Different ways of solving simultaneous equations Given \begin{align} x-2y &= 4\tag1\label1 \\ 2x+3y &= 1 \tag2\label2 \end{align}
One method I know of solving for x and y: elimination by making coefficients the same 
$2 \times \eqref{1}:$
$$2x-4y=8\tag3\label3$$
$\eqref{3}-\eqref{2}:$
$$(2x-2x)-4y-3y=8-1\tag4\label4$$
$$-7y=7$$
$$y=-1$$
Sub $y=-1$ into $\eqref{1}$ then 
$$x=2$$

What are other ways of solving these simultaneous equations? And can you demonstrate your way also.

 A: Matrices can be used to re-write $n$ simultaneous equations with $n$ unknowns. 
Let's take your example:
$$x-2y=4$$
$$2x+3y=1$$
These can be re-written in matrix form as follows:
$$\begin{bmatrix}
    1 & -2 \\
    2 & 3 \\
    \end{bmatrix} 
\begin{bmatrix}
x\\
y\\
\end{bmatrix}
=
\begin{bmatrix}
4\\
1\\
\end{bmatrix}
$$
Next, we can pre-multiply both sides by the inverse of the matrix $\begin{bmatrix}
    1 & -2 \\
    2 & 3 \\
    \end{bmatrix}$ 
leaving 
$\begin{bmatrix}
x\\
y\\
\end{bmatrix}$ on its own on the LHS. This is utilizing the rule that any matrix $A$ when multiplied by its inverse $A^{-1}$ results in a matrix known as the identity matrix, which in the case of 2x2 matrices is $\begin{bmatrix}
    1 & 0 \\
    0 & 1 \\
    \end{bmatrix}$  . The identity matrix has similar properties to that of the number $1$, where if you multiply something by it, the result is the same as the value you are multiplying.
We are now left with the follow result:
$$
\begin{bmatrix}
x\\
y\\
\end{bmatrix}
=
\begin{bmatrix}
    1 & -2 \\
    2 & 3 \\
    \end{bmatrix}^{-1} 
\begin{bmatrix}
4\\
1\\
\end{bmatrix}
$$
This can then be evaluated to give the answer:
$$
\begin{bmatrix}
x\\
y\\
\end{bmatrix}
=
\begin{bmatrix}
2\\
-1\\
\end{bmatrix}
$$
How to find the inverse of a 2x2 matrix
