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If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]$, then what is $\mathrm{B}^{-1} \mathrm{~A}^{-1}$ equal to?
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The correct answer is:
$\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]$
$|\mathrm{A}|=-1,|\mathrm{~B}|=1$
$\mathrm{~A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} \mathrm{A}=\frac{1}{-1}\left[\begin{array}{cc}1 & -1 \\ -2 & 1\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{\mid \mathrm{B}} \mid \mathrm{adj} \mathrm{B}=\frac{1}{1}\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]$
$\mathrm{B}^{-1} \mathrm{~A}^{-1}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]$
$\mathrm{~A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} \mathrm{A}=\frac{1}{-1}\left[\begin{array}{cc}1 & -1 \\ -2 & 1\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]$
$\mathrm{B}^{-1}=\frac{1}{\mid \mathrm{B}} \mid \mathrm{adj} \mathrm{B}=\frac{1}{1}\left[\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]$
$\mathrm{B}^{-1} \mathrm{~A}^{-1}=\left[\begin{array}{cc}2 & 1 \\ -1 & 0\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}-1 & 3 \\ 1 & -2\end{array}\right]$
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