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If $\mathrm{A}$ is a $3 \times 3$ matrix such that $|\mathrm{A}|=4$, then what is $\mathrm{A}(\operatorname{adj} \mathrm{A})$ equal to?
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The correct answer is:
$\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$
We know that $\mathrm{A}^{-1}=\frac{(\mathrm{adjA})}{|\mathrm{A}|}$
or, $\mathrm{AA}^{-1}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
or, $\mathrm{I}_{\mathrm{n}}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
$\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}=|\mathrm{A}|\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$
or, $\mathrm{AA}^{-1}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
or, $\mathrm{I}_{\mathrm{n}}=\frac{\mathrm{A} \cdot \mathrm{Ad}_{\mathrm{j}} \mathrm{A}}{|\mathrm{A}|}$
$\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}=|\mathrm{A}|\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$
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