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Choose the correct answer in the following questions:

    Q13.    If A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix} is such that A^2 = I

                (A)    1 + \alpha^2 + \beta \gamma = 0

                (B)    1 - \alpha^2 + \beta \gamma = 0

                (C)    1 - \alpha^2 - \beta \gamma = 0

                (D)    1 + \alpha^2 - \beta \gamma = 0

Answers (1)

best_answer

A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}

A^2 = I

\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}

\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}

\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}

Thus we obtained that

\alpha^{2} +\beta \gamma=1

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

Option C is correct.

 

 

 

 

Posted by

seema garhwal

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