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A linear system Ax=b is solvable exactly when the coefficient and augmented matrices have the same rank.
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Formula
Algebraic and dimensionless; A is m×n and b has m entries.
| Symbol | Meaning | Unit |
|---|---|---|
| Coefficient matrix of the linear system | dimensionless or model-dependent | |
| Right-hand side vector | unit matching Ax | |
| Dimension of the column or row space | dimensionless | |
| Number of unknowns | dimensionless |
\operatorname{rang}(A)=\operatorname{rang}([A|b])Rank is the number of independent rows or columns, equal to the number of pivot positions in row echelon form. The Rouché–Capelli theorem compares the rank of A with the rank of the augmented matrix. If the system is consistent, its solution is unique when the rank also equals the number n of unknowns; otherwise free variables remain.
For finite-dimensional linear systems over a field, typically the real numbers.
Row operations preserve solutions and rank; an inconsistent row appears exactly when augmentation increases the rank.
Determine consistency and uniqueness
For finite-dimensional linear systems over a field, typically the real numbers.
Solve x+y=2 and 2x+2y=4. How many solutions are there?
A=[[1,1],[2,2]], b=[2,4].
The second row is twice the first, so rank(A)=1.
The same holds for [A|b], hence rank([A|b])=1 and the system is consistent.
There are 2−1=1 free variables, and solutions satisfy x+y=2.
Check (1,1): A·x=(2,4)=b.
Confusing rank with the number of rows rather than independent pivots.
Rank is the pivot count after row reduction and is at most min(m,n).
Formula cluster
The augmented matrix directly shows whether the right-hand side is consistent with the equations.
A=[[1,1],[2,2]], b=[2,4]. The second row is twice the first, so rank(A)=1. The same holds for [A|b], hence rank([A|b])=1 and the system is consistent. There are 2−1=1 free variables, and solutions satisfy x+y=2. Check (1,1): A·x=(2,4)=b.
Checking consistency and the number of solutions, reducing model equations and finding dependencies in data sets or networks.
Here is how to work through a typical Linear systems and matrix rank (rang(A) = rang([A|b])) task step by step:
Task
Solve x+y=2 and 2x+2y=4. How many solutions are there?
Solution path
A=[[1,1],[2,2]], b=[2,4].
The second row is twice the first, so rank(A)=1.
The same holds for [A|b], hence rank([A|b])=1 and the system is consistent.
There are 2−1=1 free variables, and solutions satisfy x+y=2.
Check (1,1): A·x=(2,4)=b.
Rank counts independent directions of information. A row that is only a multiple of another adds no new condition. In row echelon form, rank is the number of pivots; geometrically it is the dimension of the column space. Each pivot column represents an independent variable direction in the image of A. Rank therefore counts information, not merely the number of written equations. Swapping rows or adding a multiple of one row to another changes neither the solution set nor rank. After comparing ranks, determine the solution type from the free variables as well. For a concrete solution, choose allowed parameter values and substitute into every original equation.
Appending b can introduce a new independent direction. If rank increases, row reduction produces an equation 0=c with c≠0, a contradiction. If the rank stays equal, the system is consistent. Gaussian elimination exposes a contradiction as 0=nonzero constant. In that case, the augmented matrix has an extra pivot in its last column. Equal ranks rule out this contradiction, but do not yet tell whether the solution is unique. The rank test is invariant under elementary row operations. You may reduce the augmented matrix as long as every operation is also applied to the right-hand side.
It must be consistent and each of the n unknowns must be determined by a pivot. This occurs exactly when rank(A)=rank([A|b])=n. A smaller rank leaves free variables and therefore infinitely many solutions if the system is consistent. Uniqueness requires a pivot for every variable, not necessarily every equation row. A system with many dependent rows can still have a unique solution if A has full column rank and the right-hand side is consistent. Fewer pivots leave free parameters. For an invertible square matrix, full rank is equivalent to a nonzero determinant; elimination reveals the rank directly.
For a consistent system with n unknowns and rank r, there are n−r free variables. In x+y=2 and 2x+2y=4, r=1, so one variable is free. Set x=t; then y=2−t. Choose a free variable as a parameter and express the pivot variables in terms of it. This gives a parameterized solution set rather than one solution. Substituting into Ax=b checks that no row operation was copied incorrectly. Free parameters describe nullspace directions along which a solution can move without changing Ax=b. This explains the count n−rank(A).
Yes. Extra equations can be redundant. With n unknowns, a unique solution is possible if all rows are consistent and rank(A)=n. Extra dependent equations do not increase rank. With measured data, equations often conflict slightly because of rounding or noise. Then Ax=b has no exact solution and one seeks an approximation, for example by least squares. Rank still matters: it shows whether parameters are identifiable or data directions are missing. If the system is inconsistent, least squares is a different problem, not proof of exact solvability. Report residuals and model assumptions separately.
Quanta exam set
Curated exam set for "Linear systems and matrix rank":
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Which formula describes Linear systems and matrix rank?
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How do you rearrange Linear systems and matrix rank for Determine consistency and uniqueness?
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Which common mistake happens with Linear systems and matrix rank?
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