(a) If A = [5, 0, 4; 2, 3, 2; 1, 2, 1] and B^-1= [1, 3, 3; 1, 4, 3; 1, 3, 4], find ( AB )^-1. Also, find |( AB )^-1|.

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 $$\mathrm{X}=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1 \\ 2 \\ \end{array}\right] $$ $$\mathrm{X}=\mathrm{A}^{-1} \mathrm{B}=\left[\begin{array}{ccc} 4 & -1 & -1 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ \end{array}\right]=\left[\begin{array}{c} -2 \\ 0 \\ \end{array}\right] $$ $\displaystyle \mathrm{x}=-2, \mathrm{y}=0, \mathrm{z}=3$