If the matrix $A = \begin{bmatrix}1&3\\2&1\end{bmatrix}$, then the value of $det (A^2 - 2A)$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 25
$A=\begin{pmatrix}1&3\\2&1\end{pmatrix}$
$A^2=\begin{pmatrix}1&3\\2&1\end{pmatrix}\begin{pmatrix}1&3\\2&1\end{pmatrix}$
$A^2=\begin{pmatrix}1+6 & 3+3\\ 2+2 & 6+1\end{pmatrix} =\begin{pmatrix}7&6\\4&7\end{pmatrix}$
$A^2-2A=\begin{pmatrix}7&6\\4&7\end{pmatrix}-\begin{pmatrix}2&6\\4&2\end{pmatrix} =\begin{pmatrix}5&0\\0&5\end{pmatrix}$
$\det(A^2-2A)=5\cdot5=25$
$\det(A^2-2A)=25$