If \( A^2=8A+kI,A=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix},I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) then value of \( k \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -7
Let us compute \( A^2 \) and compare it with \( 8A+kI \):
\( \displaystyle A^2=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-1-7 & 49\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix} \)
\( \displaystyle 8A+kI=\begin{bmatrix}8+k & 0\\-8 & 56+k\end{bmatrix} \)
Equating the (1, 1) entries: \( 1=8+k \), so k = −7. The (2, 2) entries agree: 56 − 7 = 49.
Shortcut: By the Cayley-Hamilton result for a 2 × 2 matrix, \( A^2-(\text{tr}A)A+|A|\mathrm{I}=\mathrm{O} \). Here tr A = 1 + 7 = 8 and |A| = 7 − 0 = 7, so
\( \displaystyle A^2=8A-7\mathrm{I}\ \Rightarrow\ k=-7 \)