The value of det(A^2 - 2A) , if A = […] , is

The correct answer is Option (3) → 25

Multiplying row by column:

    \( A^2 = \begin{pmatrix} 1 & 3\\ 2 & 1\end{pmatrix}\begin{pmatrix} 1 & 3\\ 2 & 1\end{pmatrix} \)

    \( = \begin{pmatrix} 1 + 6 & 3 + 3\\ 2 + 2 & 6 + 1\end{pmatrix} = \begin{pmatrix} 7 & 6\\ 4 & 7\end{pmatrix} \)

    \( 2A = \begin{pmatrix} 2 & 6\\ 4 & 2\end{pmatrix} \)

    \( A^2 - 2A = \begin{pmatrix} 5 & 0\\ 0 & 5\end{pmatrix} \)

So

    \(\det(A^2 - 2A) = 5 \times 5 - 0 = 25\)

Shortcut: \(A^2 - 2A = A(A - 2I)\), and the determinant of a product is the product of the determinants. Here \(|A| = 1 - 6 = -5\) and

    \( |A - 2I| = \begin{vmatrix} -1 & 3\\ 2 & -1\end{vmatrix} = 1 - 6 = -5 \)

so the value is \((-5)(-5) = 25\).