Question

If $A=\left[\begin{array}{cc}\mathrm{\alpha }& \mathrm{\beta }\\ \mathrm{\gamma }& -\mathrm{\alpha }\end{array}\right]$ is such that A² = I, then
(a) 1 + α² + βγ = 0
(b) 1 − α² + βγ = 0
(c) 1 − α² − βγ = 0
(d) 1 + α² − βγ = 0

Answer 1

(c) 1 − α² − βγ = 0

$$\begin{gathered} \mathrm{Here}, \\ {A}^2=I \\ \Rightarrow \left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}{\alpha }^2+\beta \gamma & \alpha \beta -\beta \alpha \\ \lambda \alpha -\alpha \gamma & \gamma \beta +{\alpha }^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}{\alpha }^2+\beta \gamma & 0\\ 0& \gamma \beta +{\alpha }^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{equal}. \\ \Rightarrow {\alpha }^2+\beta \gamma =1 \\ \Rightarrow 1-{\alpha }^2-\beta \gamma =0 \end{gathered}$$