If \(A=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(A^2 = KA - 2I\), then the value of K is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1
\( \displaystyle A^2 = \begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix} = \begin{bmatrix}1&-2\\4&-4\end{bmatrix} \)
\( \displaystyle KA - 2I = \begin{bmatrix}3K-2&-2K\\4K&-2K-2\end{bmatrix} \)
Comparing the corresponding entries:
\(3K - 2 = 1\)
\(K = 1\)
The other entries agree with K = 1: \(-2K = -2\), \(4K = 4\), \(-2K - 2 = -4\).
Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here \(\mathrm{tr}\,A = 3 - 2 = 1\) and \(|A| = -6 + 8 = 2\), so \(A^2 = A - 2I\) and K = 1.
NCERT (2026-27 edition):
Operations on Matrices
