Finding the inverse of a matrix given an equation So I've been given this equation:
$A\begin{bmatrix}
2&3&1&5\\
1&0&3&1\\
0&2&-3&2\\
0&2&3&1
\end{bmatrix} = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$
I'm supposed to find the inverse of A using this but I'm not really sure where to start. Any ideas?
 A: If $AB=C$, then $A^{-1}=BC^{-1}$.
Now if we interpret the right-hand side matrix as a change of basis matrix, such that
$$u_1=e_4,\enspace u_2=e_1,\enspace u_3=e_3, \enspace u_4=e_2,$$
then, conversely
$$ e_1=u_2,\enspace e_2=u_4,\enspace e_3=u_3,\enspace e_4=u_1,$$
so that
$$\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}^{-1}=\begin{bmatrix}
0&0&0&1\\
1&0&0&0\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}$$
and finally
$$BC^{-1}=\begin{bmatrix}
0&2&3&1\\
2&3&1&5\\
0&2&-3&2\\
1&0&3&1
\end{bmatrix}.$$
A: Bernard's answer above gives a lot of intuition about what's really going on, but in case you are not yet familiar with some of the terminology here is a more brutish answer. 
The inverse of the right hand matrix can be calculated by examination. Notice that there is only one  non-zero entry per column and row of the matrix, and these are all $1$. This has the effect that every element in a matrix multiplied with this right hand matrix appears exactly once in the product as such
$~\\ \left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 \\
 \end{array} \right)
\cdot 
\left( \begin{array}{cccc}
a_{11} & a_{12} & a_{13} & a_{14}\\
a_{21} & a_{22} & a_{23} & a_{24}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{41} & a_{42} & a_{43} & a_{44}\\
 \end{array} \right)
= 
\left( \begin{array}{cccc}
a_{21} & a_{22} & a_{23} & a_{24}\\
a_{41} & a_{42} & a_{43} & a_{44}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{11} & a_{12} & a_{13} & a_{14}\\
 \end{array} \right)\\$
So to get the identity matrix we choose $a_{21} = a_{42} = a_{33} = a_{14} = 1$ and all other entries to be zero. Then we have computed the inverse of the right hand side matrix as 
$~\\
\left( \begin{array}{cccc}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0 \\
 \end{array} \right)\\$
Hence if we multiply both sides of your equation on the right by the above matrix, on the left side we get $A$ multiplied by the product of two matrices and on the right side we get the identity matrix. Hence we have found the inverse of $A$.
