What Determines Whether a Solution Is Unique?
A consistent linear system has at least one solution. It may have exactly one, or it may have infinitely many. The distinction is visible in the pivot pattern of the reduced system: a pivot in every variable column leaves no variable free to vary, while a missing pivot leaves at least one free variable. This criterion concerns the columns for the variables, not the augmented column containing the right-hand sides.
Work over \(\mathbb{R}\), and consider a system with \(m\) equations and \(n\) variables. Write its matrix equation as \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\), \(\mathbf{x}\in\mathbb{R}^n\), and \(\mathbf{b}\in\mathbb{R}^m\). The augmented matrix is \([\mathbf{A}\mid\mathbf{b}]\). Row operations on this matrix preserve the system's solutions. A system is consistent when it has at least one solution; equivalently, its reduced form has no row of the form \([0\ \cdots\ 0\mid c]\) with \(c\ne0\).
In a reduced row echelon form (RREF), each pivot variable is determined by its row equation in terms of any free variables. Therefore, the presence of free variables does not by itself signal inconsistency: it signals freedom to choose among solutions, provided the system is consistent. The word “provided” matters. A contradictory row means there is no solution, so a pivot count cannot turn an inconsistent system into a consistent one.
The Uniqueness Criterion
Proof. Reduce the augmented matrix to RREF. Row operations preserve the solutions, so it is enough to analyze this reduced system. Consistency ensures that every row with all zero entries in the variable columns also has zero in the augmented column. Thus, the nonzero rows give equations with pivot variables, and any variables in columns without pivots may be assigned freely. Read the pivot equations from the bottom upward: for any assignment of the free variables, each pivot variable is then determined. If every variable column has a pivot, there are no free variables, and the pivot equations determine every variable. Since the system is consistent, at least one solution exists; since all its variables are determined, that solution is the only one. Conversely, if some variable column has no pivot, assign all free variables the value \(0\) to obtain one solution, and then assign one free variable the value \(1\), keeping the others at \(0\), to obtain another. The pivot variables can be determined for both assignments, and the resulting solutions differ in the chosen free variable. More generally, that variable can be assigned any real number, with the other free variables fixed, so there are infinitely many distinct solutions. Therefore the system is unique exactly when every variable column has a pivot; otherwise it has infinitely many solutions. \(\square\)
The field matters in the final step: over \(\mathbb{R}\), a free variable has infinitely many possible values. The theorem is stated here for real systems, as is standard for this course. Its condition is about the RREF of the augmented matrix, but “every variable column” excludes the augmented column. For a consistent system the augmented column cannot be a pivot column: such a pivot would come from a row with zeros in every variable column and a nonzero right-hand side, contradicting consistency.
Reading a Unique Solution from the Pivots
Worked Example: A Pivot in Every Variable Column
Solve the system represented by the augmented matrix \(\begin{bmatrix}1&1&3\\2&-1&0\end{bmatrix}\). The first two columns are variable columns and the last is the right-hand side. Reduce the augmented matrix:
There is a pivot in each of the two variable columns, so the criterion guarantees a unique solution. The reduced equations are \(x_1=1\) and \(x_2=2\).
Check: Substituting into the original equations gives \(1+2=3\) and \(2(1)-2=0\). Both equations hold, so \(\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\) is the unique solution.
In this example, no variable can be chosen independently: each has a pivot. The check confirms that the calculated vector is a solution; the pivot criterion explains why no other solution exists.
How a Free Variable Produces Infinitely Many Solutions
Worked Example: One Free Variable
Consider the system represented by \(\begin{bmatrix}1&2&-1&4\\2&4&-2&8\end{bmatrix}\), with three variables and the right-hand side in the last column. Subtract twice the first row from the second:
The zero row is not contradictory, so the system is consistent. There is a pivot in the first variable column, but none in the second or third. Set \(x_2=s\) and \(x_3=t\), where \(s,t\in\mathbb{R}\). The first row gives \(x_1+2x_2-x_3=4\), hence \(x_1=4-2s+t\). Thus all solutions are
Check: The first equation evaluates to \((4-2s+t)+2s-t=4\). The second equation is twice the first, so its left-hand side is \(2(4)=8\), as required. Choosing \((s,t)=(0,0)\) and \((s,t)=(1,0)\), for example, gives different solutions. Since \(s\) and \(t\) can take real values, the system has infinitely many solutions.
There is no need to find two particular solutions to establish that the solution set is infinite. Once consistency is known, a free variable can be varied through infinitely many real values while the pivot equations determine the remaining variables. The resulting solutions are distinct because their free-variable entries differ.
Worked Example: Redundant Equation and a Free Variable
Consider the system represented by \(\begin{bmatrix}1&1&1&2\\1&-1&1&0\\2&0&2&2\end{bmatrix}\). It has three variables. Subtract the first row from the second, and subtract twice the first row from the third:
Subtract the second row from the third, scale the second row by \(-\tfrac12\), and subtract the new second row from the first:
The zero row confirms consistency. Columns 1 and 2 are pivot columns, while column 3 is free. Set \(x_3=t\). The reduced equations give \(x_1+x_3=1\) and \(x_2=1\), so \(x_1=1-t\), \(x_2=1\), and \(x_3=t\).
Check: In the original equations, the first left-hand side is \((1-t)+1+t=2\), the second is \((1-t)-1+t=0\), and the third is \(2(1-t)+2t=2\). Each equation holds for every \(t\in\mathbb{R}\), confirming the infinite family of solutions.
This example also illustrates why consistency and uniqueness are separate questions. The zero row does not add an equation restricting the variables, and it does not create a contradiction. It records that one equation supplied no new restriction after elimination. The missing pivot in the third variable column is what allows the solution to vary.
A Common Mistake: Treating a Free Variable as Fixed
A common error is to set every free variable equal to zero and then claim that the resulting particular solution is the only one. In the three-variable example above, setting \(x_3=0\) gives \((x_1,x_2,x_3)=(1,1,0)\). That is a valid solution, but it is not the only solution. Setting \(x_3=1\) gives \((0,1,1)\), and substitution gives \(0+1+1=2\), \(0-1+1=0\), and \(2(0)+2(1)=2\). Both vectors solve the original system.
The correction is to distinguish choosing a value from proving that a value is forced. A free variable may be assigned a value to describe one solution, but unless its column has a pivot, the row equations do not force that value. To test uniqueness, count pivots in the variable columns of a consistent reduced system, not the number of solutions you happened to calculate.
It is also important not to count a pivot in the augmented column as if it were a pivot for a variable. A pivot there, with zeros in the variable columns of that row, represents a contradiction and means the system is inconsistent. The uniqueness criterion applies only after consistency has been established.
Check Your Understanding
For each question, distinguish consistency from uniqueness and use the variable-column pivot pattern when appropriate.
- A consistent system has four variables and pivots in columns 1, 2, 3, and 4. How many solutions does it have, and why?
- A consistent system has three variables and pivots in columns 1 and 3 only. Identify its free variable and state how many solutions it has.
- Explain why a row \([0\ 0\ 0\mid 5]\) prevents the uniqueness criterion from applying.
- For the system \(x_1+x_2=3,\ 2x_1+2x_2=6\), find its RREF, describe all solutions, and verify the family in the original equations.
- Why does assigning a free variable the value \(0\) produce a solution without proving that the solution is unique?
