If $A =\begin{bmatrix}4&5\\2&1\end{bmatrix}$ and $I$ is an identity matrix of order 2, then $A - 3I$ equals

$2 (I + 3A^{-1})$

The correct answer is Option (4) → $2 (I + 3A^{-1})$

$A=\begin{pmatrix}4&5\\2&1\end{pmatrix},\; I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$

$A-3I=\begin{pmatrix}4-3&5\\2&1-3\end{pmatrix}$

$A-3I=\begin{pmatrix}1&5\\2&-2\end{pmatrix}$

$A^{-1}=\frac{1}{(4)(1)-(5)(2)}\begin{pmatrix}1&-5\\-2&4\end{pmatrix}$

$A^{-1}=\frac{1}{4-10}\begin{pmatrix}1&-5\\-2&4\end{pmatrix}$

$A^{-1}=-\frac{1}{6}\begin{pmatrix}1&-5\\-2&4\end{pmatrix}$

$I+3A^{-1} =I+3\left(-\frac{1}{6}\begin{pmatrix}1&-5\\-2&4\end{pmatrix}\right)$

$I+3A^{-1} =\begin{pmatrix}1&0\\0&1\end{pmatrix} -\frac{1}{2}\begin{pmatrix}1&-5\\-2&4\end{pmatrix}$

$I+3A^{-1} =\begin{pmatrix}1-\frac12 & \frac52 \\ 1 & 1-2\end{pmatrix}$

$I+3A^{-1} =\begin{pmatrix}\frac12 & \frac52 \\ 1 & -1\end{pmatrix}$

$2(I+3A^{-1}) =\begin{pmatrix}1 & 5 \\ 2 & -2\end{pmatrix}$

$A-3I=2(I+3A^{-1})$

Thus, the required expression is $2(I+3A^{-1})$.