Three Outcomes, One Reduced Matrix
A linear system can have no solution, exactly one solution, or infinitely many solutions. Row reduction lets us distinguish these possibilities by inspecting the reduced augmented matrix. The key is to keep track of which columns represent variables and which column represents the right-hand sides: a pivot in a variable column and a pivot in the augmented column have different meanings.
Consider a system of \(m\) linear equations in \(n\) real variables \(x_1,\dots,x_n\). Its augmented matrix has the coefficients in its first \(n\) columns and the right-hand sides in its last column. Row operations preserve the system's solution set: swapping equations, multiplying an equation by a nonzero scalar, and adding a multiple of one equation to another all produce equivalent equations. We reduce the augmented matrix to reduced row echelon form (RREF), so the pivot positions and any contradictory rows are easy to identify.
In a reduced augmented matrix, a contradictory row may appear as \([0\ \cdots\ 0\mid 1]\), because the row can be scaled by its nonzero right-hand-side entry. Any such row rules out a solution. If there is no contradictory row, the system is consistent, meaning that at least one solution exists. Among consistent systems, the variable columns determine whether that solution is unique or whether free variables allow infinitely many solutions.
The Pivot-Pattern Classification
Proof. A contradictory row represents the equation \(0=c\) for some \(c\ne0\), which no values of the variables can satisfy. Since row operations preserve the solution set, the original system then has no solution.
Now suppose there is no contradictory row. Every nonzero row must have a pivot in a variable column: a row with its pivot in the augmented column would have zero entries in all earlier columns, and so would be contradictory. Thus the equations represented by the nonzero rows have leading variables in variable columns. Any variable column without a pivot corresponds to a free variable, meaning that its variable is not a leading variable in these equations.
If every variable column has a pivot, there are no free variables. In RREF, each pivot variable is specified by its row equation in terms of variables to its right, and there are no unassigned variables left. Reading the pivot equations from the last pivot row upwards determines exactly one value for every variable. This produces a solution, and any solution must have those same values, so the solution is unique.
If at least one variable column has no pivot, assign any real value to each free variable. The pivot equations then determine the pivot variables, working from the last pivot row upwards. There are no zero-coefficient rows with nonzero right-hand sides by the assumption that the system has no contradictory row, so these assignments satisfy every row. In particular, assigning one free variable two different values gives different solutions: that variable itself changes. Since a real free variable can take infinitely many values, the system has infinitely many solutions. These cases exhaust the possibilities. \(\square\)
The theorem gives a reliable order for classifying a system: first check for a contradictory row; only if none occurs, count pivots in the variable columns. In particular, do not count a pivot in the augmented column as if it belonged to a variable.
Worked Classifications
Worked Example: A Contradictory Row
Classify the system \(x+y=2\), \(2x+2y=5\) using its augmented matrix.
Subtract twice the first row from the second, then subtract twice the new second row from the first:
The last row represents \(0=1\), so it is contradictory. The system is inconsistent and has no solution. The second pivot is in the augmented column, not in a variable column.
Check: The first equation requires \(x+y=2\), while half of the second equation requires \(x+y=\tfrac52\). A sum cannot equal both \(2\) and \(\tfrac52\), confirming that no solution exists.
Worked Example: A Unique Solution
Classify and solve the system \(x+y+z=6\), \(2x-y+z=3\), \(x+2y-z=2\).
Start with its augmented matrix. Apply the indicated row operations in order:
Swap rows 2 and 3, then add three times row 2 to row 3:
Scale row 3 by \(-\tfrac17\), add twice row 3 to row 2, subtract row 3 from row 1, and finally subtract row 2 from row 1:
There is no contradictory row, and every variable column has a pivot. The system has a unique solution: \(x=1\), \(y=2\), \(z=3\).
Check: Substitution into the original equations gives \(1+2+3=6\), \(2(1)-2+3=3\), and \(1+2(2)-3=2\), as required.
Worked Example: A Free Variable and Infinitely Many Solutions
Classify and solve the system \(x+y+z=4\), \(2x+2y+2z=8\), \(x-y+z=2\).
Subtract twice row 1 from row 2 and subtract row 1 from row 3. Then swap rows 2 and 3:
Scale row 2 by \(-\tfrac12\) and subtract the new row 2 from row 1:
There is no contradictory row. Columns 1 and 2 have pivots, but variable column 3 does not. Set the free variable \(z=t\), where \(t\in\mathbb{R}\). The pivot equations give \(x=3-t\) and \(y=1\), so the solutions are \((x,y,z)=(3-t,1,t)\) for all real \(t\). The system has infinitely many solutions.
Check: Substitution gives \((3-t)+1+t=4\); \(2(3-t)+2+2t=8\); and \((3-t)-1+t=2\). All three equations hold for every real \(t\).
Read the Columns Carefully
The same number of pivots can appear in systems with different classifications, so the pivot count alone is not enough. The location of each pivot matters. In the inconsistent example, the pivot in the augmented column accompanies a contradictory row. In a consistent system, every pivot belongs to a variable column, and any variable column without a pivot marks a free variable.
| RREF pattern | Classification | Reason |
|---|---|---|
| A row has zero variable entries and a nonzero augmented entry | No solution | The row requires \(0=c\) with \(c\ne0\) |
| No contradictory row; every variable column has a pivot | One solution | There are no free variables |
| No contradictory row; at least one variable column has no pivot | Infinitely many solutions | A free variable can take infinitely many real values |
A useful way to interpret the procedure is to separate two questions. First, do the equations conflict? A contradictory row answers yes, and the system has no solution. If there is no conflict, ask whether any variable is free. No free variables means that the pivot equations determine a single solution; at least one free variable means that there are infinitely many choices.
Common Mistake: Counting the Augmented Pivot
A tempting but incorrect argument is: “The inconsistent example has two pivots for two variables, so it has a unique solution.” The reduced matrix is \(\begin{bmatrix}1&1&0\\0&0&1\end{bmatrix}\). Its second pivot is in column 3, the augmented column, not in the second variable column. The bottom row says \(0=1\), so this system has no solution. To classify a system correctly, first check for a contradictory row; then count pivots only in the variable columns.
For a consistent system, do not mistake a zero row for a contradiction. A row of all zeros represents \(0=0\), which imposes no restriction. In the infinitely many solutions example, the last row is all zero, while the missing pivot in variable column 3 leaves a free variable. The zero row and the free variable have distinct roles.
Check Your Understanding
For each question, identify any contradictory row before deciding what the variable-column pivots imply.
- A reduced augmented matrix has rows \([1\ 0\mid 3]\) and \([0\ 1\mid -2]\). How many solutions does its system have?
- A reduced augmented matrix has rows \([1\ 2\mid 4]\) and \([0\ 0\mid 1]\). Is the system consistent? Explain using the last row.
- A consistent system has three variables and pivots in variable columns 1 and 3. How many solutions does it have, and which variable is free?
- In a system with two variables, the RREF has a pivot in variable column 1, no pivot in variable column 2, and no contradictory row. Classify the system.
- Explain why a pivot in the augmented column cannot be counted as a pivot for a variable.
