Solving Systems with Gaussian Elimination

By the end of this lesson, you will be able to write systems as augmented matrices, apply row operations to reduce to row echelon form, and use back-substitution to solve any linear system.

The Big Idea

Gaussian elimination converts a system of linear equations into a simple "staircase" form (row echelon form) through three elementary row operations. It's more systematic than substitution/elimination for large systems and is the foundation for computational linear algebra.

Augmented Matrix Representation

The system $\begin{cases} 2x + y - z = 3 \ x - y + 2z = 1 \ 3x + 2y + z = 5 \end{cases}$

becomes the augmented matrix:

$$\begin{pmatrix}2&1&-1&|&3\1&-1&2&|&1\3&2&1&|&5\end{pmatrix}$$

Elementary Row Operations

1. Swap two rows: $R_i \leftrightarrow R_j$
2. Multiply a row by a nonzero scalar: $kR_i \to R_i$
3. Add a multiple of one row to another: $R_i + kR_j \to R_i$

These operations don't change the solution set.

Row Echelon Form (REF)

A matrix is in REF when:

• All zero rows are at the bottom
• Each nonzero row's first nonzero entry (leading entry/pivot) is to the right of the row above
• All entries below each pivot are zero

Reduced Row Echelon Form (RREF): additionally, each pivot is 1 and all entries above pivots are also zero.

Example 1 — Full Gaussian Elimination

Solve: $\begin{cases}x + y + z = 6\2x - y + z = 3\x + 2y - z = 2\end{cases}$

Augmented matrix:

$\begin{pmatrix}1&1&1&|&6\2&-1&1&|&3\1&2&-1&|&2\end{pmatrix}$

$R_2 - 2R_1$: $\begin{pmatrix}1&1&1&|&6\0&-3&-1&|&-9\1&2&-1&|&2\end{pmatrix}$

$R_3 - R_1$: $\begin{pmatrix}1&1&1&|&6\0&-3&-1&|&-9\0&1&-2&|&-4\end{pmatrix}$

Swap $R_2$ and $R_3$: $\begin{pmatrix}1&1&1&|&6\0&1&-2&|&-4\0&-3&-1&|&-9\end{pmatrix}$

$R_3 + 3R_2$: $\begin{pmatrix}1&1&1&|&6\0&1&-2&|&-4\0&0&-7&|&-21\end{pmatrix}$

$R_3/(-7)$: $\begin{pmatrix}1&1&1&|&6\0&1&-2&|&-4\0&0&1&|&3\end{pmatrix}$ → REF!

Back-substitution:

From R3: $z = 3$

From R2: $y - 6 = -4 \implies y = 2$

From R1: $x + 2 + 3 = 6 \implies x = 1$