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) → $-7\, I$
Given:
$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
Compute $A^2$:
$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (9-1) & (3+2) \\ (-3-2) & (-1+4) \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
Now compute $A^2 - 5A$:
$A^2 - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5+5 & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$
$A^2 - 5A = -7I$
Hence, $A^2 - 5A = -7I$