Find the inverse of the following matrix: $A$ Find the inverseof the following matrix:
$$A= \begin{bmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&1  \\
0 & 0 & 1 & 0 &0 \\0&0&0&1&0 
\end{bmatrix}$$
My attempt :  I think the inverse  will not exist because $\det A=0 $
Is it True ?
Any hints/solution will be appreciated. 

Interchange rows three times ($R_1-R_2, R_3-R_4,R_4-R_5$):
$$\det(A)= \begin{vmatrix}
0 & \color{blue}1 & 0 & 0 &0 \\
\color{red}1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&\color{green}1  \\
0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0 
\end{vmatrix}=(-1)^3\begin{vmatrix}
\color{red}1 & 0 & 0 & 0 &0 \\
0 & \color{blue}1 & 0 & 0 &0 \\
0 & 0 & \color{brown}1 & 0& 0  \\
0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1 
\end{vmatrix}=-1\ne 0.$$
Consider the expanded matrix and perform the same row interchanges:
$$\begin{vmatrix}
0 & \color{blue}1 & 0 & 0 &0 \\
\color{red}1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0&\color{green}1  \\
0 & 0 & \color{brown}1 & 0 &0 \\0&0&0&\color{purple}1&0 
\end{vmatrix}\begin{vmatrix}
1 & 0 & 0 & 0 &0 \\
0 & 1 & 0 & 0 &0 \\
0 & 0 & 1 & 0&0  \\
0 & 0 & 0 & 1 &0 \\0&0&0&0&0 
\end{vmatrix} \Rightarrow \begin{vmatrix}
\color{red}1 & 0 & 0 & 0 &0 \\
0 & \color{blue}1 & 0 & 0 &0 \\
0 & 0 & \color{brown}1 & 0& 0  \\
0 & 0 & 0 & \color{purple}1&0 \\0&0&0&0&\color{green}1 
\end{vmatrix}\begin{vmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 1&0  \\
0 & 0 & 0 & 0 &1 \\
0 & 0 & 1 & 0 &0 
\end{vmatrix} \ \ \ \ $$
Hence, the inverse matrix is:
$$A^{-1}=\begin{vmatrix}
0 & 1 & 0 & 0 &0 \\
1 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 1&0  \\
0 & 0 & 0 & 0 &1 \\
0 & 0 & 1 & 0 &0 
\end{vmatrix}$$
