If \( A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} \) , then \( A^2-5A \) is equal to : (where \( I_2 \) is identity matrix of order 2)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -7\,I_2 \)
\( \displaystyle A^2=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix} \)
\( \displaystyle A^2-5A=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-7 & 0\\0 & -7\end{bmatrix} \)
\( \displaystyle =-7\,I_2 \)
Shortcut: By the Cayley-Hamilton theorem, every 2 × 2 matrix satisfies
\( A^2-(\mathrm{tr}\,A)A+|A|\,I=O \)
Here tr A = 3 + 2 = 5 and |A| = 6 + 1 = 7, so
\( A^2-5A=-7I_2 \)
NCERT (2026-27 edition):
Matrices
