If \( A=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \) is a matrix such that \( A^2=8A+KI \), where I is an identity matrix of order 2 then value of K is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −7
Squaring A and comparing entries with 8A + KI:
\( \displaystyle A^2=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-1-7 & 49\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix} \)
\( \displaystyle 8A+KI=\begin{bmatrix}8+K & 0\\-8 & 56+K\end{bmatrix} \)
Comparing the (1, 1) entries: 8 + K = 1, so K = −7. The (2, 2) entries agree: 56 − 7 = 49.
Shortcut: By the Cayley-Hamilton theorem for a 2 × 2 matrix,
\( A^2-(\mathrm{tr}\,A)A+|A|\,I=O \).
Here tr A = 1 + 7 = 8 and |A| = 7 − 0 = 7, so
\( A^2=8A-7I \) and K = −7.
Why the other options are incorrect
Option (1) 7 — Incorrect. K = 7 has the wrong sign; in \( A^2=(\mathrm{tr}\,A)A-|A|\,I \) the determinant term is subtracted.
Option (2) 3 — Incorrect. With K = 3 the (1, 1) entry of 8A + KI would be 11, not 1.
Option (4) −2 — Incorrect. With K = −2 the (1, 1) entry of 8A + KI would be 6, not 1.