(a) If A = [1, -2, 0; 2, -1, -1; 0, -2, 1], find A^-1 and use it to solve the following system of equations: x -2 y =10,2 x - y -z=8,-2 y +z=7

$\displaystyle |\mathrm{A}|=1 \neq 0$ hence $\displaystyle \boldsymbol{\mathrm{A}}^{-1}$ exists. $$\operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right] $$ $$\mathrm{A}^{-1}=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right] $$ $\displaystyle \mathrm{AX}=\mathrm{B} \Rightarrow\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ z\end{array}\right]=\left[\begin{array}{c}10 \\ 8 \\ 7\end{array}\right]$ $$X=\mathrm{A}^{-1} \mathrm{B} \Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ z \end{array}\right]=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{c} 10 \\ 8 \\ 7 \end{array}\right]=\left[\begin{array}{c} 0 \\ -5 \\ -3 \end{array}\right] $$ $\displaystyle \Rightarrow \mathrm{x}=0, \mathrm{y}=-5, z=-3$ $$\begin{aligned} & \mathrm{A} \mathrm{A}^{-1}=I \\ & {\left[\begin{array}{ccc} -1 & \mathrm{a} & 2 \\ 1 & 2 & \mathrm{x} \\ 3 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & -1 & 1 \\ -8 & 7 & -5 \\ \mathrm{b} & \mathrm{y} & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \\ & {\left[\begin{array}{ccc} -1-8 \mathrm{a}+2 \mathrm{b} & 1+7 \mathrm{a}+2 \mathrm{y} & 5-5 \mathrm{a} \\ -15+\mathrm{b} \mathrm{x} & 13+\mathrm{x} \mathrm{y} & 3 \mathrm{x}-9 \\ -5+\mathrm{b} & 4+\mathrm{y} & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]} \end{aligned} $$ $\displaystyle -5+\mathrm{b}=0 \Rightarrow \mathrm{b}=5, \quad 5-5 \mathrm{a}=0 \Rightarrow \mathrm{a}=1$ $\displaystyle 4+\mathrm{y}=0 \Rightarrow \mathrm{y}=-4, \quad 3 \mathrm{x}-9=0 \Rightarrow \mathrm{x}=3$ $\displaystyle \therefore(\mathrm{a}+\mathrm{x})-(\mathrm{b}+\mathrm{y})=(1+3)-(5-4)=3$