If A= [ cc & & - ] is such that A^2 = I, then

Question

If $A=\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$ is such that $A^2 = I, $ then

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Answer

In the given question: $A=\left[\begin{array}{ll} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$
Calculating $A^2: A^2 = A.A$
= $\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]\left[\begin{array}{cc} {\alpha} & {\beta} \\ {\gamma} & {-\alpha} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha \cdot \alpha+\beta \cdot \gamma} & {\alpha \cdot \beta+\beta \cdot(-\alpha)} \\ {\gamma \cdot \alpha+(-\alpha) \cdot \gamma} & {\gamma \cdot \beta+(-\alpha) \cdot(-\alpha)} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {\alpha \cdot \beta-\alpha \cdot \beta} \\ {\gamma \cdot \alpha-\gamma \cdot \alpha} & {\gamma \cdot \beta+\alpha^{2}} \end{array}\right]$
= $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {0} \\ {0} & {\beta \gamma+\alpha^{2}} \end{array}\right]$
And given that $A^2 = I$
Therefore, $\left[\begin{array}{cc} {\alpha^{2}+\beta \gamma} & {0} \\ {0} & {\beta \gamma+\alpha^{2}} \end{array}\right]=\left[\begin{array}{ll} {1} & {0} \\ {0} & {1} \end{array}\right]$
Comparing corresponding elements we obtain
$\Rightarrow \alpha^{2}+\beta \gamma=1$
$\Rightarrow 1-\alpha^{2}-\beta \gamma=0$

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