Solvability: Existence & Uniqueness

The Two Questions

Given the system

A\mathbf{x} = \mathbf{b}

with

A

of size

m \times n

, two logically independent questions arise.

Existence: is there at least one vector

\mathbf{x}

satisfying

A\mathbf{x} = \mathbf{b}

? This asks whether

\mathbf{b}

lies in the column space of

A

.

Uniqueness: if a solution exists, is it the only one? This asks whether the null space of

A

is trivial.

The two questions are independent — existence can hold without uniqueness, and non-existence makes uniqueness moot. Both are answered by the rank of

A

.

Solvability Notation

Notation

Solvability Notation

Two words that classify a system before any solving happens, a rank comparison that answers the existence question without producing a single solution, and a count of solutions that is never anything but zero, one, or infinity. All catalogued among the linear algebra symbols.

A\mathbf{x} = \mathbf{b}

, the augmented matrix

[A \mid \mathbf{b}]

and its bar come from Gaussian elimination notation;

\operatorname{rank}

from rank notation; free variables and pivot positions from echelon form notation;

\operatorname{Col}

and

\operatorname{Nul}

from fundamental-space notation.

\text{consistent / inconsistent}

the system has at least one solution — or none at all

A two-word classification applied to the system, not to any solution: consistent means at least one exists, inconsistent means none does. The words settle existence only and say nothing about how many, which is why The Two Questions above keeps existence and uniqueness apart.

CasesThe marks that detect inconsistency are visual: a row reading

[\,0 \; \cdots \; 0 \mid d\,]

with

d \neq 0

asserts

0 = d

, so a single such row in the augmented matrix decides the whole system.

Do not confuse"Inconsistent" as a comment on the equations being contradictory in some loose sense. It is a precise verdict about the solution set being empty — and a system can look perfectly reasonable and still be inconsistent, which is what the rank test below is for.

\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}])

the ranks agree, so the system is consistent

Existence written as a comparison rather than a search: appending

\mathbf{b}

adds no new independent direction exactly when

\mathbf{b}

already lies in the column space. The test answers whether a solution exists while producing none — the point of The Existence Condition below.

CasesAdding the second rank gives the whole classification, the Rouché–Capelli statement below: ranks unequal means no solution, ranks equal to

n

means exactly one, ranks equal but less than

n

means infinitely many — with

n - \operatorname{rank}(A)

counting the free parameters.

Do not confuseReading

\operatorname{rank}([A \mid \mathbf{b}])

as a rank of two things. The bar is punctuation, not an operation — the argument is one matrix that happens to have been assembled from a coefficient block and a column.

0

,

1

, or

\infty

a linear system has no solutions, exactly one, or infinitely many

The count of solutions is not an arbitrary number: The Three Cases Combined below shows only these three occur. A system with two distinct solutions has infinitely many, because the whole line through them solves it as well — linearity forbids any finite count above one.

CasesThe infinite case is written as a parametrised family rather than the bare

\infty

— a particular solution plus the null space, which is why "infinitely many" still admits a precise description and a dimension.

Do not confuseExpecting "more equations than unknowns" to force zero and "fewer" to force infinitely many. Neither holds: an overdetermined system can be consistent and an underdetermined one inconsistent — the shape suggests the outcome, the ranks decide it.

All linear algebra symbols →↑ Linear Systems

The Existence Condition

The system

A\mathbf{x} = \mathbf{b}

has at least one solution if and only if

The Existence Condition

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}])

When this condition holds,

\mathbf{b}

is already a linear combination of the columns of

A

, so appending

\mathbf{b}

as an extra column does not introduce a new independent direction — the rank stays the same.

When the condition fails —

\text{rank}([A \mid \mathbf{b}]) > \text{rank}(A)

— the vector

\mathbf{b}

is not in the column space. In the echelon form of the augmented matrix, this appears as a row

[0 \; 0 \; \cdots \; 0 \mid d]

with

d \neq 0

: the equation

0 = d

has no solution, and the system is inconsistent.

Since appending one column can increase the rank by at most

1

, the only possibilities are

\text{rank}([A \mid \mathbf{b}]) = \text{rank}(A)

(consistent) or

\text{rank}([A \mid \mathbf{b}]) = \text{rank}(A) + 1

(inconsistent).

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Only two possibilities

Appending

\mathbf{b}

as an extra column can raise the rank by at most

1

. So comparing the two ranks leaves exactly two outcomes.

Existence

A\mathbf{x} = \mathbf{b}

split by $\text{rank}([A \mid \mathbf{b}])$ compared with $\text{rank}(A)$

✓

Case 1

\text{rank}([A \mid \mathbf{b}]) = \text{rank}(A)

Consistent

\mathbf{b}

is a linear combination of the columns; appending it adds no new independent direction

✗

Case 2

\text{rank}([A \mid \mathbf{b}]) = \text{rank}(A) + 1

Inconsistent

\mathbf{b}

is not in the column space; echelon form shows a row

[0 \; \cdots \; 0 \mid d]

with

d \neq 0

Takeaway

Existence is a question about

\mathbf{b}

alone: is it already a linear combination of the columns of

A

?

Existence: comparing the two ranks·/linear-algebra/linear-systems/solvabilityLearn Math Class

b in the column space: consistent

Here the columns are v₁ = (1, 0, 1) and v₂ = (0, 1, 1), and the target is w = (2, 3, 5). Reducing [V | w] produces no row of the form 0 = nonzero: the last column carries no pivot, so appending w did not raise the rank, and w is a combination of the columns — the system has a solution. Try other targets on the span and membership visualizer.

Existence is a question about b alone: is it one of the vectors the columns can reach?

The Uniqueness Condition

When a solution exists, it is unique if and only if

The Uniqueness Condition

\text{rank}(A) = n

where

n

is the number of unknowns. Full column rank means every column of

A

contains a pivot, leaving no free variables. The null space is

\{\mathbf{0}\}

, so the particular solution

\mathbf{x}_p

stands alone with nothing to add.

When

\text{rank}(A) < n

, there are

n - \text{rank}(A)

free variables. Each free variable parametrizes a direction along which the solution can move without violating the equations. The solution set is infinite — a translated copy of the null space, which has dimension

n - \text{rank}(A)

.

Pitfall: Skipping the existence check

Concluding "exactly one solution" from

\text{rank}(A) = n

alone fails for

x_1 = 1

,

x_1 = 2

: here

\text{rank}(A) = 1 = n

, yet

\text{rank}([A \mid \mathbf{b}]) = 2

and there is no solution. Uniqueness can be claimed only after

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}])

has been checked.

The rank test for uniqueness looks only at the coefficient matrix. It compares the number of pivot columns of

A

with the number of unknowns, and

\mathbf{b}

plays no part in it. The same count comes out for every right-hand side the system could be given.

Dependent columns: no uniqueness

Three columns with v₃ = v₁ + v₂. The reduction finds only two pivots, so the rank is 2 while there are 3 unknowns: one column is free. Whenever a solution exists it can be shifted along that free direction, so it is never unique. Uniqueness needs rank equal to the number of unknowns. Explore dependent sets on the span and membership visualizer.

Uniqueness is a question about A alone: whatever b is, a free column always leaves room to move.

The Three Cases Combined

Putting existence and uniqueness together, every linear system falls into exactly one of three cases.

\text{rank}(A) < \text{rank}([A \mid \mathbf{b}])

: no solution. The system is inconsistent.

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) = n

: exactly one solution. The system is consistent and fully determined.

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) < n

: infinitely many solutions. The system is consistent but underdetermined, with

n - \text{rank}(A)

free parameters.

THEOREM

There is no case with a finite number of solutions greater than one.

Fact: Exactly two solutions, in arithmetic mod 2

With arithmetic mod

2

, where

1 + 1 = 0

, the equation

x_1 + x_2 = 1

has exactly two solutions,

(1, 0)

and

(0, 1)

. The line through two solutions has one point per scalar, and mod

2

there are only two scalars.

If two distinct solutions exist, their difference lies in the null space, which is a subspace — and a nontrivial subspace contains infinitely many vectors, generating infinitely many solutions.

The argument uses only one property of the scalars: there are infinitely many of them. The points

\mathbf{x}_1 + t(\mathbf{x}_2 - \mathbf{x}_1)

all solve the system, because

A(\mathbf{x}_2 - \mathbf{x}_1) = \mathbf{b} - \mathbf{b} = \mathbf{0}

, and different values of

t

give different points since

\mathbf{x}_2 - \mathbf{x}_1 \neq \mathbf{0}

. The number of solutions is therefore at least the number of scalars available for

t

.

The three endings, side by side

Existence fails only in the middle case, where a contradiction row appears. Uniqueness fails only in the last, where a free column leaves an unknown unpinned. The two questions are independent of each other, and between them they exhaust what can happen. Build a system for each ending on the linear system solutions visualizer.

The rank conditions in the next section are these three pictures restated as arithmetic on two numbers.

The Rouché–Capelli Theorem

The existence condition

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}])

is known as the Rouché–Capelli theorem (or the Kronecker–Capelli theorem in some traditions):

Solvability Rank Criterion

A\mathbf{x} = \mathbf{b} \text{ consistent} \iff \text{rank}(A) = \text{rank}([A \mid \mathbf{b}])

Solvability rank criterion →

When consistent, the solution set has dimension

n - \text{rank}(A)

.

The theorem unifies the existence and dimension questions into a single rank comparison. It applies to every linear system regardless of the shape of

A

— square, tall, or wide. The proof follows directly from the column-space interpretation:

\mathbf{b} \in \text{Col}(A)

if and only if adding

\mathbf{b}

as a column does not increase the rank.

The rank goes up when b is appended

The off-plane preset: the target lies outside the plane of the columns. Reducing [V | w] puts a pivot in the last column — a row that reads 0 = nonzero — so rank([V | w]) is one more than rank(V). That jump is exactly the Rouché–Capelli condition failing, and the system has no solution. Move the target on and off the plane on the span and membership visualizer.

Comparing two ranks is all the theorem asks, and row reduction computes both at once.

Square Systems

When

A

is

n \times n

, the determinant provides the sharpest diagnostic.

THEOREM

If

\det(A) \neq 0

: the matrix is invertible, the rank is

n

, and the system

A\mathbf{x} = \mathbf{b}

has exactly one solution for every

\mathbf{b}

.

The solution is

\mathbf{x} = A^{-1}\mathbf{b}

, or equivalently, the solution obtained by Gaussian elimination. Existence and uniqueness both hold universally — the right-hand side does not matter.

Pitfall: Reading a zero determinant as no solution

Concluding "no solution" from

\det(A) = 0

fails for

x_1 + 2x_2 = 3

,

2x_1 + 4x_2 = 6

: the determinant is

0

, yet

x_1 = 3 - 2t

,

x_2 = t

solves it for every

t

. Only a

\mathbf{b}

outside

\text{Col}(A)

, such as

(3, 5)

, leaves no solution.

If

\det(A) = 0

: the matrix is singular, the rank is less than

n

, and the outcome depends on

\mathbf{b}

. For some

\mathbf{b}

(those in the column space), infinitely many solutions exist. For other

\mathbf{b}

(those outside the column space), no solution exists. The determinant test determines whether the coefficient matrix is adequate; the rank comparison with the augmented matrix determines whether the specific

\mathbf{b}

is reachable.

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The sharpest diagnostic

When

A

is

n \times n

, the determinant decides whether the coefficient matrix is adequate; when it is not, the rank comparison with the augmented matrix decides whether the specific

\mathbf{b}

is reachable.

Square system

A\mathbf{x} = \mathbf{b}

,

A

is

n \times n

split by $\det(A)$

1

Case 1

\det(A) \neq 0

Exactly one

solution for every

\mathbf{b}

the matrix is invertible and the rank is

n

; existence and uniqueness both hold universally

\mathbf{x} = A^{-1}\mathbf{b}

∞ or 0

Case 2

\det(A) = 0

Infinitely many or none

depends on

\mathbf{b}

the matrix is singular and the rank is less than

n

: infinitely many solutions for

\mathbf{b}

in the column space, none for

\mathbf{b}

outside it

Takeaway

With

\det(A) \neq 0

the right-hand side does not matter. With

\det(A) = 0

everything depends on whether

\mathbf{b}

lies in the column space.

Square systems: what the determinant decides·/linear-algebra/linear-systems/solvabilityLearn Math Class

Overdetermined Systems

When

m > n

— more equations than unknowns — the system is overdetermined. The coefficient matrix is tall, and generically, no solution exists.

The column space of

A

is at most

n

-dimensional inside

\mathbb{R}^m

. When

m > n

, this column space is a proper subspace — most vectors in

\mathbb{R}^m

lie outside it. A randomly chosen

\mathbf{b}

will almost certainly not be in the column space, making the system inconsistent.

Pitfall: Deciding existence from the shape alone

Concluding "no solution" because

m = 3 > n = 2

fails for

x_1 + x_2 = 2

,

x_1 - x_2 = 0

,

2x_1 = 2

: both ranks equal

2

, and

\mathbf{x} = (1, 1)

is the unique solution. Only the rank comparison decides.

A solution exists only when

\mathbf{b}

happens to lie in the column space — when the extra equations are consistent with the first

n

. When a solution does exist and

\text{rank}(A) = n

, it is unique.

When no exact solution exists, the least-squares approach finds the

\hat{\mathbf{x}}

that minimizes

\|A\mathbf{x} - \mathbf{b}\|^2

— the closest approximation. The least-squares solution satisfies the normal equations

A^T A \hat{\mathbf{x}} = A^T \mathbf{b}

and

A\hat{\mathbf{x}}

is the projection of

\mathbf{b}

onto the column space.

Extra equations rarely agree, so a tall system usually has no exact solution and least squares settles for the nearest one.

Underdetermined Systems

COROLLARY

When

m < n

— fewer equations than unknowns — the system is underdetermined. If a solution exists, it is never unique: the rank cannot exceed

m

, which is less than

n

, so at least

n - m

free variables remain.

Pitfall: Assuming a wide system always has solutions

Expecting infinitely many solutions because

m = 2 < n = 3

fails for

x_1 + x_2 + x_3 = 1

,

x_1 + x_2 + x_3 = 2

:

\text{rank}(A) = 1

but

\text{rank}([A \mid \mathbf{b}]) = 2

, so there is no solution. A wide shape rules out uniqueness, not inconsistency.

The solution set, when nonempty, is an affine subspace of dimension at least

n - m

. Multiple vectors

\mathbf{x}

satisfy the equations, and additional criteria beyond the linear system itself are needed to select a preferred one.

Common selection criteria include minimum norm (the

\mathbf{x}

closest to

\mathbf{0}

, given by the pseudoinverse

\mathbf{x} = A^+\mathbf{b}

, which equals

A^T(AA^T)^{-1}\mathbf{b}

when

A

has full row rank), sparsity (the

\mathbf{x}

with the fewest nonzero entries, central to compressed sensing), and physical constraints (bounds or nonnegativity in engineering applications).

Fact: Ten equations, a thousand unknowns

A

10 \times 1000

system whose columns are independent ten at a time has at most one solution with

5

or fewer nonzero entries. Two such solutions would differ by a null-space vector with at most

10

nonzero entries, and no

10

columns can cancel.

Sparsity deserves a closer look, because it changes how much information a solution carries. A vector in

\mathbb{R}^{1000}

with only

5

nonzero entries is described by

10

numbers: where those entries sit and what they are. Compressed sensing measures such a vector with far fewer equations than unknowns and then looks, among the infinitely many solutions, for the sparsest one.

Underdetermined systems are not defective — they arise naturally whenever a problem has more degrees of freedom than constraints. The linear system identifies the feasible set; the selection criterion picks a point within it.

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Linear systems · by shape

What the shape of A predicts

Before any reduction, the shape of $A$ narrows what can happen. It never settles the question — $\mathbf{b}$ still decides existence — but it says which outcomes are available and which are ruled out.

4shapes

Type	Condition	existence	uniqueness	practical approach
Square
Nonsingular§ 6	$m = n$ , $\det A \neq 0$	always, for every $\mathbf{b}$	always unique	factorize once, solve per $\mathbf{b}$
Singular§ 6	$m = n$ , $\det A = 0$	iff $\mathbf{b} \in \operatorname{Col}(A)$	never, when it exists	reduce $[A \mid \mathbf{b}]$ , compare ranks
More equations than unknowns
Overdetermined§ 7	$m > n$ — tall	unusual — most $\mathbf{b}$ miss $\operatorname{Col}(A)$	unique when it exists and $\operatorname{rank}(A) = n$	least squares when inconsistent
More unknowns than equations
Underdetermined§ 8	$m < n$ — wide	iff $\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}])$	never — at least $n - m$ free parameters	impose a criterion: minimum norm, sparsity

Shape constrains, rank decides. A wide matrix cannot have a unique solution whatever

\mathbf{b}

is; a tall one usually has none. What no shape determines is existence, which depends on whether

\mathbf{b}

lies in the column space — and that is a fact about

\mathbf{b}

, not about the dimensions.

What the shape of A predicts·/linear-algebra/linear-systems/solvabilityLearn Math Class

Read down the uniqueness column and the shape does more work than down the existence column. A wide matrix can never have a unique solution — rank is capped by

m

, which is below

n

, so free variables are guaranteed before

\mathbf{b}

is even mentioned. Existence is different: every row except the first says "depends on

\mathbf{b}

", because whether

\mathbf{b}

lies in the column space is not a fact about the dimensions of

A

at all.

Geometric Interpretation

Each equation

a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n = b_i

defines a hyperplane in

\mathbb{R}^n

— a flat set of dimension

n - 1

. The solution set of the full system is the intersection of all

m

hyperplanes.

When the system is inconsistent, the hyperplanes have no common point. In

\mathbb{R}^2

this means parallel lines. In

\mathbb{R}^3

it can mean parallel planes, or planes arranged in a triangular prism where each pair intersects but no point lies on all three.

When the system has a unique solution, the hyperplanes meet at a single point. This requires at least

n

independent equations — enough to cut the solution space down from

n

dimensions to

0

.

When the system has infinitely many solutions, the hyperplanes meet along a flat of dimension

n - r

, where

r = \text{rank}(A)

. Each independent equation reduces the dimension of the intersection by one, and the rank counts how many equations cut independently. The remaining

n - r

dimensions are the free directions in the solution set.

Solving a linear system is intersecting hyperplanes, and each independent equation cuts the intersection down by one dimension.

Structure of the Solution Set

When

A\mathbf{x} = \mathbf{b}

is consistent, the complete solution set has the form

Solution Structure Decomposition

\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h, \quad \mathbf{x}_h \in \text{Null}(A)

Solution structure decomposition →

where

\mathbf{x}_p

is any one particular solution and

\mathbf{x}_h

ranges over the null space. Equivalently, the solution set is

\{\mathbf{x}_p + \mathbf{x}_h : \mathbf{x}_h \in \text{Null}(A)\}

— a coset of the null space, the null space translated by

\mathbf{x}_p

. In geometric terms, it is an affine subspace: a subspace shifted away from the origin.

The dimension of the solution set equals the dimension of the null space:

n - \text{rank}(A)

. When this is

0

, the solution set is a single point

\{\mathbf{x}_p\}

. When it is

1

, the solution set is a line. When it is

2

, a plane. The shape is always a flat, and the null space determines its orientation.

Pitfall: Dropping the particular solution

Writing the solutions of

x_1 + x_2 = 2

as

\mathbf{x} = t(-1, 1)

gives only the null space: at

t = 0

it offers

\mathbf{0}

, and

0 + 0 \neq 2

. Adding

\mathbf{x}_p = (2, 0)

gives the solution set

\mathbf{x} = (2, 0) + t(-1, 1)

.

The particular solution

\mathbf{x}_p

captures the effect of the right-hand side

\mathbf{b}

. The homogeneous component

\mathbf{x}_h

captures the inherent freedom in the system — the directions along which the solution can shift without violating any equation. This decomposition into "forced" and "free" parts is one of the most fundamental structural results in linear algebra.

Component	Symbol	Where it comes from	What it captures
Particular solution	x_p	any one vector satisfying A x_p = b	the "forced" part; the response to the right-hand side b
Homogeneous part	x_h	any vector in Null(A), i.e. A x_h = 0	the "free" directions along which the solution can shift
Complete solution	x = x_p + x_h	sum over all x_h ∈ Null(A)	affine flat of dimension n − rank(A) through x_p

Summary: Solvability Questions Answered

The two foundational questions about A·x = b — existence and uniqueness — together with the structure of the solution set and the special behavior of square, overdetermined, and underdetermined shapes, can be collected as a six-row Q&A reference. The table below pairs each question a reader might bring to a linear system with the answer in terms of rank, determinant, or solution-set structure.

The first group below is the theory and the second is what to do about it. Existence and uniqueness are each a single rank comparison, and they are independent — a system can have solutions without having one solution, or have at most one without having any. The second group takes the cases where the answer is unwelcome and says what replaces the original question.

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Linear systems · solvability

Every solvability question, answered by rank

Two comparisons settle existence and uniqueness completely. Everything below them is what to do once the answer is known — and the shape of the system determines which of those situations you are in.

6questions

The two questions3

1

Does a solution exist?§ 2

equivalently

\mathbf{b} \in \operatorname{Col}(A)

\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}])

Rouché–Capelli. If the augmented column raises the rank, it contributes a direction the columns of

A

cannot reach, and the reduction reports this as a row reading

0 = d

. Existence is a question about

\mathbf{b}

, not about

A

.

2

Is the solution unique?§ 3

equivalently

\operatorname{Null}(A) = \{\mathbf{0}\}

\operatorname{rank}(A) = n

A pivot in every column means no free variables. Note this says nothing about whether a solution exists — uniqueness is a question about

A

alone, which is why the two conditions are independent and both must be checked.

3

What does the solution set look like?§ 10

when a solution exists

\mathbf{x}_p + \operatorname{Null}(A)

One particular solution plus the entire null space. The set is a point when the nullity is zero, a line when it is one, a plane when it is two — never anything else, because it is always a translated subspace.

What the shape of the system implies3

4

Square and nonsingular§ 6

m = n

and

\det A \neq 0

\mathbf{x} = A^{-1}\mathbf{b}

Both conditions hold at once and for every

\mathbf{b}

— the only case where the answer is unconditional. In practice solve by factorization rather than by forming the inverse; the formula is a statement about existence, not a method.

5

Overdetermined, inconsistent§ 7

m > n

, no exact solution

A^{\mathsf{T}}A\hat{\mathbf{x}} = A^{\mathsf{T}}\mathbf{b}

More equations than unknowns usually means no solution at all, so the question changes: least squares minimises

\|A\mathbf{x} - \mathbf{b}\|^2

instead of solving. The answer makes

A\hat{\mathbf{x}}

the projection of

\mathbf{b}

onto the column space — the closest reachable point.

6

Underdetermined§ 8

m < n

, consistent

infinitely many; choose one

Fewer equations than unknowns leaves free variables, so a solution is never unique and the algebra cannot pick between them. The criterion comes from outside: minimum norm via the pseudoinverse, sparsity, or a physical constraint the model imposes.

The whole of solvability is two numbers compared three ways:

\operatorname{rank}(A)

against

\operatorname{rank}([A \mid \mathbf{b}])

decides existence, and

\operatorname{rank}(A)

against

n

decides uniqueness. Both come from one reduction, which is why the two questions are never asked separately in practice.

Every solvability question, answered by rank·/linear-algebra/linear-systems/solvabilityLearn Math Class

Both comparisons come out of the same reduction, which is why they are never computed separately. Reduce

[A \mid \mathbf{b}]

once: a contradiction row settles existence, and the count of pivot columns settles uniqueness. Nothing else is needed, and no determinant is involved — which matters because the determinant test only applies to square systems, while these two apply to every system.

The last two rows are worth reading as a change of question rather than a failure. An overdetermined system usually has no solution, so least squares asks for the closest approximation instead and answers with a projection. An underdetermined system has too many, so a selection criterion is imposed from outside the algebra — minimum norm, sparsity, or whatever the model requires. In both cases the linear algebra is finished; what remains is deciding what you wanted.

Common Pitfalls

Five mistakes cause many wrong verdicts about linear systems. Each is marked with a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Skipping the existence check

A unique solution is announced as soon as

\text{rank}(A) = n

, without checking that a solution exists at all. Full column rank says there is at most one solution; it does not say there is one.

Wrong working: for the system

x_1 = 1

,

x_1 = 2

,

A = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad \text{rank}(A) = 1 = n \quad \Longrightarrow \quad \text{exactly one solution}

Right working: reduce the augmented matrix first.

\left(\begin{array}{c|c} 1 & 1 \\ 1 & 2 \end{array}\right) \sim \left(\begin{array}{c|c} 1 & 1 \\ 0 & 1 \end{array}\right)

The second row reads

0 = 1

, so

\text{rank}([A \mid \mathbf{b}]) = 2 > \text{rank}(A)

and there is no solution.

Why it happens: for a square matrix with

\det(A) \neq 0

, full rank does guarantee a solution, and the habit carries over to tall matrices, where it does not.

How to check: ask the existence question first. Look for a row

[0 \; \cdots \; 0 \mid d]

with

d \neq 0

in the reduced augmented matrix before counting pivot columns.

Reading a zero determinant as no solution

\det(A) = 0

is taken to mean that

A\mathbf{x} = \mathbf{b}

has no solution. A zero determinant says only that a solution, when there is one, is not unique; whether one exists depends on

\mathbf{b}

.

Wrong working:

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}, \quad \det(A) = 4 - 4 = 0 \quad \Longrightarrow \quad x_1 + 2x_2 = 3, \; 2x_1 + 4x_2 = 6 \text{ has no solution}

Right working: reduce the augmented matrix.

\left(\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 4 & 6 \end{array}\right) \sim \left(\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right)

Both ranks are

1

, so the system is consistent, with one free variable:

x_2 = t

,

x_1 = 3 - 2t

. With

\mathbf{b} = (3, 5)

instead, the second row becomes

[0 \; 0 \mid -1]

and there is no solution.

Why it happens: for a single equation

ax = b

,

a = 0

usually means no solution, and

\det(A) = 0

is read the same way. The case

0x = 0

, true for every

x

, is the one that gets forgotten.

How to check: after finding

\det(A) = 0

, compare

\text{rank}(A)

with

\text{rank}([A \mid \mathbf{b}])

. Equal ranks mean infinitely many solutions, unequal ranks mean none.

Deciding existence from the shape alone

A system with more equations than unknowns is declared inconsistent without any reduction. Most right-hand sides do make a tall system inconsistent, but some do not.

Wrong working:

x_1 + x_2 = 2, \quad x_1 - x_2 = 0, \quad 2x_1 = 2 \qquad m = 3 > n = 2 \quad \Longrightarrow \quad \text{no solution}

Right working: reduce the augmented matrix.

\left(\begin{array}{cc|c} 1 & 1 & 2 \\ 1 & -1 & 0 \\ 2 & 0 & 2 \end{array}\right) \sim \left(\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & -2 & -2 \\ 0 & 0 & 0 \end{array}\right)

Both ranks are

2 = n

, so there is exactly one solution:

x_2 = 1

,

x_1 = 1

.

Why it happens: the third equation looks like one constraint too many, and a randomly chosen right-hand side would indeed break the system. Here the third equation is the sum of the first two, so it adds nothing new.

How to check: substitute the candidate solution into every equation.

\mathbf{x} = (1, 1)

gives

2

,

0

and

2

, matching all three right-hand sides.

Assuming a wide system always has solutions

A system with fewer equations than unknowns is expected to have infinitely many solutions. The shape guarantees free variables, so a solution is never unique, but it does not guarantee that a solution exists.

Wrong working:

x_1 + x_2 + x_3 = 1, \quad x_1 + x_2 + x_3 = 2 \qquad m = 2 < n = 3 \quad \Longrightarrow \quad \text{infinitely many solutions}

Right working: reduce the augmented matrix.

\left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 \end{array}\right) \sim \left(\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right)

The second row reads

0 = 1

, so

\text{rank}(A) = 1

but

\text{rank}([A \mid \mathbf{b}]) = 2

, and there is no solution.

Why it happens: "more unknowns than equations" is remembered as "always solvable", and the condition that a solution must exist first is dropped.

How to check: before counting free variables, check that the reduction produced no row

[0 \; \cdots \; 0 \mid d]

with

d \neq 0

. Two equations with the same left side and different right sides, as here, always produce one.

Dropping the particular solution

The solution set of

A\mathbf{x} = \mathbf{b}

is written using only the free directions, as if

\mathbf{b}

were

\mathbf{0}

. That describes the null space, the solution set of the homogeneous system, not the solutions of

A\mathbf{x} = \mathbf{b}

.

Wrong working: for

x_1 + x_2 = 2

, the free variable

x_2 = t

gives the direction

(-1, 1)

, and the answer is written as

\mathbf{x} = t\begin{pmatrix} -1 \\ 1 \end{pmatrix}

Right working:

x_1 = 2 - t

, so the constant

2

stays in the answer.

\mathbf{x} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \end{pmatrix}

This is a line through

(2, 0)

, parallel to the null space and not through the origin.

Why it happens: the free directions are the part of the answer that takes work to find, and the constant column is easy to leave behind. For homogeneous systems the particular solution really is

\mathbf{0}

, which reinforces the habit.

How to check: set every parameter to

0

. The result must solve

A\mathbf{x} = \mathbf{b}

; the wrong answer gives

\mathbf{0}

, and

0 + 0 \neq 2

.

Solvability FAQ

Does having more equations than unknowns mean there is no solution?

+

No, and the converse fails too. An overdetermined system can be perfectly consistent, and an underdetermined one can be inconsistent. The shape of a system suggests an outcome without deciding it; comparing the rank of the coefficient matrix with the rank of the augmented matrix is what actually settles the question.Read more →

Is $\operatorname{rank}([A \mid \mathbf{b}])$ the rank of two things?

+

No, it is the rank of one matrix that happens to have been assembled from a coefficient block and an extra column. The bar is punctuation rather than an operation, so nothing is being combined or compared inside the brackets. The comparison happens outside, between this number and

\operatorname{rank}(A)

.Read more →

What does "inconsistent" mean exactly?

+

That the solution set is empty, which is a precise verdict rather than a loose remark about the equations disagreeing. A system can look entirely reasonable and still be inconsistent. The visual tell during elimination is a row reading all zeros to the left of the bar with a nonzero entry to its right, asserting

0 = d

.Read more →

Solvability of Linear Systems

When Solutions Exist and When They Are Unique

The Two Questions

Solvability Notation

Solvability Notation

The Existence Condition

The Uniqueness Condition

The Three Cases Combined

The Rouché–Capelli Theorem

Square Systems

Overdetermined Systems

Underdetermined Systems

What the shape of A predicts

Geometric Interpretation

Structure of the Solution Set

Summary: Solvability Questions Answered

Every solvability question, answered by rank

Common Pitfalls

Solvability FAQ

Does having more equations than unknowns mean there is no solution?

Is $\operatorname{rank}([A \mid \mathbf{b}])$ the rank of two things?

What does "inconsistent" mean exactly?

Solvability of Linear Systems

When Solutions Exist and When They Are Unique

The Two Questions

Solvability Notation

Solvability Notation

The Existence Condition

The Uniqueness Condition

The Three Cases Combined

The Rouché–Capelli Theorem

Square Systems

Overdetermined Systems

Underdetermined Systems

What the shape of A predicts

Geometric Interpretation

Structure of the Solution Set

Summary: Solvability Questions Answered

Every solvability question, answered by rank

Common Pitfalls

Solvability FAQ

Does having more equations than unknowns mean there is no solution?

Is rank⁡([A∣b])\operatorname{rank}([A \mid \mathbf{b}])rank([A∣b]) the rank of two things?

What does "inconsistent" mean exactly?

Is $\operatorname{rank}([A \mid \mathbf{b}])$ the rank of two things?