If \( A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} \) and \( I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) such that \( A^2-A+\lambda I=0 \), then value of ‘\( \lambda \)’ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
Here 0 on the right is the zero matrix O of order 2 × 2.
\( \displaystyle A^2=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} \)
\( \displaystyle A^2-A=\begin{bmatrix}1-3 & -2+2\\4-4 & -4+2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix} \)
\( \displaystyle =-2I \)
\( \displaystyle -2I+\lambda I=O\ \Rightarrow\ \lambda=2 \)
Shortcut: By the Cayley-Hamilton theorem for a 2 × 2 matrix, \( A^2-(\mathrm{tr}\,A)A+|A|I=O \). Here tr A = 3 + (−2) = 1 and |A| = −6 + 8 = 2, so \( A^2-A+2I=O \) and λ = |A| = 2.