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(B) Given: $A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$.Then,$A^{2} = A \cdot A = \begin{bmatrix} \alpha & \beta \ \gamma &

Vedclass · Accepted Answer

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

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(B) Given: $A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$.Then,$A^{2} = A \cdot A = \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \ \gamma & -\alpha \end{bmatrix}$.Calculating the product:$A^{2} = \begin{bmatrix} \alpha^{2} + \beta \gamma & \alpha \beta - \alpha \beta \ \alpha \gamma - \alpha \gamma & \beta \gamma + \alpha^{2} \end{bmatrix} = \begin{bmatrix} \alpha^{2} + \beta \gamma & 0 \ 0 & \alpha^{2} + \beta \gamma \end{bmatrix}$.Given $A^{2} = I$,where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,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 corresponding elements:$\alpha^{2} + \beta \gamma = 1$.Rearranging the equation:$1 - \alpha^{2} - \beta \gamma = 0$.Thus,the correct option is $B$.

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

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