If for A = begin{bmatrix} alpha & beta gamma | vedclass

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(C) Given the matrix $A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$.We are given that $A^2 = I$,where $I$ is the identity matr

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

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(C) Given the matrix $A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$.We are given that $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.First,calculate $A^2$:$A^2 = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & \alpha \beta - \beta \alpha \ \gamma \alpha - \alpha \gamma & \beta \gamma + \alpha^2 \end{bmatrix}$$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & 0 \ 0 & \alpha^2 + \beta \gamma \end{bmatrix}$Since $A^2 = I$,we have:$\begin{bmatrix} \alpha^2 + \beta \gamma & 0 \ 0 & \alpha^2 + \beta \gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$Comparing the elements,we get $\alpha^2 + \beta \gamma = 1$.Rearranging the terms,we get $1 - \alpha^2 - \beta \gamma = 0$.Therefore,the correct option is $C$.

If for $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$,$A^2 = I$,then . . . . . . .

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