Solve this

Question:

If $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$ is such that $A^{2}=1$, then

(a) $1+\alpha^{2}+\beta y=0$

(b) $1-\alpha^{2}+\beta y=0$

(c) $1-\alpha^{2}-\beta y=0$

(d) $1+\alpha^{2}-\beta y=0$

Solution:

(c) $1-\alpha^{2}-\beta y=0$

Here,

$A^{2}=I$

$\Rightarrow\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}\alpha^{2}+\beta \gamma & \alpha \beta-\beta \alpha \\ \lambda \alpha-\alpha \gamma & \gamma \beta+\alpha^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}\alpha^{2}+\beta \gamma & 0 \\ 0 & \gamma \beta+\alpha^{2}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

The corresponding elements of two equal matrices are equal.

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

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