In precalculus you solved systems with row operations. Here we make that tool precise, because the whole course rests on it. The first step is to stop copying the letters: in a system, only the numbers carry information.
For the system $$\begin{aligned} x + 2y - z &= 4 \\ 2x - y + 3z &= 9 \\ -x + y + z &= 0 \end{aligned}$$ the coefficient matrix collects the coefficients, one row per equation and one column per unknown, and the augmented matrix adds the constants as a last column: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 2 & -1 & 3 & 9 \\ -1 & 1 & 1 & 0 \end{array}\right]$$ The bar is a reminder that the last column holds the right-hand sides. A missing unknown has coefficient $0$, and every row must list the unknowns in the same order.
The three row operations
- Replacement: add a multiple of one row to a different row. Written $R_2 \to R_2 - 2R_1$: "replace row 2 by row 2 minus twice row 1".
- Scaling: multiply every entry of a row by a nonzero number: $R_3 \to \frac{1}{2}R_3$.
- Interchange: swap two rows: $R_1 \leftrightarrow R_2$.
Why it works
Worked example: Writing the augmented matrix
- Rewrite each equation with the unknowns in order: $-x + 3y + 0z = 2$, then $x + 0y + 4z = -1$, then $2x - y + z = 5$.
- Copy the coefficients row by row and put the constants after the bar: $\left[\begin{array}{ccc|c} -1 & 3 & 0 & 2 \\ 1 & 0 & 4 & -1 \\ 2 & -1 & 1 & 5 \end{array}\right]$
Answer: $\left[\begin{array}{ccc|c} -1 & 3 & 0 & 2 \\ 1 & 0 & 4 & -1 \\ 2 & -1 & 1 & 5 \end{array}\right]$
Worked example: Two replacements
- Row 2 minus twice row 1, entry by entry: $(2 - 2,\; -1 - 4,\; 3 + 2 \mid 9 - 8)$ $= (0, -5, 5 \mid 1)$. Rows 1 and 3 do not change: $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & -5 & 5 & 1 \\ -1 & 1 & 1 & 0 \end{array}\right]$
- Row 3 plus row 1: $(-1 + 1,\; 1 + 2,\; 1 - 1 \mid 0 + 4)$ $= (0, 3, 0 \mid 4)$: $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & -5 & 5 & 1 \\ 0 & 3 & 0 & 4 \end{array}\right]$
- The new system has the same solutions as the old one, and it is simpler: $x$ now appears only in the first equation.
Answer: $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & -5 & 5 & 1 \\ 0 & 3 & 0 & 4 \end{array}\right]$
Common mistake
Key idea