Solve this

Question:

Given $A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right], B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$. Compute $(A B)^{-1}$.

Solution:

We have,

$A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right]$

$B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$

We know $(A B)^{-1}=B^{-1} A^{-1}$

For matrix $A$,

$C_{11}=\left|\begin{array}{ll}3 & 2 \\ 2 & 1\end{array}\right|=-1, C_{12}=-\left|\begin{array}{ll}2 & 2 \\ 1 & 1\end{array}\right|=0$ and $C_{13}=\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=1$

$C_{21}=-\left|\begin{array}{ll}0 & 4 \\ 2 & 1\end{array}\right|=8, C_{22}=\left|\begin{array}{ll}5 & 4 \\ 1 & 1\end{array}\right|=1$ and $C_{23}=-\left|\begin{array}{ll}5 & 0 \\ 1 & 2\end{array}\right|=-10$

$C_{31}=\left|\begin{array}{ll}0 & 4 \\ 3 & 2\end{array}\right|=-12, C_{32}=-\left|\begin{array}{ll}5 & 4 \\ 2 & 2\end{array}\right|=-2$ and $C_{33}=\left|\begin{array}{ll}5 & 0 \\ 2 & 3\end{array}\right|=15$

Now,

$\operatorname{adj}(A)=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 8 & 1 & -10 \\ -12 & -2 & 15\end{array}\right]^{T}=\left[\begin{array}{ccc}-1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15\end{array}\right]$

and $|A|=-1$

$\therefore A^{-1}=-\left[\begin{array}{ccc}-1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15\end{array}\right]=\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right]$

So, $B^{-1} A^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]\left[\begin{array}{ccc}1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15\end{array}\right]=\left[\begin{array}{lll}-2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42\end{array}\right]$