Question

The value of det\((A^2-2A)\),If \(A=\begin{pmatrix}      1 & 3  \\[0.3em]    2  &1  \end{pmatrix}\),is

• A. \(5\)
• B. \(-5\)
• C. \(25\)
• D. \(-25\)

Answer 1

The Correct Option is C

To find the determinant of \(A^2-2A\) for the matrix \(A=\begin{pmatrix}1 & 3\\2 & 1\end{pmatrix}\), follow these steps:

1. First, compute \(A^2\):
   \[\begin{pmatrix}1 & 3\\2 & 1\end{pmatrix}\begin{pmatrix}1 & 3\\2 & 1\end{pmatrix}=\begin{pmatrix}(1\cdot1+3\cdot2) & (1\cdot3+3\cdot1)\\(2\cdot1+1\cdot2) & (2\cdot3+1\cdot1)\end{pmatrix}=\begin{pmatrix}7 & 6\\4 & 7\end{pmatrix}\]
2. Calculate \(2A\):
   \[(2)\begin{pmatrix}1 & 3\\2 & 1\end{pmatrix}=\begin{pmatrix}2 & 6\\4 & 2\end{pmatrix}\]
3. Find \(A^2-2A\):
   \[\begin{pmatrix}7 & 6\\4 & 7\end{pmatrix}-\begin{pmatrix}2 & 6\\4 & 2\end{pmatrix}=\begin{pmatrix}7-2 & 6-6\\4-4 & 7-2\end{pmatrix}=\begin{pmatrix}5 & 0\\0 & 5\end{pmatrix}\]
4. Determine the determinant of \(A^2-2A\):
   The determinant of \(\begin{pmatrix}a & b\\c & d\end{pmatrix}\) is \(ad-bc\). Thus, for \(\begin{pmatrix}5 & 0\\0 & 5\end{pmatrix}\):
   \(5\cdot5 - 0\cdot0 = 25\)

The value of \(\det(A^2-2A)\) is \(\boxed{25}\).