What Consistency Says
A linear system asks for one ordered list of variable values that satisfies every equation at once. Checking a candidate means testing that same list in each equation; meeting one constraint does not suffice. The first classification question is therefore whether any such list exists. If one does, the next question is whether there is only one or whether there are infinitely many.
Write a system of \(m\) linear equations in \(n\) variables as \(\mathbf{A}\mathbf{x}=\mathbf{b}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\) is the coefficient matrix, \(\mathbf{x}\in\mathbb{R}^n\) is the column of variables, and \(\mathbf{b}\in\mathbb{R}^m\) is the column of right-hand sides. A solution is a vector \(\mathbf{x}\) for which the matrix equation holds. Equivalently, its entries give values for all variables that satisfy every original equation.
These terms describe the number of solutions, not how difficult the equations are to solve. A system with many equations may have a solution, and a system with only two equations may have none. Also, consistency alone does not mean uniqueness: it says only that the solution set is nonempty. The examples below use real variables, so a family of distinct solutions indexed by all real numbers gives infinitely many solutions.
Three Basic Classifications
Worked Example: A System with No Solution
Classify the system
The first equation requires \(x+y=4\). Doubling that equality gives \(2x+2y=8\). The second equation instead requires \(2x+2y=10\), so no ordered pair can satisfy both equations. The system is inconsistent.
Check: If a candidate satisfies the first equation, then its value of \(2x+2y\) must be \(2(x+y)=2\cdot4=8\), not \(10\). Thus no candidate passing the first equation can pass the second.
Worked Example: A Unique Solution
Classify the system
The first equation forces \(x=3\), and the second forces \(y=-2\). There can be no other ordered pair satisfying those two equations. For these forced values, the left-hand side of the third equation is \(x+y=3+(-2)=1\), so the third equation is also satisfied. The system is consistent and has the unique solution \(\mathbf{x}=\begin{bmatrix}3\\-2\end{bmatrix}\).
Check: Substitution gives \(3=3\), \(-2=-2\), and \(3+(-2)=1\). The first two equations force the only candidate, and the check confirms it meets all three constraints.
Worked Example: Infinitely Many Solutions
Classify the system
For each real number \(t\), set \(x=t\) and \(y=4-t\). Then the first left-hand side is \(t+(4-t)=4\), while the second is \(2t+2(4-t)=2t+8-2t=8\). Thus every vector \(\begin{bmatrix}t\\4-t\end{bmatrix}\), with \(t\in\mathbb{R}\), is a solution.
These solutions are distinct for distinct values of \(t\), since their first coordinates equal \(t\). Moreover, any solution of the first equation must satisfy \(y=4-x\), so it is already included by taking \(t=x\). The system is consistent and has infinitely many solutions.
Check: For \(t=2\), the solution is \((2,2)\), and the two left-hand sides are \(2+2=4\) and \(2(2)+2(2)=8\). For \(t=0\), the solution is \((0,4)\), and the left-hand sides are \(0+4=4\) and \(0+2(4)=8\). These are different solutions, and the parameter argument verifies the entire family.
Why There Are No Other Possibilities
The examples show the three classifications in action. For systems over \(\mathbb{R}\), there cannot be a nonempty solution set containing a finite number greater than one: two different solutions generate infinitely many solutions. This fact does not require a particular solving procedure. It follows directly from the form of a linear system.
Proof. Let \(S=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{b}\}\) be the solution set. If \(S\) is empty, the system is inconsistent, and neither of the other two cases holds. Suppose instead that \(S\) is nonempty. If it has exactly one element, the system has a unique solution. If it does not have exactly one element, then it has at least two distinct elements; choose \(\mathbf{x}_1,\mathbf{x}_2\in S\) with \(\mathbf{x}_1\ne\mathbf{x}_2\). For every \(t\in\mathbb{R}\), define \(\mathbf{x}(t)=\mathbf{x}_1+t(\mathbf{x}_2-\mathbf{x}_1)\). Using \(\mathbf{A}\mathbf{x}_1=\mathbf{b}\) and \(\mathbf{A}\mathbf{x}_2=\mathbf{b}\), we obtain \[ \mathbf{A}\mathbf{x}(t) =\mathbf{A}\mathbf{x}_1+t\mathbf{A}(\mathbf{x}_2-\mathbf{x}_1) =\mathbf{b}+t(\mathbf{A}\mathbf{x}_2-\mathbf{A}\mathbf{x}_1) =\mathbf{b}+t(\mathbf{b}-\mathbf{b}) =\mathbf{b}. \] Thus \(\mathbf{x}(t)\) is a solution for every real \(t\). If \(t_1\ne t_2\), then \(\mathbf{x}(t_1)-\mathbf{x}(t_2)=(t_1-t_2)(\mathbf{x}_2-\mathbf{x}_1)\ne\mathbf{0}\), because both factors are nonzero. Therefore these give infinitely many distinct solutions. The cases are mutually exclusive and cover every possible solution set. \(\square\)
The real-number assumption matters in this theorem: the proof uses infinitely many choices of \(t\). Its conclusion concerns systems over \(\mathbb{R}\), as in the examples here.
Relating Solutions to a Homogeneous System
A system is homogeneous when its right-hand side is the zero vector, so it has the form \(\mathbf{A}\mathbf{z}=\mathbf{0}\). Every homogeneous system has at least the solution \(\mathbf{z}=\mathbf{0}\), since \(\mathbf{A}\mathbf{0}=\mathbf{0}\). Its other solutions can explain whether a nonhomogeneous consistent system has one solution or many.
Proof. First take any solution \(\mathbf{x}\) of \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Set \(\mathbf{z}=\mathbf{x}-\mathbf{x}_p\). Then \[ \mathbf{A}\mathbf{z} =\mathbf{A}(\mathbf{x}-\mathbf{x}_p) =\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{x}_p =\mathbf{b}-\mathbf{b} =\mathbf{0}, \] and \(\mathbf{x}=\mathbf{x}_p+\mathbf{z}\). This proves that every solution is in the stated set. Conversely, if \(\mathbf{A}\mathbf{z}=\mathbf{0}\), then \[ \mathbf{A}(\mathbf{x}_p+\mathbf{z}) =\mathbf{A}\mathbf{x}_p+\mathbf{A}\mathbf{z} =\mathbf{b}+\mathbf{0} =\mathbf{b}, \] so every vector in the stated set is a solution. If the homogeneous system has only \(\mathbf{z}=\mathbf{0}\), the set consists only of \(\mathbf{x}_p\), giving a unique solution. If it has a nonzero solution \(\mathbf{z}\), then \(\mathbf{x}_p\) and \(\mathbf{x}_p+\mathbf{z}\) are distinct solutions, so the system is not unique. By the preceding theorem, a consistent system that is not unique has infinitely many solutions. This proves both directions. \(\square\)
This result gives a structural way to think about classification. A particular solution establishes consistency, while the homogeneous system describes how that solution can vary without changing the right-hand side. If no nonzero homogeneous solution is available, nothing can be added to the particular solution. If there is a nonzero homogeneous solution, the consistent system cannot have a unique answer.
A Common Classification Mistake
One solution does not prove uniqueness. For the system \(x+y=4,\;2x+2y=8\), a student might verify \((x,y)=(2,2)\) and then claim it is the unique solution. The check is correct but the conclusion is not: \((1,3)\) also gives \(1+3=4\) and \(2(1)+2(3)=8\). In fact, every \((t,4-t)\) is a solution. A verified candidate proves consistency; uniqueness requires ruling out every other candidate.
The reverse mistake is also possible: if a proposed pair fails one equation, that shows only that this pair is not a solution. It does not, by itself, show that the system is inconsistent, because another pair might satisfy all equations. To classify a system, distinguish a statement about one candidate from a statement about the entire solution set.
| Solution-set description | Classification | What it tells you |
|---|---|---|
| No vectors satisfy every equation | Inconsistent | There is no solution to verify |
| Exactly one vector satisfies every equation | Consistent, unique solution | A candidate must be the only possible one |
| Infinitely many vectors satisfy every equation | Consistent, infinitely many solutions | At least one solution exists, and others are possible |
Check Your Understanding
For each system, classify its solution set and justify the classification using the equations.
- Classify \(x=2,\;y=5,\;x+y=7\). Is the solution unique?
- Classify \(x+y=6,\;2x+2y=12\). Give two distinct solutions and explain why there are infinitely many.
- Can a system over \(\mathbb{R}\) have exactly three solutions? Use the theorem about two distinct solutions to justify your answer.
- If \(\mathbf{x}_p\) solves \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and \(\mathbf{A}\mathbf{z}=\mathbf{0}\), verify directly that \(\mathbf{x}_p+\mathbf{z}\) solves \(\mathbf{A}\mathbf{x}=\mathbf{b}\).
- A proposed candidate satisfies one equation of a system but fails another. What does this establish, and what does it not establish about the system?
