If \( \mathrm{A}=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \), \( \mathrm{I}=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \), then the value of \( k \) so that \( \mathrm{A}^2=8A+k\mathrm{I} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -7
Let us first square A:
\( \displaystyle \mathrm{A}^2=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1+0 & 0+0\\-1-7 & 0+49\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix} \)
\( \displaystyle 8\mathrm{A}=\begin{bmatrix}8 & 0\\-8 & 56\end{bmatrix} \)
\( \displaystyle k\mathrm{I}=\mathrm{A}^2-8\mathrm{A} \)
\( \displaystyle =\begin{bmatrix}1-8 & 0\\-8+8 & 49-56\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-7 & 0\\0 & -7\end{bmatrix} \)
\( \displaystyle =-7\,\mathrm{I} \)
So \( k=-7 \).
Shortcut: By the Cayley-Hamilton theorem,
\( \displaystyle \mathrm{A}^2-(\mathrm{tr}\,\mathrm{A})\mathrm{A}+|\mathrm{A}|\mathrm{I}=\mathrm{O} \).
Here tr A = 1 + 7 = 8 and |A| = 7 − 0 = 7, so
\( \displaystyle \mathrm{A}^2=8\mathrm{A}-7\mathrm{I}\ \Rightarrow\ k=-7 \)