If \(A = \begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\) is such that \(A^2 = 4I\), where I is the identity matrix of order 2, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\alpha^2 + \beta\gamma = 4\)
Multiplying row by column:
\( \displaystyle A^2 = \begin{bmatrix}\alpha^2 + \beta\gamma&\alpha\beta - \alpha\beta\\\gamma\alpha - \alpha\gamma&\gamma\beta + \alpha^2\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}\alpha^2 + \beta\gamma&0\\0&\alpha^2 + \beta\gamma\end{bmatrix} \)
Comparing with \(4I = \begin{bmatrix}4&0\\0&4\end{bmatrix}\) by equality of matrices:
\(\alpha^2 + \beta\gamma = 4\)
Why the other options are incorrect
Option (1) \(\alpha^2 - \beta\gamma = 4\) — Incorrect. The diagonal element of \(A^2\) is \(\alpha \cdot \alpha + \beta \cdot \gamma\), with a plus sign.
Option (2) \(\alpha^2 + \beta\gamma = 2\) — Incorrect. With \(\alpha^2 + \beta\gamma = 2\), the matrix \(A^2\) is 2I, not 4I.
Option (3) \(\alpha^2 + \beta\gamma = -4\) — Incorrect. This would give \(A^2 = -4I\).