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Let $B=\left(\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right)$ and $C=\left(\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right)$
If a matrix $A$ is such that $\mathrm{BAC}=\mathrm{I}$, then $\mathrm{A}^{-1}=$
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If a matrix $A$ is such that $\mathrm{BAC}=\mathrm{I}$, then $\mathrm{A}^{-1}=$
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Verified Answer
The correct answer is:
$\left(\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right)$
$\begin{aligned} & \text {Given BAC }=1 \\ & \Rightarrow(B A)^{-1}=C \\ & \Rightarrow A^{-1} B^{-1}=C \Rightarrow A^{-1}=C B \\ & \Rightarrow A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]=\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\end{aligned}$
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