---

This post assumes you know about basic matrix operations. If you need to review those check this previous post

---

Elimination

Elimination is used for solving systems of equations.

The following system can be represented by matrices.

$$\begin{align} x + 2y + z & = 2\\ 3x + 8y + z & = 12\\ 4y + z & = 2 \end{align}$$

Let’s pick apart the equation $Ax = B$. $A$ is a coefficient matrix. What this means is that the coefficients from the systems are stored here. The $x$ is a vector that contains all the variables $x, y, z$. The matrix $B$ is contains everything to the left of the equals sign.

$$A= \begin{bmatrix} 1 & 2 & 1 \\ 3 & 8 & 1 \\ 0 & 4 & 1 \\ \end{bmatrix} ,\,\, x= \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} ,\,\, B= \begin{bmatrix} 2\\ 12\\ 2\\ \end{bmatrix}$$

Our goal is go from this

$$\begin{bmatrix} 1 & 2 & 1 \\ 3 & 8 & 1 \\ 0 & 4 & 1 \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} 2\\ 12\\ 2\\ \end{bmatrix}$$

to

$$\begin{bmatrix} 1 & 2 & 1 \\ 0 & 2 & -2 \\ 0 & 0 & 5 \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix} = \begin{bmatrix} 2\\ 6\\ -10\\ \end{bmatrix}$$

which is the same as

$$\begin{align} x + 2y + z & = 2\\ 2y + -2z & = 6\\ 5z & = -10 \end{align}$$

What we end up with is an upper triangle matrix. All the values below the diagonal are zero which makes substitution trivial because $z$ is solved.

The way we change indices to $0$ is by adding and subtracting rows from each other like we would with a typical system of equations.

We are going to drop the $x$ vector and incorporate it later. Lets start by adding $-3$ of the first row to the second row.

$$\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 3 + -3(1) & 8 + -3(2) & 1 + -3(1) & 12 + -3(2)\\ 0 & 4 & 1 & 2\\ \end{array} \right]$$

We are now left with the second row without an $x$ coefficient.

$$\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 0 & 2 & -2 & 6\\ 0 & 4 & 1 & 2\\ \end{array} \right]$$

Now lets add $-2$ of the second row to the third row.

$$\left[ \begin{array}{ccc|c} 1 & 2 & 1 & 2\\ 0 & 2 & -2 & 6\\ 0 & 0 & 5 & -10\\ \end{array} \right]$$

With this matrix it is easy to see that $z = -2$. Knowing that, back substitution is trivial.