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
\( \displaystyle A^2 = \begin{bmatrix}1&3\\2&1\end{bmatrix}\begin{bmatrix}1&3\\2&1\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}1+6&3+3\\2+2&6+1\end{bmatrix} = \begin{bmatrix}7&6\\4&7\end{bmatrix} \)
\( \displaystyle A^2 - 2A = \begin{bmatrix}7&6\\4&7\end{bmatrix} - \begin{bmatrix}2&6\\4&2\end{bmatrix} = \begin{bmatrix}5&0\\0&5\end{bmatrix} \)
\(\det(A^2 - 2A) = 5 \times 5 = 25\)
Shortcut: By the Cayley-Hamilton theorem, \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here \(\mathrm{tr}\,A = 2\) and \(|A| = 1 - 6 = -5\), so \(A^2 - 2A = 5I\), whose determinant is \(5^2 = 25\).