Matrices and Determinants

$\begin{aligned} & \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\ & P=A-2 A^{\top} \\ & Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\ & |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\ & \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \\ & \left|A-2 A^T\right|=4 \\ & A-2 A^T=\left[\begin{array}{lll} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2 \end{array}\right] \\ & \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\ & \Rightarrow|A|^2=16 \end{aligned}$

$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$

$\because$ Non-trivial solution

$\Rightarrow D=0$

$\begin{aligned} & \left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\ \end{aligned}$

$\begin{aligned} & -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\ & \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\ & \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\ & \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text { for } n=0 \\ & \Rightarrow \alpha=\frac{5 \pi}{24} \end{aligned}$

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$, $\mathrm{B}$, and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$, and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$, we find $\operatorname{det} \mathrm{X}$ as follows:

First, we express $|\mathrm{C}|$, the determinant of $\mathrm{C}$, in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$, using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$:

$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$

Next, we find $|\mathrm{X}|$, using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$

Substituting the expression for $|\mathrm{C}|$ obtained earlier:

$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$

The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:

Finally, substituting these values into our expression for $|\mathrm{X}|$:

$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$

Therefore, $\operatorname{det} \mathrm{X} = 729$.

$\text { Let } A=\left[\begin{array}{lll} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{array}\right]$

$\text { Given } A\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right] \quad \text{ ..... (1)}$

$\therefore\left[\begin{array}{l} \mathrm{x}_1+\mathrm{z}_1 \\ \mathrm{x}_2+\mathrm{z}_2 \\ \mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 2 \end{array}\right]$

$\begin{aligned} \therefore \mathrm{x}_1+\mathrm{z}_1 & =2 \quad \text{.... (2)}\\ \mathrm{x}_2+\mathrm{z}_2 & =0 \quad \text{.... (3)}\\ \mathrm{x}_3+\mathrm{z}_3 & =0 \quad \text{.... (4)} \end{aligned}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} -1 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} -4 \\ 0 \\ 4 \end{array}\right] \\ & \therefore\left[\begin{array}{l} -\mathrm{x}_1+\mathrm{z}_1 \\ -\mathrm{x}_2+\mathrm{z}_2 \\ -\mathrm{x}_3+\mathrm{z}_3 \end{array}\right]=\left[\begin{array}{l} 4 \\ 0 \\ 4 \end{array}\right] \end{aligned}$

$\begin{array}{r} \Rightarrow-\mathrm{x}_1+\mathrm{z}_1=-4 \quad \text{... (5)}\\ -\mathrm{x}_2+\mathrm{x}_2=0 \quad \text{... (6)}\\ -\mathrm{x}_3+\mathrm{z}_3=4 \end{array}$

$\begin{aligned} & \text { Given } A\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \\ & \therefore\left[\begin{array}{l} \mathrm{y}_1 \\ \mathrm{y}_2 \\ \mathrm{y}_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 2 \\ 0 \end{array}\right] \end{aligned}$

$\begin{aligned} & \therefore \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \therefore \text { from }(2),(3),(4),(5),(6) \text { and }(7) \\ & \mathrm{x}_1=3 \mathrm{x}, \mathrm{x}_2=0, \mathrm{x}_3=-1 \\ & \mathrm{y}_1=0, \mathrm{y}_2=2, \mathrm{y}_3=0 \\ & \mathrm{z}_1=-1, \mathrm{z}_2=0, \mathrm{z}_3=3 \end{aligned}$

$\begin{gathered} \therefore A=\left[\begin{array}{ccc} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{array}\right] \\ \therefore \text { Now }(A-31)\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} -1 \\ 2 \\ 3 \end{array}\right] \\ \therefore\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} -1 \\ 2 \\ 3 \end{array}\right] \\ {\left[\begin{array}{c} -z \\ -y \\ -x \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ {[z=-1],[y=-2],[x=-3]} \end{gathered}$

$\begin{aligned} & D=\left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right| \\ & =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\ & =\alpha \beta+3+4 \beta-18-2-3 \alpha \end{aligned}$

For infinite solutions $\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0$ and

$\begin{aligned} & \mathrm{D}_3=0 \\ & \mathrm{D}=0 \end{aligned}$

$\alpha \beta-3 \alpha+4 \beta=17$ ..... (1)

$\begin{aligned} & \mathrm{D}_1=\left|\begin{array}{ccc} -4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta \end{array}\right|=0 \\ & \mathrm{D}_2=\left|\begin{array}{ccc} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{array}\right|=0 \\ & \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\ & 13 \beta-9-36-9=0 \end{aligned}$

$13 \beta=54, \beta=\frac{54}{13}$ put in (1)

$\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17$

$54 \alpha-39 \alpha+216=221$

$15 \alpha=5 \quad \alpha=\frac{1}{3}$

Now, $12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}$

$=4+54=58$

$\begin{aligned} & x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0 \\ & \Rightarrow x, y, z \neq 0 \end{aligned}$

Also,

$\begin{aligned} & \sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{4 \pi}{3}\right)=0 \forall \theta \in \mathrm{R} \\ & \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0 \\ & \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=0 \end{aligned}$

(i) $\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}$

If $x+y+z=0$ and $x y+y z+z x=0$

$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0$

Statement (i) is False

(ii) $\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}$

Trace $(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))$

$=x y z(x+y+z) \neq 0$

Statement (ii) is also False

$\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 2 \lambda^2-\lambda-1=0 \\ & \lambda=1,-\frac{1}{2} \\ & \left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned}$

Infinite solution $\lambda=1 \& \mu=13$

For unique $\operatorname{sol}^{\mathrm{n}} \lambda \neq 1$

For no $\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu \neq 13$

If $\lambda \neq 1$ and $\mu \neq 13$

Considering the case when $\lambda=-\frac{1}{2}$ and $\mu \neq 13$ this will generate no solution case

$x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15$,

$\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \lambda \\ 1 & 3 & 4 \lambda^2 \end{array}\right|=(2 \lambda-1)^2$

For unique solution $\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)$

Let $\Delta=0, \lambda=\frac{1}{2}$

$\begin{aligned} & \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc} 4 \mu & 1 & 1 \\ 10 \mu & 2 & 1 \\ \mu^2+15 & 3 & 1 \end{array}\right| \\ & =(\mu-15)(\mu-1) \end{aligned}$

For infinite solution $\lambda=\frac{1}{2}, \mu=1$ or 15

$\begin{aligned} \left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{I}\right| & =\left|\mathrm{P}^{-1} \mathrm{AP}-2 \mathrm{P}^{-1} \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}(\mathrm{~A}-2 \mathrm{I}) \mathrm{P}\right| \\ & =\left|\mathrm{P}^{-1}\right||\mathrm{A}-2 \mathrm{I}||\mathrm{P}| \\ & =|\mathrm{A}-2 \mathrm{I}| \\ & =\left|\begin{array}{ccc} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{array}\right|=69 \end{aligned}$

So, Prime factor of 69 is 3 & 23

So, sum = 26

$\begin{aligned} & |\mathrm{A}|=\alpha^2-\beta^2 \\ & |2 \mathrm{~A}|^3=2^{21} \Rightarrow|\mathrm{A}|=2^4 \\ & \alpha^2-\beta^2=16 \\ & (\alpha+\beta)(\alpha-\beta)=16 \Rightarrow \alpha=4 \text { or } 5 \end{aligned}$

$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$

On solving given expression, we get

$\begin{aligned} & \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\ & =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}} \end{aligned}$

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

$\begin{aligned} & \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\ & \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\ & \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\ & \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\ & \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2} \end{aligned}$

Hence option (C) is correct.

$\begin{aligned} & f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(-x)=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}$

Hence statement- I is correct

Now, checking statement II

$\begin{aligned} & f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \\ & \Rightarrow f(x) \cdot f(y)=f(x+y) \end{aligned}$

Hence statement-II is also correct.

$\begin{aligned} \text{(A)}\quad & \mathrm{ax}^2+2 \mathrm{bxy}+\mathrm{cy}^2>0 \forall(\mathrm{x}, \mathrm{y}) \in \mathbb{R}^2-\{(0,0)\} \\ & \Rightarrow \mathrm{ax}+2 \mathrm{bxy}+\mathrm{cy}^2 \text { must represent pair of imaginary lines and } \mathrm{a}, \mathrm{c}>0 . \\ & \Rightarrow \mathrm{b}^2<\mathrm{ac} \end{aligned}$

$\mathrm{(B)}\quad\mathrm{b}^2<3 \times \frac{1}{12} \Rightarrow|2 \mathrm{~b}|<1$

$\mathrm{(C)}\quad$ since $\mathrm{b}^2 \neq \mathrm{ac}$

$\Rightarrow \mathrm{ax}+\mathrm{by}=1 \text { and } \mathrm{bx}+\mathrm{cy}=-1$

are not parallel lines.

$\mathrm{(D)}\quad\mathrm{ac}+\mathrm{a}+\mathrm{c}>\mathrm{b}^2 \Rightarrow$ lines are not parallel.

$\Rightarrow$ Options (B), (C), (D) are Correct.

$\alpha, \beta$ are roots of $x^2+x-1=0$

$\begin{aligned} & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right] \end{aligned}$

(P) $\quad M=\left[\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}\right] \Rightarrow 3 ! \times 2=12$

For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3! ways

$\therefore 3!\times 2=12 \text { ways }$

(Q) $\quad M=\left[\begin{array}{lll}x & a & b \\ a & y & c \\ b & c & z\end{array}\right] \Rightarrow$ Consider one such arrangement with $a=\alpha, b=\beta, c=1$

a, b, c can be arranged in 3! ways and corresponding entries can be arranged in 1 way.

(R)

$\begin{aligned} & {\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a \\ 0 \\ -c \end{array}\right]} \\ & a y+b z=a \\ & -a x+c z=0 \\ & -b x-c y=-c \end{aligned}$

It is observed that $\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0$

$\therefore$ infinite solution

(S)

$\begin{aligned} & {\left[\begin{array}{lll} 1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta \end{array}\right] } \\ & \Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad(\text { since } \alpha \beta=\alpha+\beta=-1) \end{aligned}$

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} $