If \( A=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix},\mathrm{I}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) and \( \mathrm{A}^2=8\mathrm{A}+k\mathrm{I} \), then \( k \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -7
\( \displaystyle A^2=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix} \)
\( \displaystyle 8A+kI=\begin{bmatrix}8+k & 0\\-8 & 56+k\end{bmatrix} \)
Comparing the (1, 1) entries: 1 = 8 + k, so k = −7. The (2, 2) entries give 49 = 56 + k, which agrees.
Shortcut: For a 2 × 2 matrix,
\( A^2-(\mathrm{tr}\,A)A+|A|\,I=O \)
Here tr A = 8 and |A| = 7, so
\( A^2=8A-7I \)