Two Variables
Let’s begin with the equations:
\begin{equation} \begin{cases} 2x+y = 3 \\ x - y = 0 \end{cases} \end{equation}
We will first change this into a matrix equation:
\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \end{equation}
We need to find, then, the inverse of:
\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}
Namely, we need the matrix such that:
\begin{equation} M \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = I \end{equation}
To do this, we can use row operations on both sides such that the left side becomes the identity, we are essentially inverting the process of reversing a matrix.
\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation}
Let’s begin:
\begin{align} & \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \Rightarrow\ & \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \\ \Rightarrow\ & \begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & -\frac{2}{3} \\ 0 & 1 \end{pmatrix} \\ \Rightarrow\ & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & -\frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} \end{pmatrix} \\ \Rightarrow\ & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix} \end{align}
Finally, then, we will applying this matrix to the input:
\begin{align} \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix} \begin{pmatrix} 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \end{align}
Three Variables
We do this again, but now with a much larger matrix. Namely:
\begin{equation} \begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix} \end{equation}
I spend a good two hours (yes) trying to invert this. At this point, I know its invertable but I keep making mistakes. However, a solution exists and it is of shape:
\begin{equation} \begin{pmatrix} \frac{1}{5} & \frac{1}{5} & \frac{2}{5} \\ \frac{1}{5} & \frac{1}{5} & -\frac{3}{5} \\ \frac{2}{5} & -\frac{3}{5} & \frac{4}{5} \end{pmatrix} \end{equation}
And, applying the output, we have that:
\begin{equation} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \end{equation}
So complicated of an inverse, for such a simple result…
Matrix Multiplication
Matrix multiplication is not commutative. While you can, for instance, multiply a \(2\times 3\) by a \(3\times 3\), we cannot do it the other way.
For an equation with three variables, you need three equations at a minimum to have at least one solution; you can get at most the number of equations number of solutions with fewer equations. You probably will have no solutions if you have more equations—the result is likely to be overdetermined; of course, two equations may be the same relation then in which case one is effectively nulled.