If \(A =\begin{bmatrix}3&1\\-1&2\end{bmatrix}\), then \(A^2 - 5A\) is equal to (where I is identity matrix of order 2)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-7\, I\)
\(\displaystyle A^2 = \begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}8&5\\-5&3\end{bmatrix}\)
Then
\(\displaystyle A^2 - 5A = \begin{bmatrix}8&5\\-5&3\end{bmatrix} - \begin{bmatrix}15&5\\-5&10\end{bmatrix}\)
\(\displaystyle = \begin{bmatrix}-7&0\\0&-7\end{bmatrix}\)
\(= -7I\)
Shortcut: By the Cayley-Hamilton theorem, every 2 × 2 matrix satisfies
\(A^2 - (\mathrm{tr}\,A)A + |A|\,I = O\)
Here
\(\mathrm{tr}\,A = 3 + 2 = 5\)
\(|A| = 6 + 1 = 7\)
So \(A^2 - 5A + 7I = O\), that is, \(A^2 - 5A = -7I\).
NCERT (2026-27 edition):
Matrices
