$$\begin{aligned} &\text { Given, the points are }(a+5, a-4),(a-2, a+3) \text { and }(a, a) \text {. }\\ &\therefore \quad \Delta=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{array}\right| \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =\frac{1}{2}\left|\begin{array}{rrr} 5 & -4 & 0 \\ -2 & 3 & 0 \\ a & a & 1 \end{array}\right| \quad\left[\because R_1 \rightarrow R_1-R_3 \text { and } R_2 \rightarrow R_2-R_3\right] \\ & =\frac{1}{2}[1(15-8)] \\ \Rightarrow\quad & =\frac{7}{2} \neq 0 \end{aligned}\\ &\text { Hence, given points form a triangle i.e., points do not lie in a straight line. } \end{aligned}$$

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

$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \\ \cos A-\cos C & \cos B-\cos C & 1+\cos C \\ \cos ^2 A+\cos A-\cos ^2 C-\cos C & \cos ^2 B+\cos B-\cos ^2 C-\cos C \cos ^2 C+\cos C\end{array}\right|=0\quad \left[\begin{array}{ll} \therefore & \left.C_1 \rightarrow C_1-C_3 \text { and } C_2 \rightarrow C_2-C_3\right] \end{array}\right.$

$\Rightarrow(\cos A-\cos C) \cdot(\cos B-\cos C)$

$\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A+\cos C+1 & \cos B+\cos C+1 & \cos ^2 C+\cos C\end{array}\right|=0$

[taking $(\cos A-\cos C)$ common from $C_1$ and $(\cos B-\cos C)$ common from $C_2$ ]

$$\begin{array}{lrl} \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)[(\cos B+\cos C+1)-(\cos A+\cos C+1)] & =0 \\ \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)(\cos B+\cos C+1-\cos A-\cos C-1) & =0 \\ \Rightarrow & (\cos A-\cos C) \cdot(\cos B-\cos C)(\cos B-\cos A) & =0 \\ \text { i.e., } & \cos A=\cos C \text { or } \cos B=\cos C \text { or } \cos B=\cos A \\ \Rightarrow & A=C \text { or } B=C \text { or } B=A \end{array}$$

Hence, $A B C$ is an isosceles triangle.

We have, $\quad A=\left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$

$$\begin{array}{ll} \therefore & A_{11}=-1, A_{12}=1, A_{13}=1, A_{21}=1, A_{22}=-1, A_{23}=1, A_{31}=1, A_{32}=1 \text { and } A_{33}=-1 \\ \therefore & \text { adj } A=\left|\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|^{\top}=\left|\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right| \end{array}$$

$$\begin{aligned} &\begin{aligned} \text { and }\quad & |A|=-1(-1)+1 \cdot 1=2 \\ \therefore\quad & A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{2}\left[\begin{array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\quad\text{.... (i)} \end{aligned} \end{aligned}$$

and $$ A^2=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \cdot\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]\quad\text{..... (ii)}$$

$\begin{aligned} \therefore \quad \frac{A^2-3 I}{2} & \left.=\frac{1}{2}\left\{\left|\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{array}\right|-\left|\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right|\right\}\right\}=\frac{1}{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right| \\ & =A^{-1}\end{aligned}\quad$ [using Eq. (i)]

Hence proved.

We have, $$A=\left|\begin{array}{rrr} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right|\quad\text{.... (i)}$$

$$\therefore \quad|A|=1(-3)-2(-2)+0=1 \neq 0$$

Now, $A_{11}=-3, A_{12}=2, A_{13}=2, _{21}=-2, A_{22}=1, A_{23}=1, A_{31}=-4, A_{32}=2$ and $A_{33}=3$

$$\therefore \quad \operatorname{adj}(A)=\left|\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right|^{\top}=\left|\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right|$$

$$\begin{array}{ll} \therefore & A^{-1}=\frac{\operatorname{adj} A}{|A|} \\ & =\frac{1}{1}\left|\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right| \\ \Rightarrow & A^{-1}=\left|\begin{array}{rrr} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right|\quad\text{.... (ii)} \end{array}$$

Also, we have the system of linear equations as

$$\begin{aligned} x-2 y & =10, \\ 2 x-y-z & =8 \\ \text{and }\quad -2 y+z & =7 \end{aligned}$$

In the form of $C X=D$,

$\left[\begin{array}{rrr}1 & 2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{r}10 \\ 8 \\ 7\end{array}\right]$

We know that, $$\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$$

$$ C^T=\left|\begin{array}{rrr} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right|=A$$ [using Eq. (i)]

$$\begin{array}{ll} \therefore & X=C^{-1} D \\ \Rightarrow & {\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]} \end{array}$$

$=\left[\begin{array}{c}-30+16+14 \\ -20+8+7 \\ -40+16+21\end{array}\right]=\left[\begin{array}{r}0 \\ -5 \\ -3\end{array}\right]$

$\therefore \quad x=0, y=-5$ and $z=-3$

$$\begin{array}{r} \text{Given system of equations is} \quad 3 x+2 y-2 z=3 \\ x+2 y+3 z=6 \\ \text{and}\quad 2 x-y+z=2 \end{array}$$

In the form of $A X=B$,

$$\left[\begin{array}{rrr} 3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 6 \\ 2 \end{array}\right]$$

For $A^{-1}$,

$$\begin{aligned} |A| & =|3(5)-2(1-6)+(-2)(-5)| \\ & =|15+10+10|=|35| \neq 0 \end{aligned}$$

$\therefore \quad A_{11}=5, A_{12}=5, A_{13}=-5, \quad A_{21}=0, A_{22}=7, A_{23}=7, \quad A_{31}=10, A_{32}=-11$ and $A_{33}=4$

$$\begin{array}{ll} \therefore & \operatorname{adj} A=\left|\begin{array}{rrr} 5 & 5 & -5 \\ 0 & 7 & 7 \\ 10 & -11 & 4 \end{array}\right|^T=\left|\begin{array}{rrr} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right| \\ \text { Now, } & A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{1}{35}\left|\begin{array}{rrr} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right| \end{array}$$

$$\begin{aligned} & \text { For } X=A^{-1} B, \\ & \qquad\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{35}\left[\begin{array}{rrr} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 6 \\ 2 \end{array}\right] \\ & \\ & =\frac{1}{35}\left[\begin{array}{c} 15+20 \\ 15+42-22 \\ -15+42+8 \end{array}\right]=\frac{1}{35}\left[\begin{array}{l} 35 \\ 35 \\ 35 \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \\ & \therefore \quad x=1, y=1 \text { and } z=1 \end{aligned}$$