Finding the inverse of a matrix given an equation

Question

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?

Answer 1

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 $$BC^{-1}=\begin{bmatrix} 0&2&3&1\\ 2&3&1&5\\ 0&2&-3&2\\ 1&0&3&1 \end{bmatrix}.$$

Answer 2

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 $A$.