$(i)$ If $A=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1\end{array}\right]$, find $A^{-1} .$ Using $A^{-1}$, solve these stem of linear equations : $x-2 y=10,2 x+y+3 z=8,-2 y+z=7$ (ii) $A=\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$, find $A^{-1}$ and hence solve thesystem of linear equations: $3 x-4 y+2 z=-1,2 x+3 y+5 z=7, x+z=2$

$(i)$ If $A=\left[\begin{array}{ccc}1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1\end{array}\right]$ , find $A^{-1} .$ Using $A^{-1}$ , solve these stem of linear equations : $x-2 y=10,2 x+y+3 z=8,-2 y+z=7$
(ii) $A=\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]$ , find $A^{-1}$ and hence solve thesystem of linear equations: $3 x-4 y+2 z=-1,2 x+3 y+5 z=7, x+z=2$

(i) Solution:

Given that
$\begin{array}{l} A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{array}\right] \\ |A|=1(1+6)+2(2-0)+0 \\ =7+4 \\ =11 \end{array}$
Now, the cofactors of $\mathrm{A}$
$C_{11}=(-1)^{1+1} 1+6=7$
$\begin{array}{l} C_{21}=(-1)^{2+1}-2-0=2 \\ C_{31}=(-1)^{3+1}-6-0=-6 \\ C_{12}=(-1)^{1+2} 2-0=-2 \\ C_{22}=(-1)^{2+1} 1-0=1 \\ C_{32}=(-1)^{3+1} 3-0=-3 \\ C_{13}=(-1)^{1+2}-4-0=-4 \\ C_{23}=(-1)^{2+1}-2-0=2 \\ C_{33}=(-1)^{3+1} 1+4=5 \\ \text { Adj } A=\left[\begin{array}{ccc} 7 & 2 & -4 \\ -2 & 1 & -3 \\ -6 & 2 & 5 \end{array}\right]^{T}=\left[\begin{array}{ccc} 7 & -2 & -6 \\ 2 & 1 & 2 \\ -4 & -3 & 5 \end{array}\right] \\ A^{-1=} \frac{1}{|A|} \operatorname{adj} A \\ A^{-1}=\frac{1}{11}\left[\begin{array}{ccc} 7 & -2 & -6 \\ 2 & 1 & 2 \\ -4 & -3 & 5 \end{array}\right] \end{array}$
Now we can write the given equation as:
$\begin{array}{l} {\left[\begin{array}{lll} 1 & -2 & 0 \\ 2 & -1 & 3 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{Z} \end{array}\right]=\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]} \\ \mathrm{AX}=\mathrm{B} \end{array}$
Or, $X=A^{-1} B$
$\begin{array}{l} =\frac{1}{11}\left[\begin{array}{ccc} 7 & -2 & -6 \\ 2 & 1 & 2 \\ -4 & -3 & 5 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right] \\ {\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} 70+16-42 \\ -20+8-21 \\ -40+16+35 \end{array}\right]} \\ \mathrm{X}={ }^{\frac{1}{11}\left[\begin{array}{c} 44 \\ -33 \\ 11 \end{array}\right]} \end{array}$
$\mathrm{X}=\left[\begin{array}{c} 4 \\ -3 \\ 1 \end{array}\right]$
As a result, $x=4, y=-3$ and $z=1$

(ii) Solution:

Given that
$\begin{array}{l} \quad A=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right] \\ |A|=3(3-0)+4(2-5)+2(0-3) \\ =9-12-6 \\ =-9 \end{array}$
Now, the cofactors of $\mathrm{A}$
$\begin{aligned} C_{11} &=(-1)^{1+1} 3-0=3 \\ C_{21} &=(-1)^{2+1}-4-0=4 \\ C_{31} &=(-1)^{3+1}-20-6=-26 \\ C_{12} &=(-1)^{1+2} 2-5=3 \\ C_{22} &=(-1)^{2+1} 3-2=1 \\ C_{32} &=(-1)^{3+1} 15-4=-11 \\ C_{13} &=(-1)^{1+2} 0-3=-3 \\ C_{23} &=(-1)^{2+1} 0+4=-4 \\ C_{33} &=(-1)^{3+1} 9+8=17 \\ \operatorname{Adj} A=\left[\begin{array}{ccc}3 & 3 & -3 \\ 4 & 1 & -4 \\ -26 & -4 & 27\end{array}\right]^{T}=\left[\begin{array}{ccc}3 & 4 & -26\\ 3 & 1 & -11 \\ -3 & -4 & 17\end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj} A \end{aligned}$
$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
$A^{-1}=\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & -11 \\ -3 & -4 & 17 \end{array}\right]$
Now we can write the given equation as:
$\begin{array}{l} {\left[\begin{array}{ccc} 3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{Z} \end{array}\right]=\left[\begin{array}{c} -1 \\ 7 \\ 2 \end{array}\right]} \\ \mathrm{A} X=\mathrm{B} \end{array}$
Or, $X=A^{-1} B$
$\begin{array}{l} =\frac{1}{-9}\left[\begin{array}{ccc} 3 & 4 & -26 \\ 3 & 1 & 11 \\ -3 & -4 & 17 \end{array}\right]\left[\begin{array}{c} -1 \\ 7 \\ 2 \end{array}\right] \\ {\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{-9}\left[\begin{array}{c} -3+28-52 \\ 21+7+22 \\ 3-28+34 \end{array}\right]} \\ \mathrm{X}=\stackrel{\frac{1}{-9}\left[\begin{array}{c} -27 \\ -18 \\ 9 \end{array}\right]} \\ \end{array}$
$X=\left[\begin{array}{c} 3 \\ 2 \\ -1 \end{array}\right]$
As a result, $x=3, y=2$ and $z=-1$