(a)OR(b)

Given $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}-4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1\end{array}\right]$ and $\displaystyle \mathrm{B}=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3\end{array}\right]$, find AB . Hence, solve the system of linear equations: \[\begin{aligned} & \mathrm{x}-\mathrm{y}+\mathrm{z}=4 \\ & \mathrm{x}-2 \mathrm{y}-2 \mathrm{z}=9 \\ & 2 \mathrm{x}+\mathrm{y}+3 \mathrm{z}=1 \end{aligned} \]

\[\mathrm{A} \mathrm{B}=\left[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right]=8 I \] The system of equations is equivalent to the matrix equation: \[\mathrm{B} X=C \text {, where } C=\left[\begin{array}{l} 4 \\ 9 \\ \end{array}\right], X=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] \] $\displaystyle \Rightarrow X=\mathrm{B}^{-1} C$ $\displaystyle \mathrm{A} \mathrm{B}=8 I$ \[\begin{aligned} & \Rightarrow \mathrm{B}^{-1}=\frac{1}{8} \mathrm{A} \\ & X=\frac{1}{8}\left[\begin{array}{ccc} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{array}\right]\left[\begin{array}{l} 4 \\ 9 \\ \end{array}\right]=\frac{1}{8}\left[\begin{array}{c} 24 \\ -16 \\ -8 \end{array}\right]=\left[\begin{array}{c} 3 \\ -2 \\ -1 \end{array}\right] \end{aligned} \] $\displaystyle \therefore \mathrm{x}=3, \mathrm{y}=-2, \mathrm{z}=-1$ $\displaystyle |\mathrm{A}|=1 \neq 0 \Rightarrow \mathrm{A}^{-1}$ exists. \[\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \]