(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 the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + 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 the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) $A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]$, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right] \\ \left|A\right|\text{=1}\left(1+6\right)+2\left(2-0\right)+0\left(-4-0\right) \\ =7+4+0 \\ =11 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A=}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}1& 3\\ -2& 1\end{array}\right|=7,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 3\\ 0& 1\end{array}\right|=-2,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 1\\ 0& -2\end{array}\right|=-4 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-2& 0\\ -2& 1\end{array}\right|=2,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& -2\\ 0& -2\end{array}\right|=2 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-2& 0\\ 1& 3\end{array}\right|=-6,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 0\\ 2& 3\end{array}\right|=-3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& -2\\ 2& 1\end{array}\right|=5 \\ \therefore \mathrm{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}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ \mathrm{or},AX=B \\ \mathrm{where},A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \mathrm{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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44\\ -33\\ 11\end{array}\right] \\ \therefore x=4,y=-3\mathrm{and}z=1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\text{Here,} \\ A=\left[\begin{array}{ccc}3& -4& 2\\ 2& 3& 5\\ 1& 0& 1\end{array}\right] \\ \left|A\right|\text{=3}\left(3-0\right)+4\left(2-5\right)+2\left(0-3\right) \\ =9-12-6 \\ =-9 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A=}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}3& 5\\ 0& 1\end{array}\right|=3,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& 5\\ 1& 1\end{array}\right|=3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 3\\ 1& 0\end{array}\right|=-3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-4& 2\\ 0& 1\end{array}\right|=4,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& 2\\ 1& 1\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}3& -4\\ 1& 0\end{array}\right|=-4 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-4& 2\\ 3& 5\end{array}\right|=-26,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& 2\\ 2& 5\end{array}\right|=-11,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}3& -4\\ 2& 3\end{array}\right|=17 \\ \mathrm{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}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{-9}\left[\begin{array}{ccc}3& 4& -26\\ 3& 1& -11\\ -3& -4& 17\end{array}\right] \\ AX\mathit{=}B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}3& -4& 2\\ 2& 3& 5\\ 1& 0& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{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}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-9}\left[\begin{array}{c}-27\\ -18\\ 9\end{array}\right] \\ \therefore x=3,y=2\mathrm{and}z=-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\text{Here,} \\ A=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\mathrm{and}B=\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ AB=\left[\begin{array}{ccc}1& -2& 0\\ 2& 1& 3\\ 0& -2& 1\end{array}\right]\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ \Rightarrow AB=\left[\begin{array}{ccc}7+4+0& 2-2+0& -6+6+0\\ 14-2-12& 4+1+6& -12-3+15\\ 0+4-4& 0-2+2& 0+6+5\end{array}\right] \\ =\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& 11\end{array}\right] \\ \Rightarrow AB=11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ AB=11I_3 \\ \Rightarrow \frac{1}{11}AB=I_3 \\ \Rightarrow \left(\frac{1}{11}B\right)A=I_3 \\ \Rightarrow A^{-1}=\frac{1}{11}B \\ \Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right] \\ X=A^{-1}B \\ X=\frac{1}{11}\left[\begin{array}{ccc}7& 2& -6\\ -2& 1& -3\\ -4& 2& 5\end{array}\right]\left[\begin{array}{c}10\\ 8\\ 7\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}70+16-42\\ -20+9-21\\ -40+16+35\end{array}\right] \\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}44\\ -33\\ 11\end{array}\right] \\ \therefore x=4,y=-3\mathrm{and}z=1 \end{gathered}$$