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Solution:

$($ i) Here,

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

$|A|=1(1+6)+2(2-0)+0(-4-0)$

$=7+4+0$

$=11$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}1 & 3 \\ -2 & 1\end{array}\right|=7, C_{12}=(-1)^{1+2}\left|\begin{array}{ll}2 & 3 \\ 0 & 1\end{array}\right|=-2, C_{13}=(-1)^{1+3}\left|\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right|=-4$

$C_{21}=(-1)^{2+1}\left|\begin{array}{ll}-2 & 0 \\ -2 & 1\end{array}\right|=2, C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1, C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & -2 \\ 0 & -2\end{array}\right|=2$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-2 & 0 \\ 1 & 3\end{array}\right|=-6, C_{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 0 \\ 2 & 3\end{array}\right|=-3, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & -2 \\ 2 & 1\end{array}\right|=5$

$\therefore$ adj $A=\left[\begin{array}{ccc}7 & -2 & -4 \\ 2 & 1 & 2 \\ -6 & -3 & 5\end{array}\right]^{T}$

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

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

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

or, $A X=B$

where, $A=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

Now,

$\therefore X=A^{-1} B$

$\Rightarrow X=\frac{1}{11}\left[\begin{array}{ccc}7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5\end{array}\right]\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

$\Rightarrow X=\frac{1}{11}\left[\begin{array}{c}70+16-42 \\ -20+8-21 \\ -40+16+35\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44 \\ -33 \\ 11\end{array}\right]$

$\therefore x=4, y=-3$ and $z=1$

$($ ii $)$ Here,

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

$\begin{aligned}|A| &=3(3-0)+4(2-5)+2(0-3) \\ &=9-12-6 \\ &=-9 \end{aligned}$

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right] .$ Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|=3, C_{12}=(-1)^{1+2}\left|\begin{array}{ll}2 & 5 \\ 1 & 1\end{array}\right|=3, C_{13}=(-1)^{1+3}\left|\begin{array}{ll}2 & 3 \\ 1 & 0\end{array}\right|=-3$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-4 & 2 \\ 0 & 1\end{array}\right|=4, C_{22}=(-1)^{2+2}\left|\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right|=1, C_{23}=(-1)^{2+3}\left|\begin{array}{cc}3 & -4 \\ 1 & 0\end{array}\right|=-4$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-4 & 2 \\ 3 & 5\end{array}\right|=-26, C_{32}=(-1)^{3+2}\left|\begin{array}{ll}3 & 2 \\ 2 & 5\end{array}\right|=-11, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}3 & -4 \\ 2 & 3\end{array}\right|=17$

$\operatorname{adj} A=\left[\begin{array}{ccc}3 & 3 & -3 \\ 4 & 1 & -4 \\ -26 & -11 & 17\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{ccc}3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-9}\left[\begin{array}{ccc}3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}-1 \\ 7 \\ 2\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow X=\frac{1}{-9}\left[\begin{array}{ccc}3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17\end{array}\right]\left[\begin{array}{c}-1 \\ 7 \\ 2\end{array}\right]$

$\Rightarrow X=\frac{1}{-9}\left[\begin{array}{c}-3+28-52 \\ -3+7-22 \\ 3-28+34\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-9}\left[\begin{array}{c}-27 \\ -18 \\ 9\end{array}\right]$

$\therefore x=3, y=2$ and $z=-1$

(iii) Here,

$A=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5\end{array}\right]$

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

$\begin{aligned} \Rightarrow A B &=\left[\begin{array}{ccc}7+4+0 & 2-2+0 & -6+6+0 \\ 14-2-12 & 4+1+6 & -12-3+15 \\ 0+4-4 & 0-2+2 & 0+6+5\end{array}\right] \\ &=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \end{aligned}$

$\Rightarrow A B=11\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$A B=11 I_{3}$

$\Rightarrow \frac{1}{11} A B=I_{3}$

$\Rightarrow\left(\frac{1}{11} B\right) A=I_{3}$

$\Rightarrow A^{-1}=\frac{1}{11} B$

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

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}70+16-42 \\ -20+9-21 \\ -40+16+35\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44 \\ -33 \\ 11\end{array}\right]$

$\therefore x=4, y=-3$ and $z=1$

$($ iv $)$ Here,

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

$|A|=1(-1-2)+2(2)$

$=-3+4$

$=1$

Let $C_{i j}$ be the cofactors of the elements $a_{i j}$ in $A=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{rr}-1 & -2 \\ -1 & 1\end{array}\right|=-3, C_{12}=(-1)^{1+2}\left|\begin{array}{rr}-2 & -2 \\ 0 & 1\end{array}\right|=2, C_{13}=(-1)^{1+3}\left|\begin{array}{rr}-2 & -1 \\ 0 & -1\end{array}\right|=2$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}2 & 0 \\ -1 & 1\end{array}\right|=-2, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=1, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right|=1$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}2 & 0 \\ -1 & -2\end{array}\right|=-4, C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & 0 \\ -2 & -2\end{array}\right|=2, C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 2 \\ -2 & -1\end{array}\right|=3$

$\therefore$ adj $A=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]^{T}$

$=\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$

$=\frac{1}{1}\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]$

$=\left[\begin{array}{ccc}-3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{array}\right]$

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

Here, $C=A^{T}$

i. e.,$C=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right]$

$\therefore C^{-1}=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]$

or, $C X=B$

where, $C=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

Now,

$\therefore X=C^{-1} B$

$\Rightarrow X=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{c}-30+16+14 \\ -20+8+7 \\ -40+16+21\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}0 \\ -5 \\ -3\end{array}\right]$

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

$(\mathrm{v})$ Here,

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

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

$\Rightarrow B A=\left[\begin{array}{ccc}2+4+0 & 2-2+0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10\end{array}\right]$

$=\left[\begin{array}{lll}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$

$\Rightarrow B A=6\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow B A=6 I_{3}$

$\Rightarrow B\left(\frac{1}{6} A\right)=I_{3}$

$\Rightarrow B^{-1}=\frac{1}{6} A$

$\Rightarrow B^{-1}=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$

Now, $B X=C$

where, $B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $C=\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$

$\therefore X=B^{-1} C$

$\Rightarrow X=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6+34-28 \\ -12+34-28 \\ 6-17+35\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12 \\ -6 \\ 24\end{array}\right]$\

$\therefore x=2, y=-1$ and $z=4$.

(vi) We have, $A=\left[\begin{array}{rrr}2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1\end{array}\right]$.

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

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

$=-8-15+7$

$=-16 \neq 0$

So, $A$ is invertible.

Let $\mathrm{C}_{i j}$ be the co-factors of the elements $a_{i j}$ in $A\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}2 & 2 \\ 1 & -1\end{array}\right|=-2-2=-4$

$C_{12}=(-1)^{1+2}\left|\begin{array}{cc}1 & 2 \\ -3 & -1\end{array}\right|=-1(-1+6)=-5$

$C_{13}=(-1)^{1+3}\left|\begin{array}{cc}1 & 2 \\ -3 & 1\end{array}\right|=1+6=7$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}3 & 1 \\ 1 & -1\end{array}\right|=3+1=4$

$C_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & 1 \\ -3 & -1\end{array}\right|=-2+3=1$

$C_{23}=(-1)^{2+3}\left|\begin{array}{cc}2 & 3 \\ -3 & 1\end{array}\right|=-1(2+9)=-11$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}3 & 1 \\ 2 & 2\end{array}\right|=6-2=4$

$C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & 1 \\ 1 & 2\end{array}\right|=-1(4-1)=-3$

$C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & 3 \\ 1 & 2\end{array}\right|=4-3=1$

$\therefore \operatorname{Adj} A=\left[\begin{array}{ccc}-4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1\end{array}\right]^{T}=\left[\begin{array}{ccc}-4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{\operatorname{Adj} A}{|A|}=\frac{1}{-16}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1\end{array}\right]$

Now, the given system of equations is expressible as

$\left[\begin{array}{ccc}2 & 1 & -3 \\ 3 & 2 & 1 \\ 1 & 2 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}13 \\ 4 \\ 8\end{array}\right]$

Or $A^{\mathrm{I}} X=B$, where $X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right], B=\left[\begin{array}{c}13 \\ 4 \\ 8\end{array}\right]$

Now, $\left|A^{T}\right|=|A|=-16 \neq 0$

So, the given system of equations is consistent with a unique solution given by

$X=\left(A^{T}\right)^{-1} B=\left(A^{-1}\right)^{T} B$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{ccc}-4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1\end{array}\right]^{T}\left[\begin{array}{c}13 \\ 4 \\ 8\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{ccc}-4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1\end{array}\right]\left[\begin{array}{c}13 \\ 4 \\ 8\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-52-20+56 \\ 52+4-88 \\ 52-12+8\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=-\frac{1}{16}\left[\begin{array}{c}-16 \\ -32 \\ 48\end{array}\right]$

 

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

Hence, $x=1, y=2$ and $z=-3$ is the required solution.

(vii) Suppose, $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right] B=\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$

$A \times B=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$

$=\left[\begin{array}{ccc}-2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8\end{array}\right]$

 

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

Since, $A \times B=I$,

$\therefore B=A^{-1}$                ....(1)

Now, the given system of equations is

$x+3 z=9$

$-x+2 y-2 z=4$

 

$2 x-3 y+4 z=-3$

This can also be represented as,

$\left[\begin{array}{ccc}1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

Here, we can observe that $\left[\begin{array}{ccc}1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4\end{array}\right]=A^{T}$

So, $A^{T}\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

Multiply the above expression by $\left(A^{T}\right)^{-1}$.

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left(A^{T}\right)^{-1}\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left(A^{-1}\right)^{T}\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

 

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=B^{T}\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$                [Using (1)]

$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]^{T}\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

$=\left[\begin{array}{ccc}-2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2\end{array}\right]\left[\begin{array}{c}9 \\ 4 \\ -3\end{array}\right]$

$=\left[\begin{array}{c}-18+36-18 \\ 0+8-3 \\ 9-12+6\end{array}\right]$

 

$=\left[\begin{array}{l}0 \\ 5 \\ 3\end{array}\right]$

Hence, $x=0, y=5$ and $z=3$.