If \( A=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} \) and \( I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \), then for what value of \( k \), \( A^2=5A+kI \) ?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -7
Squaring A:
\( \displaystyle A^2=\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix} \)
The right side is
\( \displaystyle 5A+kI=\begin{bmatrix}15+k & 5\\-5 & 10+k\end{bmatrix} \)
Equating the (1, 1) entries:
\( \displaystyle 8=15+k \)
\( \displaystyle k=-7 \)
The (2, 2) entries agree: 10 + (−7) = 3.
Shortcut: By the Cayley–Hamilton result for a 2 × 2 matrix,
\( \displaystyle A^2-(\mathrm{tr}\,A)A+|A|\,I=O \)
Here tr A = 3 + 2 = 5 and |A| = 6 + 1 = 7, so
\( \displaystyle A^2=5A-7I \)
which gives k = −7.
Why the other options are incorrect
Option (1) 5 — Incorrect. With k = 5 the (1, 1) entry of 5A + kI is 20, not 8.
Option (2) -5 — Incorrect. With k = −5 the (1, 1) entry of 5A + kI is 10, not 8.
Option (3) 7 — Incorrect. This is |A| = 7 with the wrong sign: \( A^2=5A-|A|\,I \), so k = −7.
NCERT (2026-27 edition):
Matrices
