(a)OR(b)

If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1\end{array}\right]$, find $\displaystyle \mathrm{A}^{-1}$ and use it to solve the following system of equations : \[\mathrm{x}-2 \mathrm{y}=10,2 \mathrm{x}-\mathrm{y}-z=8,-2 \mathrm{y}+z=7 \]

$\displaystyle |\mathrm{A}|=1 \neq 0$ 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}$\displaystyle -5+\mathrm{b}=0 \Rightarrow \mathrm{b}=5, \quad 5-5 \mathrm{a}=0 \Rightarrow \mathrm{a}=1$ $\displaystyle 4+\mathrm{y}=0 \Rightarrow \mathrm{y}=-4, \quad 3 \mathrm{x}-9=0 \Rightarrow \mathrm{x}=3$ $\displaystyle \therefore(\mathrm{a}+\mathrm{x})-(\mathrm{b}+\mathrm{y})=(1+3)-(5-4)=3$