Solving Systems with Gaussian Elimination

What is Gaussian elimination?

Gaussian elimination is a systematic procedure that uses elementary row operations to transform the augmented matrix $[A \mid \mathbf{b}]$ of a linear system into row-echelon form (upper triangular), from which the solution is read off by back substitution.

The three elementary row operations

1. Swap two rows: $R_i \leftrightarrow R_j$.
2. Scale a row: $R_i \to c \, R_i$ with $c \ne 0$.
3. Combine: replace $R_i$ with $R_i + c \, R_j$.

Each operation preserves the solution set — the system before and after has the same solutions.

The augmented matrix

$[A \mid \mathbf{b}] = \left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array}\right]$

Step-by-step solution

Step 1 — Create a zero in column 1, row 2. Use $R_2 \to R_2 + \dfrac{3}{2} R_1$: $R_2 = (-3, -1, 2, -11) + \tfrac{3}{2}(2, 1, -1, 8) = (0, \tfrac{1}{2}, \tfrac{1}{2}, 1)$

Step 2 — Create a zero in column 1, row 3. Use $R_3 \to R_3 + R_1$: $R_3 = (-2, 1, 2, -3) + (2, 1, -1, 8) = (0, 2, 1, 5)$

Matrix now: $\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & 1/2 & 1/2 & 1 \\ 0 & 2 & 1 & 5 \end{array}\right]$

Step 3 — Create a zero in column 2, row 3. Use $R_3 \to R_3 - 4 R_2$: $R_3 = (0, 2, 1, 5) - 4(0, \tfrac{1}{2}, \tfrac{1}{2}, 1) = (0, 0, -1, 1)$

Upper triangular form: $\left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\ 0 & 1/2 & 1/2 & 1 \\ 0 & 0 & -1 & 1 \end{array}\right]$

Step 4 — Back substitution.

From row 3: $-z = 1 \implies z = -1$.

From row 2: $\tfrac{1}{2} y + \tfrac{1}{2}(-1) = 1 \implies y = 3$.

From row 1: $2x + 3 - (-1) = 8 \implies 2x = 4 \implies x = 2$.

Solution: $\boxed{(x, y, z) = (2, 3, -1)}$.

Verification

• Eq 1: $2(2) + 3 - (-1) = 4 + 3 + 1 = 8$ ✓
• Eq 2: $-3(2) - 3 + 2(-1) = -6 - 3 - 2 = -11$ ✓
• Eq 3: $-2(2) + 3 + 2(-1) = -4 + 3 - 2 = -3$ ✓

Common mistakes

• Forgetting to update the right-hand side. Every row operation applies to the full augmented row, not just the coefficient part.
• Not following a consistent pivot order. Work column by column from left to right, top to bottom, to avoid re-introducing non-zero entries.
• Dividing by zero. If the pivot element is 0, swap rows first to get a non-zero pivot.

Try it in the visualization

Each row operation is animated as a transformation of the matrix. Step through operations to see the zeros appearing in the lower triangle and the back-substitution unwinding from bottom to top.