If \(A = \begin{bmatrix}3&1\\2&2\end{bmatrix}\) and \(A^2 = kA - 4I\), where I is the identity matrix of order 2, then the value of \(k\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
Squaring A by row-by-column multiplication:
\( \displaystyle A^2 = \begin{bmatrix}9 + 2&3 + 2\\6 + 4&2 + 4\end{bmatrix} = \begin{bmatrix}11&5\\10&6\end{bmatrix} \)
The right side is
\( \displaystyle kA - 4I = \begin{bmatrix}3k - 4&k\\2k&2k - 4\end{bmatrix} \)
Comparing the elements in the first row, second column:
\(k = 5\)
The other elements agree: \(3(5) - 4 = 11\), \(2(5) = 10\), \(2(5) - 4 = 6\).
Shortcut: by the Cayley-Hamilton theorem, a \(2 \times 2\) matrix satisfies \(A^2 - (\text{tr}\,A)A + |A|\,I = O\). Here tr A = 3 + 2 = 5 and |A| = 6 − 2 = 4, so \(A^2 = 5A - 4I\) and k = 5.