Question
If matrix $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I,$ then $X=$

Answer

We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I$
$ \Rightarrow X=A^{-1} I$
$ \because |A|=3-8=-5 \neq 0 \Rightarrow A^{-1} \text { exists. } $
$A^{-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]  $
$ \therefore X=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\
0 & 1 \end{array}\right]=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right] $

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