Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Choose the correct answer in the following questions:

Question:

Choose the correct answer in the following questions:

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

A. $1+\alpha^{2}+\beta \gamma=0$

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

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

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

Solution:

$A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$

$\therefore A^{2}=A \cdot A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]$

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

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

Now, $A^{2}=I \Rightarrow\left[\begin{array}{lc}\alpha^{2}+\beta \gamma & 0 \\ 0 & \beta \gamma+\alpha^{2}\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

On comparing the corresponding elements, we have:

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

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

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