Question

If for the matrix, $A=\left[\begin{array}{cc}1& -\alpha \\ \alpha & \beta \end{array}\right],A{A}^T=I_2,$ then the value of ${\alpha }^4+{\beta }^4$ is:

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

Given, $A{A}^T=I_2$
$\left[\begin{array}{cc}1& -\alpha \\ \alpha & \beta \end{array}\right]\left[\begin{array}{cc}1& \alpha \\ -\alpha & \beta \end{array}\right]=\left[\begin{array}{cc}1+{\alpha }^2& \alpha -\alpha \beta \\ \alpha -\alpha \beta & {\alpha }^2+{\beta }^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

So, $1+{\alpha }^2=1\Rightarrow {\alpha }^2=0,$
${\alpha }^2+{\beta }^2=1\Rightarrow {\beta }^2=1$
$\Rightarrow {\alpha }^4=0,{\beta }^4=1$
$\therefore {\alpha }^4+{\beta }^4=1$