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If $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$, then what is $A(\operatorname{adj} A)$ equal to?
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The correct answer is:
$\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
Let $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$
We have If $A$ is a square matrix of order $n$ then $A(\operatorname{adj} A)=|A| . \mathrm{I}_{\mathrm{n}}$
Here, $n=2$ $\therefore A(\operatorname{adj} A)=\mathrm{I}_{2}|A|$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left|\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right|=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](12-2)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}10 & 0 \\ 0 & 10\end{array}\right]$
We have If $A$ is a square matrix of order $n$ then $A(\operatorname{adj} A)=|A| . \mathrm{I}_{\mathrm{n}}$
Here, $n=2$ $\therefore A(\operatorname{adj} A)=\mathrm{I}_{2}|A|$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left|\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right|=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](12-2)=10\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}10 & 0 \\ 0 & 10\end{array}\right]$
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