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

MCQ

If $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ 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 $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ and $A^2 = I,$ then
$\text{A}^2=\text{I}$
$\Rightarrow\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\alpha^2+\beta\gamma=1$
$\Rightarrow1-\alpha^2-\beta\gamma=0$

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