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Question: Answered & Verified by Expert
Choose the correct answer in the following questions:
If $\mathbf{A}=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ is such that $\mathbf{A}^2=I$, then
(a) $1+\alpha^2+\beta \gamma=0$
(b) $1-\alpha^2+\beta \gamma=0$
(c) $1-\alpha^2-\beta \gamma=0$
(d) $1+\alpha^2-\beta \gamma=0$
MathematicsMatrices
Solution:
1480 Upvotes Verified Answer
$$
\begin{aligned}
&\mathrm{A}^2=\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right] \\
&=\left[\begin{array}{cc}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}
$$
$\therefore \mathrm{A}^2=\mathrm{I} \quad \alpha^2+\beta \gamma=1 \quad$ or $\quad 1-\alpha^2-\beta \gamma=0$
Option (c) is correct.

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