Now we can finally restate Gaussian elimination in matrix form. First let’s simplify the notation even further, dropping the variables completely: $\begin{aligned} x + 2y + 3z &= 6 \\ 3x + 4y + z &= 8 \implies \begin{pmatrix}1&2&3&6\\3&4&1&8\\2&-1&1&2\end{pmatrix}, \\ 2x - y + z &= 2 \end{aligned}$ where the last column represents the RHS of each equation. Now we follow this algorithm:

• Determine what's called a “pivot,” i.e. the top leftmost nonzero number that isn’t in a row/column already used. (If necessary, scale this whole row to make the pivot entry 1). Here we have the pivots in blue.

$\begin{aligned} \begin{pmatrix} \color{#3D99F6}{1}&2&3&6\\3&4&1&8\\2&-1&1&2 \end{pmatrix} \end{aligned}$

• For each row $r$ below the pivot row $p,$ multiply the pivot row by the leading entry and subtract: $r \mapsto r- (\text{leading entry of}\ r) \cdot p$

$\begin{aligned} r_{2} &\rightarrow r_{2}-3 r_{1} \\ r_{3} &\rightarrow r_{3}-2 r_{1} \end{aligned}$ $\begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&-2&-8&-10\\0&-5&-5&-10\end{pmatrix} \end{aligned}$

Each row below the pivot row will be left with zeros in that column.

• Repeat until there are no more pivots. At this point, the matrix will have all 0s in the lower left triangle.

$\begin{aligned} \ \ r_{2} & \rightarrow -\frac{1}{2} r_{2} \\ \ r_{3} & \rightarrow r_{3} + 5 r_{2} \end{aligned}$

$\begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&\color{#3D99F6}{1}&4&5\\0&0&15&15\end{pmatrix} \end{aligned}$

Now we need pivot in $r_3$:

$\begin{aligned} r_{3} \rightarrow \frac{1}{15} r_{3} \end{aligned}$

$\begin{aligned} \begin{pmatrix}\color{#3D99F6}{1}&2&3&6\\0&\color{#3D99F6}{1}&4&5\\0&0&\color{#3D99F6}{1}&1\end{pmatrix} \end{aligned}$

Now let’s look at that last row--the equation it represents is $z = 1$. Then we can find $y$ and $x$ directly.