Question

If $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$ and $A^2=I,$ then find $x.$

Answer 1

Given $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]and{A}^2=I$

Consider, $A^2=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}{x}^2+1& x+0\\ x+0& 1+0\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right]$

Given, $A^2=1$

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

On comparing elements, we get

$x^2+1=1$

$\Rightarrow x^2=0$

$\Rightarrow x=0.$