I have this assignment:
Given:
$A = \begin{pmatrix} 2 & 4 \\ 0 & 3 \end{pmatrix}$
$C = \begin {pmatrix} -1 & 2 \\ -6 & 3 \end{pmatrix}$
Find all B that satisfy $AB = C$.
I know that one option is to say $B = \left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) $ and multiply it with $A$. By making each member equal to the one in $C$, I have a system of linear equations which I can solve.
However, I also know that I can set up a system like this:
$ \left( \begin{array} {cc|cc} 2 & 4 & -1 & 2 \\ 0 & 3 & -6 & 3 \end{array} \right)$
If I manipulate it like I would a system of linear equations (for example, by swapping rows, or adding a multiple of a row to another) to get the identity matrix $\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$, then what I'm looking for (matrix $B$) will appear in the right hand side, like this:
$ \left( \begin{array} {cc|cc} 1 & 0 & 7/2 & -1 \\ 0 & 1 & -2 & 1 \end{array} \right)$
In this case, $B = \left( \begin{smallmatrix} 7/2 & -1 \\ -2 & 1\end{smallmatrix} \right)$.
My question is, quite simply, how does this work? It looks like magic to me right now.