Question

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

• A. $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

$A=\left[\begin{array}{cc}x& 1\\ 1& 0\end{array}\right]$

Given: $A^2=I$ where $I$ is $2\times 2$ identity matrix

Let us find $A^2$

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

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

Given $A^2=I$

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

Equating,we get

$x^2+x=1,x=0$

Put $x=0$ in $A$

$A=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

We have $A^2=I$

Pre-multiply $A^{-1}$ both sides,we get

$A^{-1}{A}^2=A^{-1}I$

$\Rightarrow A=A^{-1}$

Hence,$A^{-1}=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$