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If $2 A=\left(\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right)$, then what is $A^{-1}$ equal to?
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Given, $2 \mathrm{~A}=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]$
$\mathrm{A}=\left(\begin{array}{cc}1 & \frac{1}{2} \\ \frac{3}{2} & 1\end{array}\right)$
$|\mathrm{A}|=\frac{1}{4}$
$\operatorname{adj} \mathrm{A}=\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)$
$\mathrm{A}^{-1}=4\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)=\left(\begin{array}{cc}4 & -2 \\ -6 & 4\end{array}\right)$
$\mathrm{A}=\left(\begin{array}{cc}1 & \frac{1}{2} \\ \frac{3}{2} & 1\end{array}\right)$
$|\mathrm{A}|=\frac{1}{4}$
$\operatorname{adj} \mathrm{A}=\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)$
$\mathrm{A}^{-1}=4\left(\begin{array}{cc}1 & -\frac{1}{2} \\ \frac{-3}{2} & 1\end{array}\right)=\left(\begin{array}{cc}4 & -2 \\ -6 & 4\end{array}\right)$
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