If A = & &- is such that A^2 = I, then: - Vidyadip

MCQ

If $A = \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}$is such that $A^2 = I,$ then:

A
$1 + \alpha^2 + \beta\gamma = 0$
B
$1 - \alpha^2 + \beta\gamma = 0$
✓
$1 - \alpha^2 - \beta\gamma = 0$
D
$1 + \alpha^2 - \beta\gamma = 0$

✓

Answer

Correct option: C.

$1 - \alpha^2 - \beta\gamma = 0$

Given: $\text{A}=\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix} \text{and}\ \text{A}^{2}=\text{I}$
$\Rightarrow\ \begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\ \gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}\alpha^{2}+\beta\gamma&\alpha\beta-\alpha\beta\\ \alpha\gamma-\gamma\alpha&\beta\gamma+\alpha^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}\alpha^{2}+\beta\gamma&0\\0&\beta\gamma+\alpha^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
Equating corresponding entries, we have
$\alpha^{2}+\beta\gamma=1$
$\Rightarrow 1-\alpha^{2}-\beta\gamma=0$
Therefore, option $(C)$ is correct.

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