Two questions come before solving: does a solution exist (is the system consistent)? And if so, is it unique? Once a matrix is in echelon form, both answers can be read from the pivot positions, often without finishing the solution. This builds on free variables.
The theorem
- A system is consistent exactly when the last column of the augmented matrix is not a pivot column, that is, when the echelon form has no row $\left[\,0\ \cdots\ 0 \mid b\,\right]$ with $b \neq 0$.
- If it is consistent, the solution is unique when every column of the coefficient matrix is a pivot column (no free variables), and there are infinitely many solutions when at least one variable is free.
Why it works
A row $\left[\,0\ 0\ \cdots\ 0 \mid b\,\right]$ stands for the equation $0x_1 + \cdots + 0x_n = b$, which says $0 = b$. If $b \neq 0$, no choice of unknowns makes it true, and since row operations keep the solution set, the original system has no solution either. If there is no such row, back substitution never gets stuck: give the free variables any values and solve upward. No free variables means one solution; any free variable means one solution for each of its infinitely many values.
Worked example: Reading three echelon forms
How many solutions? $$\text{(a) } \left[\begin{array}{ccc|c} 1 & 4 & -2 & 3 \\ 0 & 2 & 1 & 5 \\ 0 & 0 & 3 & 6 \end{array}\right] \quad \text{(b) } \left[\begin{array}{ccc|c} 1 & 4 & -2 & 3 \\ 0 & 2 & 1 & 5 \\ 0 & 0 & 0 & 6 \end{array}\right] \quad \text{(c) } \left[\begin{array}{ccc|c} 1 & 4 & -2 & 3 \\ 0 & 2 & 1 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right]$$
- (a) No row of the form $[\,0\ 0\ 0 \mid b\,]$. Pivots in all three variable columns: no free variables. Exactly one solution.
- (b) Row 3 says $0 = 6$. No solution.
- (c) Row 3 says $0 = 0$, which is always true; it just tells us one equation was redundant. Column 3 has no pivot, so $x_3$ is free: infinitely many solutions.
Answer: (a) one, (b) none, (c) infinitely many.
Worked example: A parameter in the system
For which $h$ is the system with augmented matrix $\left[\begin{array}{cc|c} 1 & -3 & 2 \\ -2 & 6 & h \end{array}\right]$ consistent?
- $R_2 \to R_2 + 2R_1$ gives $\left[\begin{array}{cc|c} 1 & -3 & 2 \\ 0 & 0 & h + 4 \end{array}\right]$.
- Row 2 says $0 = h + 4$. That is possible only when $h = -4$.
- For $h = -4$ the second row is all zeros and $y$ is free, so there are infinitely many solutions. For every other $h$, none.
Answer: Only $h = -4$.
Common mistake
"There is a row of zeros, so there is no solution." Not so. A row that is zero everywhere, including after the bar, says $0 = 0$: harmless. Only a row that is zero on the left and nonzero after the bar means no solution. Check: read the row aloud as an equation.
| Shape of the coefficient matrix ($m$ equations, $n$ unknowns) | What it tells you |
|---|---|
| A pivot in every row | Consistent for every right-hand side |
| A pivot in every column | At most one solution (unique if consistent) |
| More unknowns than equations ($n > m$) | Some variable is free: never a unique solution |
| More equations than unknowns ($m > n$) | Some right-hand sides give no solution |
Key idea
Look at the pivots. A pivot in the last (augmented) column means no solution. Otherwise there is a solution, and it is unique exactly when every variable column has a pivot.