Matrices and Determinants

Given the matrices $A$ and $B$:

$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} $

We have the condition:

$ (A+B)(A-B) = A^2 - B^2 $

This implies:

$ AB - BA = 0 \quad \Rightarrow \quad AB = BA $

Let's compute $AB$ and $BA$:

$ AB = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} x + 2 & y + 4 \\ 2x + 1 & 2y + 2 \end{bmatrix} $

$ BA = \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 1 + 4 & 2 + 2 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 5 & 4 \end{bmatrix} $

Equating $AB$ and $BA$, we have:

$ \begin{bmatrix} x + 2 & y + 4 \\ 2x + 1 & 2y + 2 \end{bmatrix} = \begin{bmatrix} x + 2y & 2x + y \\ 5 & 4 \end{bmatrix} $

From this, we derive:

$ \begin{aligned} 2x + 1 &= 5 \\ 2y + 2 &= 4 \\ \end{aligned} $

Thus:

$ \begin{aligned} 2x &= 4 \quad \Rightarrow \quad x = 2 \\ 2y &= 2 \quad \Rightarrow \quad y = 1 \\ \end{aligned} $

Given matrix $C$:

$ C = \begin{bmatrix} x & 2 \\ 1 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix} $

The trace of matrix $C$, denoted as $\operatorname{tr}(C)$, is:

$ \operatorname{tr}(C) = x + y = 2 + 1 = 3 $

To solve the given determinant equation, we start by simplifying the expression:

$\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \\ x-4 & 3(x-3) & 5(x-5) \\ x-8 & 3(x-9) & 5(x-25)\end{array}\right| = 0$

This simplifies to:

$\left|\begin{array}{ccc}x-2 & x-1 & x-1 \\ x-4 & x-3 & x-5 \\ x-8 & x-9 & x-25\end{array}\right| = 0$

Next, perform column operations to simplify the determinant. Subtract the second column from the first:

$C_1 \rightarrow C_1 - C_2 \implies \left|\begin{array}{ccc} -1 & x-1 & x-1 \\ -1 & x-3 & x-5 \\ 1 & x-9 & x-25 \end{array}\right| = 0 $

Now, apply row operations:

Add the third row to the first row: $ R_1 \rightarrow R_1 + R_3 $

Add the third row to the second row: $ R_2 \rightarrow R_2 + R_3 $

This results in:

$\left|\begin{array}{ccc} 0 & 2x-10 & 2x-26 \\ 0 & 2x-12 & 2x-30 \\ 1 & x-9 & x-25 \end{array}\right| = 0 $

To solve, expand the determinant by the first column:

$ (2x-30)(2x-10) - (2x-12)(2x-26) = 0 $

This simplifies to:

$ 2(x-15) \cdot 2(x-5) - 2(x-6) \cdot 2(x-13) = 0 $

Further simplifying, we have:

$(x-15)(x-5) - (x-6)(x-13) = 0$

Solving this, we calculate:

$x^2 - 20x + 75 - (x^2 - 19x + 78) = 0$

This leads to:

$-x - 3 = 0 \implies x = -3$

Therefore, $x = -3$ satisfies the equation $x^2 + 2x - 3 = 0$.

We start with the well-known formula for any invertible matrix $ A $:

$ A \cdot \operatorname{adj}(A) = |A| I. $

Since $ A $ is non-singular ($|A| \neq 0$), we can write

$ \operatorname{adj}(A) = |A|\, A^{-1}. $

Now, replace $ A $ with $ A^{-1} $. Then, applying the same formula for $ A^{-1} $, we have:

$ A^{-1} \cdot \operatorname{adj}(A^{-1}) = |A^{-1}|\, I. $

Recall that $ |A^{-1}| = \frac{1}{|A|} $. Thus,

$ A^{-1} \cdot \operatorname{adj}(A^{-1}) = \frac{1}{|A|}\, I. $

Multiplying both sides on the left by $ A $ (which is valid since $ A $ is invertible) gives:

$ A \cdot A^{-1} \cdot \operatorname{adj}(A^{-1}) = A \cdot \frac{1}{|A|}\, I. $

Since $ A \cdot A^{-1} = I $, this simplifies to:

$ \operatorname{adj}(A^{-1}) = \frac{1}{|A|}\, A. $

On the other hand, we earlier had:

$ \operatorname{adj}(A) = |A|\, A^{-1}. $

Taking the inverse of both sides (and recalling that for any nonzero scalar $ c $ and invertible matrix $ B $, $(cB)^{-1} = \frac{1}{c}\,B^{-1}$), we get:

$ (\operatorname{adj}(A))^{-1} = (|A|\, A^{-1})^{-1} = \frac{1}{|A|}\, (A^{-1})^{-1} = \frac{1}{|A|}\, A. $

Thus, we obtain:

$ \operatorname{adj}(A^{-1}) = (\operatorname{adj}(A))^{-1}. $

Therefore, the correct option is:

Option A – $(\operatorname{adj} A)^{-1}$.

$\left|\begin{array}{rrr}1 & -2 & 3 \\\\ 2 & 4 & -5 \\\\ 3 & \lambda & \mu\end{array}\right|=0$

$ \begin{aligned} & 1(4 \mu+5 \lambda)+2(2 \mu+15)+3(2 \lambda-12)=0 \\\\ & \Rightarrow 4 \mu+5 \lambda+4 \mu+30+6 \lambda-36=0 \\\\ & 11 \lambda+8 \mu-6=0 \\\\ & \therefore \quad 8 \mu+11 \lambda=6 \end{aligned} $

$\frac{x^2}{2 x^4+7 x^2+6}=\frac{x^2}{\left(x^2+2\right)\left(2 x^2+3\right)}$

$ =\frac{A x+B}{x^2+a}+\frac{C x+D}{a x^2+3} $

which gives, $a=2$

$ \begin{aligned} & \frac{x^2}{\left(x^2+2\right)\left(2 x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{2 x^2+3} \\\\ & x^2=(A x+B)\left(2 x^2+3\right)+(C x+D)\left(x^2+2)\right. \\\\ & 0=2 A+C \\\\ & 1=2 B+D \end{aligned} $

On solving these equations, we get .

$ \begin{aligned} 0 & =3 A+2 C \\\\ 0 & =3 B+2 D \\\\ A & =C=0 \\\\ B & =2, D=-3 \\\\ \therefore \quad A+ & B+C-2 D=0+2+0-2(-3) \\\\ & =2+6=8 \\\\ & =4(2)=4 a \end{aligned} $

Let's consider the $3 \times 3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ where each element is a positive integer and each row sums to 6, with $a_{22} = 2$.

The matrix properties provide the following conditions:

$a_{22} = a_{23} + a_{32}$ implies $a_{23} + a_{32} = 2$.

$a_{11} = a_{12} + a_{21}$.

$a_{33} = a_{31} + a_{13}$.

Substituting the conditions about row sums, we get:

$a_{11} + a_{12} + a_{13} = 6$.

$a_{21} + a_{22} + a_{23} = 6$.

$a_{31} + a_{32} + a_{33} = 6$.

From $a_{23} + a_{32} = 2$, if $a_{23} = 1$, then $a_{32} = 1$.

From the second row sum, we have: $a_{21} + 2 + 1 = 6$ which gives $a_{21} = 3$.

Given $a_{11} = a_{12} + a_{21}$ and the first row sum,

with $2a_{12} + a_{13} = 3$,

choosing $a_{12} = 1$ and $a_{13} = 1$, results in $a_{11} = 4$.

Finally, using the third row equation $2a_{31} + a_{32} + a_{13} = 6$,

substituting $a_{32} = 1$ and $a_{13} = 1$ results in $a_{31} = 2$.

Now, from $a_{31} + a_{32} + a_{33} = 6$, with $a_{31} = 2$ and $a_{32} = 1$,

results in $a_{33} = 3$.

Matrix $A$ is:

$ A = \begin{bmatrix} 4 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix} $

To find the determinant $|A|$, we calculate:

$ |A| = 4(2 \cdot 3 - 1 \cdot 1) - 1(3 \cdot 3 - 1 \cdot 2) + 1(3 \cdot 1 - 2 \cdot 2) $

$ = 4(6 - 1) - 1(9 - 2) + 1(3 - 4) $

$ = 4 \cdot 5 - 1 \cdot 7 + 1 \cdot (-1) $

$ = 20 - 7 - 1 = 12 $

Thus, the determinant $|A| = 12$.

Given that $ |\operatorname{adj} A| = x $ and $ |\operatorname{adj} B| = y $, we are tasked with finding $ \left|(\operatorname{adj}(A B))^{-1}\right| $.

First, recognize that the determinant of the adjugate of a matrix can be expressed with respect to the determinants of the matrix itself. Using this relation, we have:

$ \left|\operatorname{adj}(AB)\right| = \left|\operatorname{adj}(\operatorname{adj}(A))\right| \cdot \left|\operatorname{adj}(\operatorname{adj}(B))\right| $

The inverse of the adjugate adjoint product can be decomposed as:

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

Because $ |\operatorname{adj} A| = x $ and $ |\operatorname{adj} B| = y $, the properties of determinants give us:

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

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

Thus, the determinant of the inverse of the product is:

$ \left|(\operatorname{adj}(A B))^{-1}\right| = \frac{1}{x} \times \frac{1}{y} = \frac{1}{xy} $

Therefore, the result is $ \frac{1}{xy} $.

Given, $ x+3 b y+b z=0 $

$ x+2 a y+a z=0 $

$ x+4 c y+c z=0 $

$ \Rightarrow \Delta=\left|\begin{array}{lll}1 & 3 b & b \\ 1 & 2 a & a \\ 1 & 4 c & c\end{array}\right| $

$ \Delta=2 a c+3 a b-2 a b+4 b c-4 a c-3 b c $

$ \Delta=-2 a c+a b+b c $

For non-zero solution $ \Delta=0 $

$ \Rightarrow 2 a c=a b+b c $

To evaluate the determinant of the matrix:

$ \left|\begin{array}{ccc}\frac{-bc}{a^{2}} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^{2}} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^{2}}\end{array}\right| $

we can simplify the process by factoring out $ \frac{1}{a^{2}}, \frac{1}{b^{2}}, $ and $ \frac{1}{c^{2}} $ from rows $ R_1, R_2, $ and $ R_3 $ respectively. This allows us to focus on a simpler determinant:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}}\left|\begin{array}{ccc} -bc & ac & ab \\ bc & -ac & ab \\ bc & ac & -ab \end{array}\right| $

Next, apply row operations to simplify further: replace $ R_2 $ with $ R_2 + R_1 $ and $ R_3 $ with $ R_3 + R_1 $:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}}\left|\begin{array}{ccc} -bc & ac & ab \\ 0 & 0 & 2ab \\ 0 & 2ac & 0 \end{array}\right| $

This matrix is now triangular, allowing us to compute the determinant by multiplying the diagonal elements:

$ \Rightarrow \quad \frac{1}{a^{2} b^{2} c^{2}} \times (4a^{2}b^{2}c^{2}) = 4 $

Thus, the value of the determinant is 4.

Given, $5 A^{-1}=\left[\begin{array}{ccc}-3 & 2 & 2 \\\\ 2 & -3 & 2 \\\\ 2 & 2 & -3\end{array}\right]$

and

$ A=\left[\begin{array}{lll} x & y & y \\\\ y & x & y \\\\ y & y & x \end{array}\right] $

$ \Rightarrow \quad 5 A^{-1} A=5 I $

$ \left[\begin{array}{ccc} -3 & 2 & 2 \\\\ 2 & -3 & 2 \\\\ 2 & 2 & -3 \end{array}\right]\left[\begin{array}{lll} x & y & y \\\\ y & x & y \\\\ y & y & x \end{array}\right]=\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right] $

$ \left[\begin{array}{ccc} -3 x+4 y & -y+2 x & -y+2 x \\\\ -y+2 x & -3 x+4 y & -y+2 x \\\\ -y+2 x & -y+2 x & -3 x+4 y \end{array}\right] $

$ =\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right] $

$ \begin{aligned} & \Rightarrow \quad y=2 x \text { and }-3 x+4 y=5 \\\\ & \Rightarrow \quad 5 x=5 \\\\ & \Rightarrow \quad x=1 \end{aligned} $

Put $x=1$, then $y=2$

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

$ \begin{aligned} A^2 & =\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 9 & 8 & 8 \\\\ 8 & 9 & 8 \\\\ 8 & 8 & 9 \end{array}\right] \\\\ A^2-4 A & =\left[\begin{array}{lll} 9 & 8 & 8 \\\\ 8 & 9 & 8 \\\\ 8 & 8 & 9 \end{array}\right]-4\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\ 2 & 2 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 5 & 0 & 0 \\\\ 0 & 5 & 0 \\\\ 0 & 0 & 5 \end{array}\right]=5 I \end{aligned} $

$ \begin{aligned} & \text { Given, } A=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right] \\\\ & \text { and } A A^T-A^2=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{array}\right] \end{aligned} $

Now,

$ A^2=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]=\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] $

$ \begin{aligned} & A A^T-A^2=\left[\begin{array}{lll} 9 & 3 & 0 \\\\ 1 & 5 & 8 \\\\ 7 & 6 & 2 \end{array}\right]\left[\begin{array}{lll} 9 & 1 & 7 \\\\ 3 & 5 & 6 \\\\ 0 & 8 & 2 \end{array}\right] \\\\ & -\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] \\\\ & =\left[\begin{array}{lll} 90 & 24 & 81 \\\\ 24 & 90 & 53 \\\\ 81 & 53 & 89 \end{array}\right]-\left[\begin{array}{lll} 84 & 42 & 24 \\\\ 70 & 76 & 56 \\\\ 83 & 63 & 52 \end{array}\right] \\\\ & =\left[\begin{array}{ccc} 6 & -18 & 57 \\\\ -46 & 14 & -3 \\\\ -2 & -10 & 37 \end{array}\right] \\\\ & \sum a_{i j}=6-18+57-46+14 \\ & 1 \leq i \leq 3 \\ & 1 \leq j \leq 3 \end{aligned} $

Given,

$ \begin{array}{l} \Delta_{1}=\left|\begin{array}{lll} 1 & a^{2} & b c \\\\ 1 & b^{2} & c a \\\\ 1 & c^{2} & a b \end{array}\right| \\\\ R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1} \\\\ \Delta_{1}=\left|\begin{array}{ccc} 1 & a^{2} & b c \\\\ 0 & b^{2}-a^{2} & c(a-b) \\\\ 0 & c^{2}-a^{2} & b(a-c) \end{array}\right| \\\\ =(a-b)(c-a)\left|\begin{array}{ccc} 1 & a^{2} & b c \\\\ 0 & -(a+b) & c \\\\ 0 & c+a & -b \end{array}\right| \\\\ =(a-b)(c-a)\left(a b+b^{2}-c^{2}-a c\right) \\\\ =(a-b)(c-a)(a(b-c)+(b-c)(b+c) \\\\ \Delta_{1}=(a-b)(b-c)(c-a)(a+b+c) \ldots \\\\ \text { and } \Delta_{2}=\left|\begin{array}{ccc} 1 & 1 & 1 \\\\ a^{2} & b^{2} & c^{2} \\\\ a^{3} & b^{3} & c^{3} \end{array}\right| \\\\ C_{2} \rightarrow C_{2}-C_{1} \text { and } C_{3} \rightarrow C_{3}-C_{1} \\\\ =\left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2} \\\\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right| \\\\ =-(a-b)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & a+b & c+a \\\\ a^{3} & \left(a^{2}+b^{2}+a b\right) & c^{2}+a^{2}+a c \end{array}\right| \\\\ C_{2} \rightarrow C_{2}-C_{3} \\\\ =-(a-b)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & b-c & c+a \\\\ a^{3} & \left(b^{2}-c^{2}\right)+a(b-c) & c^{2}+a^{2}+a c \end{array}\right| \\\\ =-(a-b)(b-c)(c-a) \\\\ \left|\begin{array}{ccc} 1 & 0 & 0 \\\\ a^{2} & 1 & c+a \\\\ a^{3} & b+c+a & c^{2}+a^{2}+a c \end{array}\right| \end{array} $

$ \begin{array}{l} =-(a-b)(b-c)(c-a) \\\\ {\left[c^{2}+a^{2}+a c-b c-c^{2}-a c-a b-a c-a^{2}\right]} \\\\ =-(a-b)(b-c)(c-a)[-b c-a b-a c] \\\\ \Delta_{2}=(a-b)(b-c)(c-a)[a b+b c+c a] \end{array} $

From Ist and 2nd equation replace $ \Delta_{1} $ and $ \Delta_{2} $

$ \begin{array}{l} \frac{\Delta_{1}}{\Delta_{2}}=\frac{6}{11} \Rightarrow \frac{a+b+c}{a b+b c+c a}=\frac{6}{11} \\\\ \therefore 11(a+b+c)=6(a b+b c+c a) \end{array} $

Given, $ x+3 y+7=0 $

$ \begin{array}{c} 3 x+10 y-3 z+18=0 \\\\ 3 y-9 z+2=0 \\\\ D=\left|\begin{array}{ccc} 1 & 3 & 0 \\\\ 3 & 10 & -3 \\\\ 0 & 3 & -9 \end{array}\right|=0 \\\\ D_{x}=\left|\begin{array}{ccc} -7 & 3 & 0 \\\\ -18 & 10 & -3 \\\\ -2 & 3 & -9 \end{array}\right|=99 \\\\ D_{y}=\left|\begin{array}{ccc} 1 & -7 & 0 \\\\ 3 & -18 & -3 \\\\ 0 & -2 & -9 \end{array}\right|=-33 \\\\ D_{z}=\left|\begin{array}{ccc} 1 & 3 & -7 \\\\ 3 & 10 & -18 \\\\ 0 & 3 & -2 \end{array}\right|=-11 \end{array} $

As $ D=0 $, and none $ D_{x}, D_{y} $ and $ D_{z} $ is zero, the given system of equation has no solution.

We have, $\alpha, \beta, \gamma$ are the roots of the equation $\left|\begin{array}{ccc}x & 2 & 2 \\\\ 2 & x & 2 \\\\ 2 & 2 & x\end{array}\right|=0$ and

$ \min (\alpha, \beta, \gamma)=\alpha $

Now, $\quad\left|\begin{array}{lll}x & 2 & 2 \\\\ 2 & x & 2 \\\\ 2 & 2 & x\end{array}\right|=0$

$ \Rightarrow\left|\begin{array}{ccc} x-2 & 0 & 2 \\\\ 0 & x-2 & 2 \\\\ 2-x & 2-x & x \end{array}\right|=0 $

$\begin{gathered}{\left[C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right]} \\\\ \Rightarrow \quad(x-2)^2\left|\begin{array}{ccc}1 & 0 & 2 \\\\ 0 & 1 & 2 \\\\ -1 & -1 & x\end{array}\right|=0 \\\\ {\left[C_1 \rightarrow \frac{1}{x-2} C_1 \text { and } C_2 \rightarrow \frac{1}{x-2} C_2\right]} \\\\ \Rightarrow(x-2)^2[1(x+2)-0+(-1)(0-2)]=0 \\\\ \Rightarrow(x-2)^2(x+4)=0 \Rightarrow x=-4,2,2 \\\\ \therefore \quad \alpha=-4, \beta=2 \text { and } \gamma=2 \\\\ \text { Now, } 2 \alpha+3 \beta+4 \gamma \\\\ =2(-4)+3(2)+4(2) \\\\ =-8+6+8=6\end{gathered}$

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & \begin{array}{l} A=\left[\begin{array}{lll} 1 & 2 & 2 \\\\ 3 & 2 & 3 \\\\ 1 & 1 & 2 \end{array}\right] \text { and } A^{-1}=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{array}\right] \\\\ |A|=1(4-3)-2(6-3)+2(3-2)=1-6+2=-3\begin{array}{l} \end{array} \\\\ C_{11}=\left|\begin{array}{ll} 2 & 3 \\\\ 1 & 2 \end{array}\right|=1 \\\\ C_{12}=-\left|\begin{array}{ll} 3 & 3 \\\\ 1 & 2 \end{array}\right|=-3 \\\\ C_{13}=\left|\begin{array}{ll} 3 & 2 \\\\ 1 & 1 \end{array}\right|=1 \\\\ C_{21}=-\left|\begin{array}{ll} 2 & 2 \\\\ 1 & 2 \end{array}\right|=-2 \\\\ C_{22}=\left|\begin{array}{ll} 1 & 2 \\\\ 1 & 2 \end{array}\right|=0 \\\\ C_{23}=-\left|\begin{array}{ll} 1 & 2 \\\\ 1 & 1 \end{array}\right|=1 \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & C_{31}=\left|\begin{array}{ll} 2 & 2 \\\\ 2 & 3 \end{array}\right|=2, C_{32}=-\left|\begin{array}{ll} 1 & 2 \\\\ 3 & 3 \end{array}\right|=3 \\\\ & C_{33}=\left|\begin{array}{ll} 1 & 2 \\\\ 3 & 2 \end{array}\right|=-4 \\\\ & \operatorname{adj} A=\left[\begin{array}{ccc} 1 & -3 & 1 \\\\ -2 & 0 & 1 \\\\ 2 & 3 & -4 \end{array}\right]^T=\left[\begin{array}{ccc} 1 & -2 & 2 \\\\ -3 & 0 & 3 \\\\ 1 & 1 & -4 \end{array}\right] \\\\ & \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-3}\left[\begin{array}{ccc} 1 & -2 & 2 \\\\ -3 & 0 & 3 \\\\ 1 & 1 & -4 \end{array}\right] \end{aligned} $

Now,

$ \sum_{\substack{1 \leq i \leq 3 \\\\ 1 \leq j \leq 3}} a_{i j}=\left(-\frac{1}{3}\right) $

$ \begin{aligned} & {[1+(-2)+2+(-3)+0+3+1+1+(-4)]} \\\\ & \quad=\left(-\frac{1}{3}\right)(-1)=\frac{1}{3} \end{aligned} $

We have,

$A X=D$, where

$ \begin{aligned} A & =\left[\begin{array}{ccc} 3 & -4 & 7 \\\\ 5 & 2 & -4 \\\\ 8 & -6 & -1 \end{array}\right], X=\left[\begin{array}{l} x \\\\ y \\\\ z \end{array}\right] \text { and } D=\left[\begin{array}{c} -6 \\\\ -9 \\\\ -5 \end{array}\right] \\\\ |A| & =3(-2-24)+4(-5+32)+7(-30-16) \\\\ & =-78+108-322 \\\\ & =-292 \neq 0 \end{aligned} $

$\therefore$ Rank $(A)=$ Order of $A=3$

$ A D=\left[\begin{array}{ccc} 3 & -4 & 7 \\\\ 5 & 2 & -4 \\\\ 8 & -6 & -1 \end{array}\right]\left[\begin{array}{l} -6 \\\\ -9 \\\\ -5 \end{array}\right]=\left[\begin{array}{c} -17 \\\\ -28 \\\\ 11 \end{array}\right] $

Clearly, Rank $([A D])=3$

Hence, Rank $(A)=\operatorname{Rank}([A D])=3$

We have,

$(x, y, z)=(\alpha, \beta, \gamma)$ is a unique solution of the system of simultaneous linear equations

$ \begin{aligned} 3 x-4 y+z & =-7 \\\\ 2 x+3 y-z & =10 \\\\ x-2 y-3 z & =3 \end{aligned} $

so, by cramer's rule,

$ \begin{aligned} & x=\alpha=\frac{\left|\begin{array}{ccc} -7 & -4 & 1 \\\\ 10 & 3 & -1 \\\\ 3 & -2 & -3 \end{array}\right|}{\left|\begin{array}{ccc} 3 & -4 & 1 \\\\ 2 & 3 & -1 \\\\ 1 & -2 & -3 \end{array}\right|} \\\\ & =\frac{-7(-9-2)+4(-30+3)+(-20-9)}{3(-9-2)+4(-6+1)+(-4-3)} \\\\ & =\frac{77-108-29}{-33-20-7} \\\\ & =\frac{-60}{-60}=1 \end{aligned} $

Given,

$\alpha, \beta, \gamma$ are the roots of the equation

$ \begin{gathered} 2 x^3-5 x^2+4 x-3=0 \\\\ \therefore \alpha+\beta+\gamma=\frac{-(-5)}{2}=\frac{5}{2} \quad \ldots \text { (i) } \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{4}{2}=2 \quad \ldots \text { (ii) } \\\\ \alpha \beta \gamma=\frac{-(-3)}{2}=\frac{3}{2} \quad \ldots \text { (iii) } \end{gathered} $

Now, $\Sigma \alpha \beta(\alpha+\beta)$

$ \begin{aligned} & =\alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\alpha+\gamma) \\\\ & =\alpha^2 \beta+\alpha \beta^2+\beta^2 \gamma+\beta \gamma^2+\alpha^2 \gamma+\alpha \gamma^2 \\\\ & =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma \\\\ & =\left(\frac{5}{2}\right)(2)-3\left(\frac{3}{2}\right) \end{aligned} $

[using Eqs. (i), (ii) and (iii)]

$ =\frac{10}{2}-\frac{9}{2}=\frac{1}{2} $

$\because A+B$ is symmecric

$ \begin{aligned} & \Rightarrow \quad(A+B)^T=A+B \\\\ & \Rightarrow \quad A^T+B^T=A+B \quad \ldots \text { (i) } \end{aligned} $

and $A-B$ is skew-symmetric

$ \begin{aligned} \Rightarrow \quad\left(A-A^T\right. & =-(A-B) \\\\ A^T-B^T & =-A+B \quad \ldots \text { (ii) } \end{aligned} $

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

$ B=A^T $

$ \Rightarrow B=\left[\begin{array}{ccc} -1 & 2 & 3 \\\\ 4 & 3 & -2 \\\\ 3 & -4 & 5 \end{array}\right]^r=\left[\begin{array}{ccc} -1 & 4 & 3 \\\\ 2 & 3 & -4 \\\\ 3 & -2 & 5 \end{array}\right] $

Given $D$ is the transpose of $C$

$ \begin{aligned} & \Rightarrow \quad=C^r=\left[\begin{array}{ccc} 0 & 1 & -2 \\\\ 2 & -1 & 0 \\\\ 0 & 2 & 1 \end{array}\right]^T \\\\ & D=\left[\begin{array}{ccc} 0 & 2 & 0 \\\\ 1 & -1 & 2 \\\\ -2 & 0 & 1 \end{array}\right] \\\\ & 50, B+D= {\left[\begin{array}{ccc} -1 & 4 & 3 \\\\ 2 & 3 & -4 \\\\ 3 & -2 & 5 \end{array}\right]+\left[\begin{array}{ccc} 0 & 2 & 0 \\\\ 1 & -1 & 2 \\\\ -2 & 0 & 1 \end{array}\right] } \\\\ &= {\left[\begin{array}{ccc} -1 & 6 & 3 \\\\ 3 & 2 & -2 \\\\ 1 & -2 & 6 \end{array}\right] } \end{aligned} $

We have, $A^2+I=2 A \Rightarrow A^2=2 A-I$

$ \begin{aligned} & \because A^9=A^2 \cdot A^2 A^2 A^2 A \\\\ & =(2 A-I)(2 A-I)(2 A-I)(2 A-I) A \\\\ & \qquad \begin{aligned} \Rightarrow A^9= & \left(4 A^2+I-4 A\right)\left(4 A^2+I-4 A\right) A \\\\ & =(4 A-3 I)(4 A-3 I) A \\\\ & =\left(16 A^2+9 I-24 A\right) A \\\\ & =(8 A-7 I) A \\\\ & =8 A^2-7 A \\\\ & =16 A-8 I-7 A \\\\ & =9 A-8 I \end{aligned} \end{aligned} $

Let $\Delta=\left[\begin{array}{ccc}\frac{a^2+b^2}{c} & c & c \\\\ a & \frac{b^2+c^2}{a} & a \\\\ b & b & \frac{c^2+a^2}{b}\end{array}\right]$

On multiply $R_1, R_2$ and $R_3$ by $c, a$ and $b$, respectively

$ \Rightarrow \Delta=\frac{1}{a b c}\left[\begin{array}{ccc} a^2+b^2 & c^2 & c^2 \\\\ a^2 & b^2+c^2 & a^2 \\\\ b^2 & b^2 & c^2+a^2 \end{array}\right] $

Apply $R_1 \rightarrow R_1-\left(R_2+R_3\right)$

$ \begin{aligned} \therefore \Delta= & \frac{1}{a b c}\left[\begin{array}{ccc} 0 & -2 b^2 & -2 a^2 \\\\ a^2 & b^2+c^2 & a^2 \\\\ b^2 & b^2 & c^2+a^2 \end{array}\right] \\\\ & =\frac{-2}{a b c}\left[\begin{array}{ccc} 0 & b^2 & a^2 \\\\ a^2 & b^2+c^2 & a^2 \\\\ b^2 & b^2 & c^2+a^2 \end{array}\right] \end{aligned} $

Apply $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$, we get

$ \begin{aligned} & =\frac{-2}{a b c}\left[\begin{array}{ccc} 0 & b^2 & a^2 \\\\ a^2 & c^2 & 0 \\\\ b^2 & 0 & c^2 \end{array}\right] \\\\ & =\frac{-2}{a b c}\left[0-b^2\left(a^2 c^2-0\right)+a^2\left(0-b^2 c^2\right)\right] \\\\ & =\frac{-2}{a b c}\left(-2 a^2 b^2 c^2\right)=4 a b c \end{aligned} $

We have,

$ \begin{aligned} & x-2 y+3 z=4 \\\\ & 3 x+y-2 z=7 \\\\ & 2 x+3 y+z=6 \end{aligned} $

Here,

$ \begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 3 \\\\ 3 & 1 & -2 \\\\ 2 & 3 & 1 \end{array}\right| \\\\ &=1(1+6)+2(3+4)+3(9-2)=42 \\\\ & D_x=\left|\begin{array}{ccc} 4 & -2 & 3 \\\\ 7 & 1 & -2 \\\\ 6 & 3 & 1 \end{array}\right| \\\\ &=4(1+6)+2(7+12)+3(21-6)=111 \\\\ & D_y=\left|\begin{array}{ccc} 1 & 4 & 3 \\\\ 3 & 7 & -2 \\\\ 2 & 6 & 1 \end{array}\right|=3 \\\\ & D_z=\left|\begin{array}{ccc} 1 & -2 & 4 \\\\ 3 & 1 & 7 \\\\ 2 & 3 & 6 \end{array}\right|=21 \end{aligned} $

So, system of linear equations has unique solution and $z=\frac{D_z}{D}=\frac{21}{42}=\frac{1}{2}$

$ \begin{aligned} &\text { We have, } A=\left[\begin{array}{lll} 83 & 74 & 41 \\ 93 & 96 & 31 \\ 24 & 15 & 79 \end{array}\right]\\ &\begin{aligned} A^T & =\left[\begin{array}{lll} 83 & 93 & 24 \\ 74 & 96 & 15 \\ 41 & 31 & 79 \end{array}\right] . \\ \therefore A-A^T & =\left[\begin{array}{lll} 83 & 74 & 41 \\ 93 & 96 & 31 \\ 24 & 15 & 79 \end{array}\right]-\left[\begin{array}{lll} 83 & 93 & 24 \\ 74 & 96 & 15 \\ 41 & 31 & 79 \end{array}\right] \\ & =\left[\begin{array}{ccc} 0 & -19 & 17 \\ 19 & 0 & 16 \\ -17 & -16 & 0 \end{array}\right] \end{aligned}\\ &\begin{aligned} & \therefore \operatorname{det}\left(A-A^T\right)=\left|\begin{array}{ccc} 0 & -19 & 17 \\ 19 & 0 & 16 \\ -17 & -16 & 0 \end{array}\right| \\ & 0(0+16 \times 16)+19(9 \times 0+16 \times 17) \\ & +17(-19 \times 16-0) \\ & =19 \times 16 \times 17-17 \times 19 \times 16 \\ & =0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, }\left|\begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array}\right|>0 \\ & =\{(a b c-a)-1(c-1)+1(1-b)\}>0 \\ & =(a b c-a-c+1+1-b)>0 \\ & =(a b c-(a+b+c)+2)>0 \\ & =a b c+2>(a+b+c) \\ & \begin{array}{l} a b c+2>3(a b c)^{\frac{1}{3}} \end{array} \\ & {\left[\because A m>G m \Rightarrow \frac{a+b+c}{3}>(a+b)^{1 / 3}\right]} \\ & \text { let } x=(a b c)^{1 / 3} \Rightarrow a b c=x^3 \\ & \qquad x^3+2>3 x \\ & \qquad \begin{array}{l} x^3-3 x+2>0 \\ (x-1)^2(x+2)>0 \\ (x+2)>0=x>-2 \\ \quad 1 \\ \quad=(a b c)^{\frac{1}{3}}>-2=(a b c)>-8 \end{array} \end{aligned} \end{aligned} $

The given system of equations are homogeneous and has only trival solution, then

$ \left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|=0 $

$\operatorname{Rank}[A]=n$, where $n$ is the order of square matrix and

$ A=\left[\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right] $

$\therefore \operatorname{Rank}[A]=3$

Given,

$ A=\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right] $

and $B$ is a matrix such that $A B=B A$. And $A B$ is not an identity matrix.

Let

$ B=\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right] $

Now, $A B=B A$

$ \begin{aligned} & {\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]}=\left[\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 2 \\ 2 & 3 & 0 \\ 4 & 0 & 3 \end{array}\right] \\ & {\left[\begin{array}{lll} b_1+2 c_1 & b_2+2 c_2 & b_3+2 c_3 \\ 2 a_1+3 b_1 & 2 a_2+3 b_2 & 2 a_3+3 b_3 \\ 4 a_1+3 c_1 & 4 a_2+3 c_2 & 4 a_3+3 c_3 \end{array}\right]} =\left[\begin{array}{llll} 2 a_2+4 a_3 & a_1+3 a_2 & 2 a_1+3 a_3 \\ 2 b_2+4 b_3 & b_1+3 b_2 & 2 b_1+3 b_3 \\ 2 c_2+4 c_3 & c_1+3 c_2 & 2 c_1+3 c_3 \end{array}\right] \end{aligned} $

On comparing two matrix and solve for $a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2$ and $c_3$, we get

$ \begin{aligned} & a_1=9, a_2=-3 \text { and } a_3=-6 \\ & b_1=-6, b_2=-8 \text { and } b_3=4 \\ & c_1=-12 c_2=4 \text { and } c_3=-2 \end{aligned} $

So, $B=\left[\begin{array}{ccc}9 & -3 & -6 \\ -6 & -8 & 4 \\ -12 & 4 & -2\end{array}\right]$

Let $A=\left[\begin{array}{ccc}x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2 x-1\end{array}\right]$ is a singular matrix.

So, $|A|=0$

$ \left|\begin{array}{ccc} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2 x-1 \end{array}\right|=0 $

$ \begin{aligned} & \Rightarrow(x-2)[(x+3)(2 x-1)-0]-0 \\ & \Rightarrow \quad(x-2)(x+3)(2 x-1)-2(x+3)]=0 \\ & \Rightarrow \quad(x+3)[(x-2)(2 x-1)-2]=0 \\ & \Rightarrow \quad(x+3)\left[2 x^2-x-4 x+2-2\right]=0 \\ & \Rightarrow \quad(x+3)\left[2 x^2-5 x\right]=0 \\ & \Rightarrow \quad x(x+3)(2 x-5)=0 \\ & \Rightarrow \quad x=-3,0,5 / 2 \end{aligned} $

So, $\alpha=-3, \beta=0$ and $\gamma=5 / 2$

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ [given } \alpha<\beta<\gamma \text { ] } $

Now,

$ \begin{aligned} 2 \alpha+3 \beta+4 \gamma & =2 \times(-3)+3 \times 0+4 \times \frac{5}{2} \\ & =-6+10=4 \end{aligned} $

To determine the nature of the solutions for the system of linear equations:

$ \begin{aligned} x + 2y + z &= -3, \\ 3x + 3y - 2z &= -1, \\ 2x + 7y + 7z &= -4, \end{aligned} $

we begin by calculating the determinant $\Delta$ of the coefficient matrix:

$ \Delta = \left| \begin{array}{ccc} 1 & 2 & 1 \\ 3 & 3 & -2 \\ 2 & 7 & 7 \end{array} \right| $

Calculating this determinant:

$ \Delta = 1(21 + 14) - 2(21 + 4) + 1(21 - 6) $

$ = 1 \times 35 - 2 \times 25 + 1 \times 15 $

$ = 35 - 50 + 15 = 0 $

Since $\Delta = 0$, the system of equations is either inconsistent (no solution) or dependent (infinitely many solutions). To resolve this, we check $\Delta_x$, the determinant of the matrix when the first column is replaced by the constants from the right side of the equations:

$ \Delta_x = \left| \begin{array}{ccc} -3 & 2 & 1 \\ -1 & 3 & -2 \\ -4 & 7 & 7 \end{array} \right| $

Calculating $\Delta_x$:

$ \Delta_x = (-3)(35) - 2(-15) + 1(5) $

$ = -105 + 30 + 5 = -70 $

Since $\Delta = 0$ and $\Delta_x \neq 0$, the system has no solution. Hence, the equations are inconsistent.

To determine the condition for $a$ and $b$ so, that the set of equations

$ \begin{aligned} & x+2 y+3 z=6 \\ & x+3 y+5 z=9 \end{aligned} $

$2 x+5 y+a z=b$ has a unique solution. We need to ensure that the coefficient matrix has a non zero determinant.

Let $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right)$

Then, $\operatorname{det}(A)$

$ \begin{aligned} & =1(3 a-25)-2(\cdot a-10)+3(5-6) \\ & =1(3 a-25)-2(a-10)+3(-1) \\ & =3 a-25-2 a+20-3 \\ & =a-8 \end{aligned} $

For system to have a unique solution. $\operatorname{det}(A)$ must be non-zero.

i.e. $a-8 \neq 0$

Thus, $a \neq 8$

So, $b$ can be other number which should be a non-zero real number and therefore $b$ must be a real number.

Thus, $a \neq 8, b \in R$.

Given, $ |PQ|=1 $ and $ |P|=9 $.

For any $ n \times n $ matrix $ A $, the determinant of $\operatorname{adj}(A)$ is $|A|^{n-1}$, where $ n $ is the dimension of the square matrix.

If $ B = kA $, where $ k $ is a scalar, then $|B| = k^n|A|$ for an $ n \times n $ matrix $ A $.

Also, $|PQ| = |P||Q|$.

Utilizing these properties, for a $ 3 \times 3 $ matrix we have:

$ |3Q| = 3^3|Q| = 27|Q|. $

Now, considering the matrix $ \operatorname{adj}(3Q) $:

$ \operatorname{Det}(\operatorname{adj}(3Q)) = (27|Q|)^2. $

Since $ |PQ| = 1 $ and $ |P| = 9 $, we have:

$ \begin{aligned} 1 & = 9 \cdot |Q|, \\ |Q| & = \frac{1}{9}. \end{aligned} $

Finally, compute $\operatorname{det}(\operatorname{adj}(3Q))$:

$ \begin{aligned} \operatorname{det}(\operatorname{adj}(3Q)) & = (27|Q|)^2 \\ & = (27 \times \frac{1}{9})^2 \\ & = 3^2 \\ & = 9. \end{aligned} $

For $\operatorname{adj}(P \cdot \operatorname{adj}(3Q))$:

$ \begin{aligned} \operatorname{adj}(P \cdot \operatorname{adj}(3Q)) & = (9 \times 9)^{3-1} \\ & = (9^2)^2 \\ & = 9^4. \end{aligned} $

We know that for a square matrix the product $A \cdot \operatorname{adj}(A)=\operatorname{det}(A)$. I where I is the identity matrix of the same order as $A$.

If $A$ is a $3 \times 3$ matrix, then the product should be : $A \cdot \operatorname{adj}(A)=\operatorname{det}(A) \cdot I_3$

Given matrix

$ \begin{aligned} A & =\left[\begin{array}{lll} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{array}\right] \\ \operatorname{adj}(A) & =\left[\begin{array}{ccc} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{array}\right] \end{aligned} $

Let's multiply $A$ with adj $(A)$

$ A \cdot \operatorname{adj}(A)=\left[\begin{array}{lll} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{array}\right] $

$ =\left[\begin{array}{ccc} 7 a-1 & b+1 & 7c \\ -a+3 & -3 b+17 & -c \\ -5 a+15 & 5 b+5 & -5+20 \end{array}\right] $

Setting these elements equal to identity matrix adjusted by determinant

We get, $-a+3=0$

$ \begin{aligned} \Rightarrow \quad a & =3, b+1=0 \Rightarrow b=-1 \\ 7 c & =0 \Rightarrow c=0 \end{aligned} $

Thus, $\quad a=3, b=-1$ and $c=0$

Hence, $a^2+b^2+c^2$

$ =3^2+(-1)^2+0=9+1=10 $

$ \begin{aligned} &\text { Given that } 3 A=\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\\ &\begin{aligned} \Rightarrow A & =\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{a}{3} & \frac{2}{3} & \frac{b}{3} \end{array}\right] \\ \therefore A^T & =\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{a}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{b}{3} \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Now, } A A^T=I \Rightarrow\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{a}{3} & \frac{2}{3} & \frac{b}{3} \end{array}\right]\left[\begin{array}{ccc} \frac{1}{3} & \frac{2}{3} & \frac{a}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{2}{3} & \frac{b}{3} \end{array}\right] =\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} \frac{1}{9}+\frac{4}{9}+\frac{4}{9} & \frac{2}{9}+\frac{2}{9}-\frac{4}{9} & \frac{a}{9}+\frac{4}{9}+\frac{2 b}{9} \\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9} & \frac{4}{9}+\frac{1}{9}+\frac{4}{9} & \frac{2 a}{9}+\frac{2}{9}-\frac{2 b}{9} \\ \frac{a}{9}+\frac{4}{9}+\frac{2 b}{9} & \frac{2 a}{9}+\frac{2}{9}-\frac{2 b}{9} & \frac{a^2}{9}+\frac{4}{9}+\frac{b^2}{9} \end{array}\right] =\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow\left[\begin{array}{ccc} 1 & 0 & \frac{a+4+2 b}{9} \\ \frac{a+4+2 b}{9} & \frac{2 a+2-2 b}{9} & \frac{\frac{a^2+4+b^2}{9}}{0} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \quad a^2+b^2+4=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\\ & \Rightarrow \quad a^2+b^2=5 \\ & \text { and } 2 a+2-2 b=0 \\ & \Rightarrow \quad b=a+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } a+4+2 b=0 \\ & \Rightarrow \quad a+2 b=-4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { On solving Eqs. (ii) and (iii), we get }\\ &a+2 b=-4 \end{aligned} $

$ \begin{aligned} \Rightarrow & & a+2(a+1) & =-4 \\ \Rightarrow & & a+2 a+2 & =-4 \\ \Rightarrow & & 3 a & =-6 \\ \Rightarrow & & a & =-2 \text { and } b=-2+1 \\ \Rightarrow & & b & =-1 \end{aligned} $

$ \begin{aligned} &\text { }\\ &Now,\,\,\,\,\,\,\,\,\,\,\frac{a}{b}+\frac{b}{a}=\frac{-2}{-1}+\frac{-1}{-2}=2+\frac{1}{2}=\frac{5}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right| \\ & \text { Applying } C_1 \rightarrow C_1+C_2+C_3 \\ & \quad=\left|\begin{array}{ccc} 2(a+b+c) & a & b \\ 2(b+c+a) & b+c+2 a & b \\ 2(c+a+b) & a & c+a+2 b \end{array}\right| \end{aligned} \end{aligned} $

$ \begin{aligned} &=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{array}\right|\\ &\text { Applying } R_1 \rightarrow R_1-R_2 \text { and } R_2 \rightarrow R_2-R_3\\ &\begin{aligned} & =2(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & a+b+c & -(a+b+c) \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right| \\ & =2(a+b+c)^3[1(1-0)]=2(a+b+c)^3 \end{aligned} \end{aligned} $

To clarify:

When we have a matrix $ B $ of order 3 with a determinant $ |B| = 6 $, the formula to find the determinant of its adjugate matrix, $|\operatorname{adj}(B)|$, is given by:

$ |\operatorname{adj}(B)| = |B|^{n-1} $

where $ n $ is the order of the matrix $ B $.

In this case, since $ n = 3 $, we have:

$ |\operatorname{adj}(B)| = 6^{3-1} = 6^2 = 36 $

This confirms that the assertion is true.

However, the reason provided, $ |\operatorname{adj}(B)| = |B|^n $, is incorrect because it states that the determinant of the adjugate matrix is raised to the power of $ n $ rather than $ n-1 $. The correct relationship involves raising the determinant of $ B $ to the power of $ n-1 $.

Thus, the assertion is true, but the given reason is false.

Given the matrix $ A = \left[\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right] $, which is singular, we need to find the quadratic equation with roots $ k $ and $ \frac{1}{k} $.

Since $ A $ is singular, its determinant is zero:

$ |A| = \left|\begin{array}{lll} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{array}\right| = 2(5k - 2) - 3(5 - 8) + 4(1 - 4k) = 0 $

Expanding this:

$ 2(5k - 2) = 10k - 4 $

$ -3(5 - 8) = -3 \cdot 3 = -9 $

$ 4(1 - 4k) = 4 - 16k $

Putting it all together, we get:

$ 10k - 4 - 9 + 4 - 16k = 0 $

Simplifying:

$ -6k + 9 = 0 $

Solving for $ k $:

$ k = \frac{3}{2} $

Thus, the roots of the quadratic equation are $ \frac{3}{2} $ and $ \frac{2}{3} $. Using the standard form of a quadratic equation $ x^2 - (\alpha + \beta)x + \alpha \beta = 0 $ where the roots are $ \alpha = \frac{3}{2} $ and $ \beta = \frac{2}{3} $, we calculate:

$ \alpha + \beta = \frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} $

$ \alpha \beta = \frac{3}{2} \cdot \frac{2}{3} = 1 $

Substitute these into the quadratic equation:

$ x^2 - \left(\frac{13}{6}\right)x + 1 = 0 $

To eliminate fractions, multiply through by 6:

$ 6x^2 - 13x + 6 = 0 $

This is the quadratic equation with the specified roots.

Given that $|P| = \left|\frac{A}{2}\right|$, we start by using the property of the adjoint matrix, where $|\text{adj} (A)| = |A|^{n-1}$ for an $ n \times n $ matrix. Here, since $ A $ is a $ 4 \times 4 $ matrix, $|\text{adj} (A)| = |A|^{3}$.

Substituting the given condition, we have:

$ |A|^{3} = \left(\frac{1}{2}\right)^{4} |A| $

Simplifying, this becomes:

$ |A|^{3} = \frac{1}{16} |A| $

Assuming $|A| \neq 0$, divide both sides by $|A|$:

$ |A|^{2} = \frac{1}{16} $

Taking the square root on both sides gives:

$ |A| = \pm \frac{1}{4} $

To find $|A^{-1}|$, we use the property $|A^{-1}| = \frac{1}{|A|}$:

$ |A^{-1}| = \pm 4 $

Therefore, the absolute determinant of the inverse of $ A $ is $\pm 4$.

We have,

$ \begin{array}{r} x+2 y+3 z=4 \\ 4 x+5 y+3 z=5 \\ 3 x+4 y+3 z=\lambda \end{array} $

Write the equation in matrix form $\left[A_1^1 B\right]$

$ \left[\begin{array}{lll:l} 1 & 2 & 3 & 4 \\ 4 & 5 & 3 & 5 \\ 3 & 4 & 3 & \lambda \end{array}\right] $

$R_2 \rightarrow R_2-4 R_1$ and $R_3 \rightarrow R_3-3 R_1$

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & -2 & -6 & \lambda-12 \end{array}\right] $

$ \sim\left[\begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & -3 & -9 & -11 \\ 0 & 0 & 0 & 3 \lambda-14 \end{array}\right] $

$ R_3 \leftrightarrow 3 R_3-2 R_2 $

For consistency $3 \lambda-14=0 \Rightarrow 3 \lambda=14$

Given, $3 \lambda=n+100$

Therefore, $n=14-100$

$ n=-86 $

So, the value of $n$ is -86 .

Let $\Delta=\left|\begin{array}{ccc}a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|\quad \ldots \text { (i) }$

Applying, $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2+R_3$ we get

$ \Delta=\left|\begin{array}{ccc} a+1 & b+1 & 0+1 \\ a^2+1 & b^2+1 & c^2+1 \\ 1 & 1 & 1 \end{array}\right| $

Applying $C_1 \rightarrow C_1-C_2$ and $C_2 \rightarrow C_2-C_3$, we get

$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} a-b & b-c & c+1 \\ a^2-b^2 & b^2-c^2 & c^2+1 \\ 0 & 0 & 1 \end{array}\right| \\ & =1\left[(a-b)\left(b^2-c^2\right)-(b-c)\left(a^2-b^2\right)\right] \\ & =1\left[a b^2-a c^2-b^3+b c^2-a^2 b+b^3+a^2 c-b^2 c\right] \end{aligned} $

$ =a b^2-a^2 b-a c^2+a^2 c-b^2 c+b c^2 $

Thus, option (a) and (b) are correct. Now, option (c),

$ \left|\begin{array}{ccc} a(a+1) & b(b+1) & c(c+1) \\ a+1 & b+1 & c+1 \\ -1 & -1 & -1 \end{array}\right| $

Applying, $C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$

$ =\left|\begin{array}{ccc} a(a+1) & b^2+b-a^2-a & c^2+c-a^2-a \\ a+1 & b-a & c-a \\ -1 & 0 & 0 \end{array}\right| $

$ =-1\left[\left(b^2+b-a^2-a\right)(c-a)-\left(c^2+c-a^2-a\right)(b-a)\right] $

$ =\left(c^2+c-a^2-a\right)(b-a)-\left(b^2+b-a^2-a\right)(c-a) $

$ =c^2 b-a c^2+b c-a c-a^2 b+a^3-a b+a^2-b^2 c+a b^2-b c+a b+a^2 c -a^3+a c-a^2 $

Hence, option (d) is correct required answer.

Given, $A B C D E=I$

$\Rightarrow A^{-1}(A B C D E) E^{-1}=A^{-1} I E^{-1}$

( $\because$ Pre multiply by $A^{-1}$ and post multiply by $E^{-1}$, respectively)

$ \begin{array}{ll} \Rightarrow & I B C D I=A^{-1} E^{-1} \\ \Rightarrow & B C D=A^{-1} E^{-1} \end{array} $

Now, pre multiply by $B^{-1}$ and post multiply by $D^{-1}$, we get

$ \begin{array}{l} \Rightarrow & B^{-1}(B C D) D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & B^{-1} B C D D^{-1} & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & I C I & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \Rightarrow & C & =B^{-1} A^{-1} E^{-1} D^{-1} \\ \therefore & C^{-1} & =\left(B^{-1} A^{-1} E^{-1} D^{-1}\right)^{-1} \\ & & =\left(D^{-1}\right)^{-1}\left(E^{-1}\right)^{-1}\left(A^{-1}\right)^{-1}\left(B^{-1}\right)^{-1} \\ & & \text { (using reversal law) } \end{array} $

$\begin{array} & &=D E A B \\ \therefore \quad C^{-1} & =D E A B \end{array}$

Given, $A=\left[a_{i j}\right], 1 \leq i, j \leq n$ with $n \geq 2$ and $a_{i j}=i+j$

$\therefore$ The matrix is

$A=\left[\begin{array}{ccccc}2 & 3 & 4 & \cdots & n+1 \\ 3 & 4 & 5 & \cdots & n+2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & n+3 & \cdots & n+n\end{array}\right]$

$ =\left[\begin{array}{cccc} 2 & 3 & \cdots & n+1 \\ 3 & 4 & \cdots & n+2 \\ 4 & 5 & \cdots & n+3 \\ \vdots & \vdots & \cdots & \vdots \\ n+1 & n+2 & \cdots & 2 n \end{array}\right] $

Use elementary row operation to subtract 1st row from the $i$ th row for $2 \leq i \leq n$, we get

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 2 & 2 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ n-1 & n-1 & n-1 & \cdots & n-1 \end{array}\right] $

Applying, $R_3 \rightarrow R_3-2 R_2$ and $R_4 \rightarrow R_4-3 R_2$ and so on

$ A=\left[\begin{array}{ccccc} 2 & 3 & 4 & \cdots & n+1 \\ 1 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{array}\right] $

$\therefore$ Rank $A=2$

Given the matrix $ A $:

$ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

We are tasked to find $ A^2 - 5A + 6I $.

Calculate $ A^2 $:

$ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} $

Perform the matrix multiplication:

$ A^2 = \begin{bmatrix} 1\cdot1 + 0\cdot2 + 2\cdot3 & 1\cdot0 + 0\cdot1 + 2\cdot2 & 1\cdot2 + 0\cdot3 + 2\cdot4 \\ 2\cdot1 + 1\cdot2 + 3\cdot3 & 2\cdot0 + 1\cdot1 + 3\cdot2 & 2\cdot2 + 1\cdot3 + 3\cdot4 \\ 3\cdot1 + 2\cdot2 + 4\cdot3 & 3\cdot0 + 2\cdot1 + 4\cdot2 & 3\cdot2 + 2\cdot3 + 4\cdot4 \end{bmatrix} $

$ = \begin{bmatrix} 1 + 0 + 6 & 0 + 0 + 4 & 2 + 0 + 8 \\ 2 + 2 + 9 & 0 + 1 + 6 & 4 + 3 + 12 \\ 3 + 4 + 12 & 0 + 2 + 8 & 6 + 6 + 16 \end{bmatrix} $

$ = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} $

Calculate $ 5A $:

$ 5A = 5 \cdot \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} $

Calculate $ 6I $:

$ 6I = 6 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

Combine the results to find $ A^2 - 5A + 6I $:

$ A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} $

$ = \begin{bmatrix} 7 - 5 + 6 & 4 - 0 + 0 & 10 - 10 + 0 \\ 13 - 10 + 0 & 7 - 5 + 6 & 19 - 15 + 0 \\ 19 - 15 + 0 & 10 - 10 + 0 & 28 - 20 + 6 \end{bmatrix} $

$ = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

Thus, the resulting matrix is:

$ \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} $

We have, $\left|\begin{array}{ccc}x^2+2 x & x+2 & 1 \\ 2 x+1 & x-1 & 1 \\ x+2 & -1 & 1\end{array}\right|=0$

$ \begin{aligned} & \Rightarrow\left(x^2+2 x\right)\left|\begin{array}{cc} x-1 & 1 \\ -1 & 1 \end{array}\right|-(x+2)\left|\begin{array}{cc} 2 x+1 & 1 \\ x+2 & 1 \end{array}\right|+1 \\ & \Rightarrow\left(x^2+2 x\right)(x-1+1)-(x+2)(2 x+1-x-2) \\ & \quad+1\left(-2 x-1-x^2-x+2\right)=0 \\ & \Rightarrow x\left(x^2+2 x\right)-(x+2)(x-1)+1 \\ & \quad\left(-x^2-3 x+1\right)=0 \\ & \Rightarrow x^3+2 x^2-x^2-x+2-x^2-3 x+1=0 \\ & \Rightarrow x^3-4 x+3=0 \end{aligned} $

On putting, $x=1$, we get

$ (1)^3-4 \times 1+3=0 $

Thus, $x=1$ is a root of the given equation.

Thus, $(x-1)\left(x^2+x-3\right)=0$

Solve $x^2+x-3=0$

$ \begin{aligned} x & =\frac{-1 \pm \sqrt{1^2-4 \times 1 \times(-3)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{1+12}}{2}=\frac{-1 \pm \sqrt{13}}{2} \end{aligned} $

Hence, positive roots are 1 and $\frac{-1+\sqrt{13}}{2}$

Sum of positive roots $=1+\left(\frac{-1+\sqrt{13}}{2}\right)$

$ =\frac{1+\sqrt{13}}{2} $

Given system of equation

$ \begin{aligned} x+y-z & =6 \\ 3 x+2 y-z & =5 \\ 2 x-y-2 z & =-3 \end{aligned} $

We can represent this as

$ A \cdot X=B $

Where

$ \begin{aligned} & A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 3 & 2 & -1 \\ 2 & -1 & -2 \end{array}\right], X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 6 \\ 5 \\ -3 \end{array}\right] \\ & \text { Now, }[A / B]=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 3 & 2 & -1 \\ 2 & -1 & -2 \end{array}\right]\left[\begin{array}{c} 6 \\ 5 \\ -3 \end{array}\right] \end{aligned} $

$ \begin{aligned} & R_2 \rightarrow R_2-3 R_1=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 2 \\ 2 & -1 & -2 \end{array}\right]\left[\begin{array}{c} 6 \\ -13 \\ -3 \end{array}\right] \\ & R_3 \rightarrow R_3-2 R_1=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -1 & 2 \\ 0 & -3 & 0 \end{array}\right]\left[\begin{array}{c} 6 \\ -13 \\ -15 \end{array}\right] \\ & \begin{aligned} R_2 \rightarrow-R_2 & =\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & -3 & 0 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ -15 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & R_3 \rightarrow R_3+3 R_2=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & -6 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ 24 \end{array}\right] \\ & R_3 \rightarrow \frac{R_3}{-6} \\ &=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 13 \\ -4 \end{array}\right] \end{aligned} $

$ \begin{aligned} R_2 \rightarrow R_2+2 R_3 & =\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} 6 \\ 5 \\ -4 \end{array}\right] \end{aligned} $

Now, we get

$ \begin{aligned} x+y-z & =6 \\ y & =5 \\ z & =-4 \\ x+5+4 & =6 \\ x+9 & =6 \\ x & =-3 \end{aligned} $

Thus, $x=\alpha=-3$

$ \begin{aligned} & y=\beta=5 \\ & z=\gamma=-4 \end{aligned} $

then, $\alpha+\beta=-3+5=2$

We have, $\left|\begin{array}{ccc}1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3\end{array}\right|$

$ \begin{aligned} & C_2 \rightarrow C_2-C_1 \text { and } C_3 \rightarrow C_3-C_1 \\ & =\left|\begin{array}{ccc} 1 & 0 & 0 \\ a^2 & b^2-a^2 & c^2-a^2 \\ a^3 & b^3-a^3 & c^3-a^3 \end{array}\right| \end{aligned} $

$\begin{aligned} & =(b-a)(c-a) \\ & \left|\begin{array}{ccc}1 & 0 & 0 \\ a^2 & (b+a) & (c+a) \\ a^3 & \left(b^2+a^2+a b\right) & \left(c^2+a^2+a c\right)\end{array}\right| \\ & =(b-a)(c-a) \\ & \left|\begin{array}{cc}b+a & c+a \\ b^2+a^2+a b & c^2+a^2+a c\end{array}\right| \\ & =\left[(b-a)(c-a)\left[b c^2+a^2 b+a b c+a c^2+a^3\right.\right. \\ & \left.+a^2 c-b^2 c-a^3-a^2 b-a^2 c-a b c-a b^2\right] \\ & =(b-a)(c-a)\left[b c^2+a c^2-b^2 c-a b^2\right] \\ & =(b-a)(c-a)[b c(c-b) \\ & +a(c-b)(c+b)] \\ & =(a-b)(b-c)(c-a)(a b+b c+c a)\end{aligned}$

To solve the given equation $\alpha A^2 + \beta A = 2I$ where $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$, we first need to compute $A^2$:

$ A^2 = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} $

Calculating the matrix multiplication:

$ A^2 = \begin{bmatrix} (1)(1) + (2)(-2) & (1)(2) + (2)(-5) \\ (-2)(1) + (-5)(-2) & (-2)(2) + (-5)(-5) \end{bmatrix} = \begin{bmatrix} -3 & -8 \\ 8 & 21 \end{bmatrix} $

Next, substitute $A^2$ into the equation $\alpha A^2 + \beta A = 2I$:

$ \alpha \begin{bmatrix} -3 & -8 \\ 8 & 21 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $

Break this down into:

$ \begin{bmatrix} -3\alpha & -8\alpha \\ 8\alpha & 21\alpha \end{bmatrix} + \begin{bmatrix} \beta & 2\beta \\ -2\beta & -5\beta \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $

Equate the corresponding elements from both sides:

$-3\alpha + \beta = 2$

$-8\alpha + 2\beta = 0$

$8\alpha - 2\beta = 0$

$21\alpha - 5\beta = 2$

From the second equation $-8\alpha + 2\beta = 0$, we solve for $\beta$:

$ 2\beta = 8\alpha \Rightarrow \beta = 4\alpha $

Substitute $\beta = 4\alpha$ into the first equation:

$ -3\alpha + 4\alpha = 2 \Rightarrow \alpha = 2 $

Using $\beta = 4\alpha$, we find:

$ \beta = 4(2) = 8 $

Finally, calculate $\alpha + \beta$:

$ \alpha + \beta = 2 + 8 = 10 $

To find the value of $ a $ that makes the system of equations have no solution, we need to ensure that the determinant of the coefficient matrix is zero. This will indicate that the equations are linearly dependent, leading to no unique solutions.

Given the system of equations:

$ \begin{align*} x + 2y + 3z &= 6, \\ x + 3y + 5z &= 9, \\ 2x + 5y + az &= 12, \end{align*} $

we consider the coefficient matrix:

$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{bmatrix} $

Calculate the determinant of this matrix:

$ D = \begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{vmatrix} $

The determinant is calculated as follows:

$ \begin{align*} D &= 1 \cdot (3a - 25) - 2 \cdot (a - 10) + 3 \cdot (5 - 6) \\ &= 3a - 25 - 2a + 20 - 3 \\ &= 3a - 2a - 25 + 20 - 3 \\ &= a - 8. \end{align*} $

Setting the determinant $ D = 0 $ gives:

$ a - 8 = 0, $

which simplifies to:

$ a = 8. $

Therefore, the system of equations has no solution when $ a = 8 $.

We have.

$ \left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$ R_1 \rightarrow R_1+R_2+R_y \text { we get } $

$ \Rightarrow\left|\begin{array}{ccc} -x & -x & -x \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$ \Rightarrow-x\left|\begin{array}{ccc} 1 & 1 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0 $

$C_2 \rightarrow C_2-C_1$ and $C_3 \rightarrow C_3-C_1$, we get

$ \begin{aligned} & \Rightarrow-x\left|\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 6-x & 0 \\ 1 & -3 & -x \end{array}\right|=0 \\ & \Rightarrow \quad(-x)(-x)(6-x)=0 \\ & \quad x^{\prime}-6 x^2=0 \end{aligned} $

Sum of product of zero

$ =\frac{\text { Coefficient of } x}{\text { Cocflicient of } x^3}=0 $

Hence, $\alpha \beta+\beta y+p^2=0$

We are given that the determinant of a 3rd order matrix $ A $ is $ K $. Hence, $|A| = K$.

For a matrix $ A $, the determinant properties used here are:

The determinant of the transpose of a matrix is the same as the determinant of the matrix itself. So, $|A^T| = |A| = K$.

The determinant of the product of a matrix with its transpose is $|AA^T| = |A||A^T| = K \times K = K^2$.

Now, we are asked to find the sum of the determinants of matrices $ A^4 $ and $ (A - A^4) $.

Since the determinant of a product is the product of the determinants and given $|A| = K$, we have:

$|A^4| = |A|^4 = K^4$.

To find $|A - A^4|$, note that:

By matrix properties, $|(A - A^4)| = \text{simplify or evaluate separately}$.

Given the expresion $|(AA^T) + (A - A^T)|$ in the context, we simplify:

$|A A^T| + |(A - A^4)| = K^2 + |A| - |A^T| = K^2 + K - K = K^2$.

Thus the result for the sum of these determinants aligns with the property calculations stated, giving us the simplified determinant result of $ K^2 $.

We have,

$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & -1 & 2 \\ -1 & 1 & 5 \end{array}\right| \\ &=1(-5-2)-1(10+2)+1(2-1) \\ &=-7-12+1 \\ &=-18 \neq 0 \\ & \Delta_1=\left|\begin{array}{ccc} 5 & 1 & 1 \\ 4 & -1 & 2 \\ 11 & 1 & 5 \end{array}\right| \\ &=5(-5-2)-1(20-22)+1(4+11) \\ &=-35+2+15 \\ &=-18 \\ & \Delta_3=\left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & -1 & 4 \\ -1 & 1 & 11 \end{array}\right| \\ &=1(-11-4)-1(22+4)+5(2-1) \\ &=-15-26+5=-36 \\ & \therefore \quad \alpha=\frac{\Delta_1}{\Delta}=\frac{-18}{-18}=1, \\ & \therefore \quad \beta=\frac{\Delta_3}{\Delta}=\frac{-36}{-18}=2 \end{aligned} $

Hence, $\alpha^2+\beta^2=1^2+2^2=5$.