If \( A=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix},I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) and \( A^2=\alpha A+\beta I \) , for some constant \( \alpha \) and \( \beta \) then the value of \( \alpha \) and \( \beta \) respectively:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4, −1
\( \displaystyle A^2=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix} \)
\( \displaystyle \alpha A+\beta I=\begin{bmatrix}2\alpha+\beta & 3\alpha\\\alpha & 2\alpha+\beta\end{bmatrix} \)
Comparing corresponding entries:
\( \displaystyle 3\alpha=12\ \Rightarrow\ \alpha=4 \)
\( \displaystyle 2\alpha+\beta=7 \)
\( \displaystyle 8+\beta=7\ \Rightarrow\ \beta=-1 \)
Shortcut: By the Cayley-Hamilton theorem for a 2 × 2 matrix,
\( \displaystyle A^2-(\mathrm{tr}\,A)A+|A|\,I=O \)
Here tr A = 4 and |A| = 4 − 3 = 1, so \( A^2=4A-I \), giving \( \alpha=4,\ \beta=-1 \).
NCERT (2026-27 edition):
Adjoint and Inverse of a Matrix
