diff --git a/mechanical/mmme1026_maths_for_engineering.md b/mechanical/mmme1026_maths_for_engineering.md
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+++ b/mechanical/mmme1026_maths_for_engineering.md
@@ -390,3 +390,109 @@ $$z^3 = 8i$$
 > 4. There are three solutions
 
 </details>
+
+# Matrices (and Simultaneous Equations)
+
+## Gaussian Elimination
+
+Gaussian eliminiation can be used when the number of unknown variables you have is equal to the
+number of equations you are given.
+
+I'm pretty sure it's the name for the method you use to solve simultaneous equations in school.
+
+For example if you have 1 equation and 1 unknown:
+
+\begin{align*}
+2x &= 6 \\
+x &= 3
+\end{align*}
+
+### Number of Solutions
+
+Let's generalise the example above to
+
+$$ax = b$$
+
+There are 3 possible cases:
+
+\begin{align*}
+a \ne 0 &\rightarrow x = \frac b a \\
+a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\
+a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$}
+\end{align*}
+
+### 2x2 System
+
+A 2x2 system is one with 2 equations and 2 unknown variables.
+
+<details>
+<summary>
+
+#### Example 1
+
+\begin{align*}
+3x_1 + 4x_2 &= 2 &\text{(1)} \\
+x_1 + 2x_2 &= 0 &\text{(2)} \\
+\end{align*}
+
+</summary>
+
+\begin{align*}
+3\times\text{(2)} = 3x_1 + 6x_2 &= 0 &\text{(3)} \\
+\text{(3)} - \text{(1)} = 0x_1 + 2x_2 &= -2 \\
+x_2 &= -1
+\end{align*}
+
+We've essentially created a 1x1 system for $x_2$ and now that's solved we can back substitute it
+into equation (1) (or equation (2), it doesn't matter) to work out the value of $x_1$:
+
+\begin{align*}
+3x_1 + 4x_2 &= 2 \\
+3x_1 - 1 &= 2 \\
+3x_1 &= 6 \\
+x_1 &= 2
+\end{align*}
+
+You can check the values for $x_1$ and $x_2$ are correct by substituting them into equation (2).
+
+</details>
+
+### 3x3 System
+
+A 3x3 system is one with 3 equations and 3 unknown variables.
+
+<details>
+<summary>
+
+#### Example 1
+
+\begin{align*}
+2x_1 + 3x_2 - x_3 &= 5 &\text{(1)} \\
+4x_1 + 4x_2 - 3x_3 &= 5 &\text{(2)} \\
+2x_1 - 3x_2 + x_3 &= 5 &\text{(3)} \\
+\end{align*}
+
+</summary>
+
+The first step is to eliminate $x_1$ from (2) and (3) using (1):
+
+\begin{align*}
+\text{(2)}-2\times\text{(1)} = -2x_2 -x_3 &= -1 &\text{(4)} \\
+\text{(3)}-\text{(1)} = -6x_2 + 3x_3 &= -6 &\text{(5)} \\
+\end{align*}
+
+This has created a 2x2 system of $x_2$ and $x_3$ which can be solved as any other 2x2 system.
+I'm too lazy to type up the working, but it is solved like any other 2x2 system.
+
+\begin{align*}
+x_2 &= -2
+x_3 &= 5
+\end{align*}
+
+These values can be back-substituted into any of the first 3 equations to find out $x_1$:
+
+\begin{align*}
+-2x_1 + 3x_2 - x_3 = 2x_1 + 6 - 3 = 5 \rightarrow x_1 = 1
+\end{align*}
+
+</details>