If $A=\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\end{array}\right]$ and $I=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$, then $A^2-6\mathrm{A}$ is equal to

${\mathrm{A}}^2=\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\end{array}\right]\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\end{array}\right]=\left[\begin{array}{l}19\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\\ 18\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\end{array}\right]$

So, ${\mathrm{A}}^2-6\mathrm{A}=\left[\begin{array}{l}19\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\\ 18\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\end{array}\right]-6\left[\begin{array}{l}4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\end{array}\right]$

$=\left[\begin{array}{l}19\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\\ 18\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\end{array}\right]-\left[\begin{array}{r}24\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\\ 18\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}12\end{array}\right]=\left[\begin{array}{r}-5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-5\end{array}\right]$

$=-5\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=-5I$

$\therefore$Option (b) is the correct answer.