What Makes a Solution Unique?
A system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) asks for a vector \(\mathbf{x}\) whose matrix product with \(\mathbf{A}\) equals the given vector \(\mathbf{b}\). A system may have no solution, exactly one solution, or more than one solution. Here the goal is to identify and solve systems in the middle case: exactly one solution. This can happen for a square matrix, but it can also happen when there are more equations than variables.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{b}\in\mathbb{R}^{m}\). The system has \(n\) variables, the entries of the column vector \(\mathbf{x}\in\mathbb{R}^{n}\), and \(m\) equations. Its augmented matrix is \([\mathbf{A}\mid\mathbf{b}]\). Row reduction applies the same elementary row operations to the entire augmented matrix, including its final column.
A pivot in every variable column means every variable is determined. It does not require a pivot in every row: in a system with extra equations, some rows may reduce to zeros. The other requirement for a solution is consistency, meaning that no contradiction row appears.
Proof. First, row operations on an augmented matrix preserve its solution set. Swapping two rows just changes the order of two equations. Multiplying a row by a nonzero scalar gives an equivalent equation, because the operation can be reversed by dividing by that scalar. Adding a multiple of one row to another replaces an equation by a combination of the equations; subtracting that same multiple reverses the operation. Therefore the original system and its RREF describe exactly the same solutions.
If the RREF has a contradiction row, its equation is \(0=c\) with \(c\ne0\), so there is no solution. If there is no contradiction row, each nonzero row has a pivot in a variable column. When every variable column has a pivot, each variable is a pivot variable. In RREF, its pivot is the only nonzero entry in its column, so the pivot equation determines that variable from the right-hand side. Thus there is exactly one solution.
Conversely, suppose there is no pivot in some variable column. Assign the corresponding non-pivot variable the value \(0\), and then the value \(1\), keeping any other non-pivot variables fixed at \(0\). In each assignment, the pivot equations determine the pivot variables: each pivot variable has coefficient \(1\) in its pivot row and coefficient \(0\) in the other pivot rows. The zero rows impose no further conditions, and the assumption of no contradiction row ensures no impossible equation remains. Both assignments therefore give solutions, and they differ in the chosen variable. There cannot be exactly one solution. Together, these cases prove the criterion. \(\square\)
A Square System with One Solution
Worked Example: Solve a 3 by 3 System
Solve the system represented by \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}=\begin{bmatrix}1&1&1\\2&-1&1\\1&2&-1\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}2\\-1\\6\end{bmatrix}\).
Reduce the augmented matrix. First apply \(R_2\to R_2-2R_1\) and \(R_3\to R_3-R_1\):
Swap rows 2 and 3, then apply \(R_3\to R_3+3R_2\):
Scale row 3 by \(-\tfrac17\), then use \(R_2\to R_2+2R_3\) and \(R_1\to R_1-R_3\). Finally apply \(R_1\to R_1-R_2\):
There is a pivot in each of the three variable columns and no contradiction row. The criterion gives a unique solution, \(\mathbf{x}=\begin{bmatrix}1\\2\\-1\end{bmatrix}\).
Check by substitution: Multiply the original matrix by the answer, not the reduced matrix:
A Rectangular System with More Equations
A rectangular system can still have exactly one solution. When there are more equations than variables, some equations may provide information already implied by the others. The decisive test remains a pivot in every variable column together with consistency—not a pivot in every row.
Worked Example: Three Equations and Two Variables
Solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}=\begin{bmatrix}1&1\\1&-1\\2&0\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}3\\1\\4\end{bmatrix}\).
Apply \(R_2\to R_2-R_1\) and \(R_3\to R_3-2R_1\):
Apply \(R_3\to R_3-R_2\), scale row 2 by \(-\tfrac12\), and apply \(R_1\to R_1-R_2\):
Both variable columns have pivots, while the last row is a zero row rather than a contradiction. Hence the unique solution is \(\mathbf{x}=\begin{bmatrix}2\\1\end{bmatrix}\).
Check by substitution:
The zero row records that one equation became redundant during reduction. It does not create another variable choice, because both variable columns still contain pivots.
A Taller Rectangular System
Worked Example: Four Equations and Three Variables
Solve the system with augmented matrix shown below. Its fourth equation is compatible with the first three, but that must be confirmed by reduction.
Use \(R_4\to R_4-R_1\), then \(R_4\to R_4-R_2\), and finally \(R_4\to R_4-R_3\):
All three variable columns have pivots and there is no contradiction row. The unique solution is \(\mathbf{x}=\begin{bmatrix}2\\-1\\3\end{bmatrix}\). The zero row confirms that the fourth equation adds no further restriction.
Check by substitution:
This example also illustrates why the number of equations alone does not determine uniqueness. There are four equations, but there are only three variable columns to check for pivots.
How the Number of Variables Affects the Test
A system with fewer equations than variables cannot have a unique solution. Indeed, an \(m\times n\) matrix has at most \(m\) pivot positions, one per row. If \(m<n\), at least one of the \(n\) variable columns has no pivot. The unique-solution criterion then rules out exactly one solution: the system is either inconsistent or, if consistent, has multiple solutions. This is why square systems and rectangular systems with extra equations can have unique solutions, while systems with more variables than equations cannot.
For a square system, having the right number of equations and variables is not by itself enough. Row reduction must still show both consistency and a pivot in every variable column. For a rectangular system with extra equations, a zero row is allowed, but a contradiction row is not. In either case, direct substitution is a useful final check: it confirms that the computed vector satisfies the original equation, while the pivot test establishes that no second solution exists.
Common Pitfall: Assuming a Square System Is Automatically Unique
A common mistake is to assume that a square system must have one solution simply because its numbers of equations and variables match. For example, take \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}3\\6\end{bmatrix}\). The two proposed vectors \(\begin{bmatrix}3\\0\end{bmatrix}\) and \(\begin{bmatrix}1\\1\end{bmatrix}\) both satisfy the system:
The correction is to check the augmented matrix rather than infer uniqueness from its shape. Subtracting \(2R_1\) from \(R_2\) gives
The second variable column has no pivot, so the system does not have a unique solution. The two direct checks above verify that it has at least two solutions. Always use the pivot-and-consistency test to establish uniqueness, then substitute the resulting vector into the original \(\mathbf{A}\mathbf{x}=\mathbf{b}\).
Check Your Understanding
For each system, use the augmented matrix to decide whether the available information establishes a unique solution. When a solution is given or obtained, verify it by multiplication.
- What two conditions on the RREF of an augmented matrix guarantee exactly one solution?
- Can a system with two equations and three variables have exactly one solution? Explain using pivot columns.
- A consistent system has a zero row and a pivot in every variable column. Must its solution be unique? Why?
- Reduce the augmented matrix \(\begin{bmatrix}1&1&|&5\\2&-1&|&1\\3&0&|&6\end{bmatrix}\), then state and check its solution.
- Explain why a contradiction row rules out a unique solution.
