$2 x+p y+6 z=8, x+2 y+q z=5$ and $x+y+3 z=4$ may have no solution.

$ \begin{aligned} & x=\frac{\Delta x}{\Delta}, y=\frac{\Delta y}{\Delta}, z=\frac{\Delta z}{\Delta} \\ & \Delta=\left|\begin{array}{lll} 2 & p & 6 \\ 1 & 2 & q \\ 1 & 1 & 3 \end{array}\right|=(p-2)(q-3) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Delta_x=\left|\begin{array}{lll} 8 & p & 6 \\ 5 & 2 & q \\ 4 & 1 & 3 \end{array}\right|=(15-4 q)(2-p) \\ & \Delta_y=\left|\begin{array}{lll} 2 & 8 & 6 \\ 1 & 5 & q \\ 1 & 4 & 3 \end{array}\right|=0 \\ & \Delta_z=\left|\begin{array}{lll} 2 & p & 8 \\ 1 & 2 & 5 \\ 1 & 1 & 4 \end{array}\right|=p-2 \end{aligned}\\ &\text { For no solution }\\ &\begin{aligned} & (p-2)(q-3)=0 \\ & p-2 \neq 0 \\ & q=3 \\ & p \neq 2 \end{aligned} \end{aligned} $

A is det of order 3 with entries 0 or 1 .

$\therefore B$ and $C$ has same number of elements.