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If matrix $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ is such that $\mathrm{AX}=\mathrm{I}$, where $\mathrm{I}$ is $2 \times 2$ unit matrix, then $X=$
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The correct answer is:
$\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
$\mathrm{AX}=\mathrm{I} \Rightarrow \mathrm{X}=\mathrm{A}^{-1}=\mathrm{A}^{-1}$
$\Rightarrow X=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]^{-1}=\frac{-1}{5}\left[\begin{array}{cc}3 & -2 \\ -4 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
$\Rightarrow X=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]^{-1}=\frac{-1}{5}\left[\begin{array}{cc}3 & -2 \\ -4 & 1\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}-3 & 2 \\ 4 & -1\end{array}\right]$
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