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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