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

$1$

Explanation for the correct option:

Step 1. Find the value of ${\alpha }^4+{\beta }^4$:

Given, $A=\left[\begin{array}{cc}1& -\alpha \\ \alpha & \beta \end{array}\right]$

then, $A^T=\left[\begin{array}{cc}1& \alpha \\ -\alpha & \beta \end{array}\right]$

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

Step 2. Compare the corresponding values with identity matrix:

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

${\alpha }^2=0$

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

${\beta }^2=1$

$\therefore {\alpha }^4=0$ and ${\beta }^4=1$

Thus, ${\mathit{\alpha }}^{\mathbf{4}}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{4}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘A’ is Correct.