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If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$, find $(A B)^{-1}$.
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Verified Answer
$\begin{aligned}
&|\mathrm{B}|=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]=1 \neq 0 \\
&\therefore \mathrm{B}^{-1}=\text { exists } \\
&\mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|}(\text { adj } \mathrm{B})=\left[\begin{array}{ccc}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]
\end{aligned}$
Now $(\mathrm{AB})^{-1}=\mathrm{B}^{-1}$ $\mathrm{~A}^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
$=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]$
&|\mathrm{B}|=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]=1 \neq 0 \\
&\therefore \mathrm{B}^{-1}=\text { exists } \\
&\mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|}(\text { adj } \mathrm{B})=\left[\begin{array}{ccc}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]
\end{aligned}$
Now $(\mathrm{AB})^{-1}=\mathrm{B}^{-1}$ $\mathrm{~A}^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
$=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]$
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