Question

If $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$ & $A^2$ is the identity matrix, Then $x=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B

$0$

Explanation for the correct option:

Finding the value of X:

Given, $A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$ & $A^2$ is the identity matrix

Now,

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

But $A^2=I$

Therefore,

$\begin{array}{rcl}\left[\begin{array}{cc}{x}^2+1& x\\ x& 1\end{array}\right]& =& \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right][\sin ce\mathrm{I}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]]\\ \therefore x& =& 0\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$.