(a)OR(b)

If $\displaystyle \mathrm{A}=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right]$ and $\displaystyle \mathrm{B}^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$, find $\displaystyle (\mathrm{AB})^{-1}$. Also, find $\displaystyle \left|(\mathrm{AB})^{-1}\right|$.

We know that (AB) $\displaystyle { }^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$|\mathrm{A}|=5(-1)+4(1)=-1 \neq 0 \text {. Hence, } \mathrm{A}^{-1} \text { exists. }Cofactors of the elements of A are: $\displaystyle \mathrm{A}_{11}=-1, \mathrm{A}_{12}=0, \mathrm{A}_{13}=1$ $\displaystyle \mathrm{A}_{21}=8, \mathrm{~A}_{22}=1, \mathrm{~A}_{23}=-10$ $\displaystyle \mathrm{A}_{31}=-12, \mathrm{A}_{32}=-2, \mathrm{A}_{33}=15$\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} -1 & 8 & -12 0 & 1 & -2 1 & -10 & 15 \end{array}\right] \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\left[\begin{array}{ccc} 1 & -8 & 12 0 & -1 & 2 -1 & 10 & -15 \end{array}\right]$\displaystyle (\mathrm{A} \mathrm{B})^{-1}=\mathrm{B}^{-1} \mathrm{A}^{-1}=\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]\left[\begin{array}{ccc} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{array}\right]=$\left[\begin{array}{lll} -2 & 19 & -27 -2 & 18 & -25 -3 & 29 & -42 \end{array}\right]$\displaystyle \left|(\mathrm{A} \mathrm{B})^{-1}\right|=\left|\mathrm{B}^{-1} \mathrm{A}^{-1}\right|=\left|\mathrm{B}^{-1}\right|\left|\mathrm{A}^{-1}\right|$ $\displaystyle =1 \times \frac{1}{-1}=-1$\begin{aligned} & |\mathrm{A}|=1(4)-1(2)+1(-1)=1 & \text { Cofactors of the elements of } \mathrm{A} \text { are: } & \mathrm{A}_{11}=4, \mathrm{A}_{12}=-2, \mathrm{A}_{13}=-1 & \mathrm{A}_{21}=-1, \mathrm{A}_{22}=1, \mathrm{A}_{23}=0 & \mathrm{A}_{31}=-1, \mathrm{A}_{32}=0, \mathrm{A}_{33}=1 & \therefore \operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} 4 & -1 & -1 -2 & 1 & 0 -1 & 0 & 1 \end{array}\right] & \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\left[\begin{array}{ccc} 4 & -1 & -1 -2 & 1 & 0 -1 & 0 & 1 \end{array}\right] \end{aligned}Given system of equations can be written as $\displaystyle \mathrm{AX}=\mathrm{B}$, where\mathrm{X}=\left[\begin{array}{l} \mathrm{x} \mathrm{y} \mathrm{z} \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1 2 \end{array}\right] \mathrm{X}=\mathrm{A}^{-1} \mathrm{B}=\left[\begin{array}{ccc} 4 & -1 & -1 -2 & 1 & 0 -1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} 1 2 \end{array}\right]=\left[\begin{array}{c} -2 0 \end{array}\right]$\displaystyle \mathrm{x}=-2, \mathrm{y}=0, \mathrm{z}=3$