When the reduced echelon form has a column without a pivot, the system does not pin down every unknown. This topic is about describing all the solutions in that case, clearly and completely.
An unknown whose column holds a pivot is a basic variable. The others are free variables. Each nonzero row of the RREF contains exactly one basic variable, so each row can be solved for its basic variable in terms of the free ones.
Why it works
Why may a free variable take any value? Its column has no pivot, so no row "belongs" to it: no equation forces its value. Whatever numbers you choose for the free variables, each pivot row then gives exactly one value for its basic variable. So every choice of the free variables gives one solution, and every solution arises this way.
Worked example: One free variable
The augmented matrix of a system in $x_1, x_2, x_3$ reduces to $\left[\begin{array}{ccc|c} 1 & 0 & -2 & 3 \\ 0 & 1 & 1 & -1 \end{array}\right]$. Describe all solutions.
- Pivots are in columns 1 and 2, so $x_1, x_2$ are basic and $x_3$ is free. Call it $x_3 = t$.
- Row 1 says $x_1 - 2x_3 = 3$, so $x_1 = 3 + 2t$.
- Row 2 says $x_2 + x_3 = -1$, so $x_2 = -1 - t$.
- All solutions: $(x_1, x_2, x_3)$ $= (3 + 2t,\; -1 - t,\; t)$ for every real $t$.
- In vector form: $\mathbf{x} = \begin{bmatrix} 3 \\ -1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$, a line in $\mathbb{R}^3$ through $(3, -1, 0)$ in the direction $(2, -1, 1)$.
- Spot check with $t = 1$: $(5, -2, 1)$ gives $5 - 2 = 3$ ✓ and $-2 + 1 = -1$ ✓.
Answer: $\mathbf{x} = (3 + 2t,\; -1 - t,\; t)$
Worked example: Two free variables
Describe all solutions of the system whose augmented matrix reduces to $\left[\begin{array}{cccc|c} 1 & 3 & 0 & -2 & 4 \\ 0 & 0 & 1 & 5 & -1 \end{array}\right]$ (unknowns $x_1, \dots, x_4$).
- Pivots in columns 1 and 3: $x_1, x_3$ are basic; $x_2 = s$ and $x_4 = t$ are free.
- Row 1: $x_1 + 3x_2 - 2x_4 = 4$, so $x_1 = 4 - 3s + 2t$.
- Row 2: $x_3 + 5x_4 = -1$, so $x_3 = -1 - 5t$.
- $\mathbf{x}$ $= \begin{bmatrix} 4 \\ 0 \\ -1 \\ 0 \end{bmatrix} + s\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} 2 \\ 0 \\ -5 \\ 1 \end{bmatrix}$: a plane of solutions inside $\mathbb{R}^4$.
Answer: $(4 - 3s + 2t,\; s,\; -1 - 5t,\; t)$
Common mistake
Setting the free variables to $0$ and reporting that single point as "the solution". It is a solution (for the line above, $(4, 0)$), but the question asks for all of them, and there are infinitely many. Check: if your answer has no letter in it while some column had no pivot, something is missing.
The constant vector (here $(3, -1, 0)$) is the solution with every free variable equal to $0$. The vectors multiplying the parameters will turn out to solve the system with all constants replaced by $0$. That pattern is coming soon, in homogeneous systems.
Key idea
Basic variables (pivot columns) are solved for; free variables (no pivot) are parameters that can be anything. The general solution writes every unknown in terms of the parameters, often as $\mathbf{x} = \mathbf{p} + t\mathbf{v}$ (+ more terms for more free variables).