Linear systems: row reduction, rank and nullity

Objectives, prerequisites and scope

Represent a linear system as Ax=b, reduce its augmented matrix, test consistency and describe every solution using free parameters. Prerequisites: matrix operations and determinants. Rank and nullity are supporting tools under the official Matrices and System of linear equations headings. Work over R; solution counts over finite fields require different interpretation.

1. The augmented matrix keeps the right-hand side

A system of m equations in n unknowns is Ax=b with A of size m×n, x of size n×1 and b of size m×1. The augmented matrix [A|b] has n+1 columns. Row operations must act on the entire augmented row, including b. Allowed operations are swapping equations, multiplying an equation by a nonzero scalar, and adding a multiple of another equation. Each is reversible and preserves the solution set.

Column operations are different: they change the variables or their order and do not preserve the same coordinate solution set unless that change is tracked. Do not silently use column operations to solve Ax=b as though they were row operations.

2. Pivots and row-echelon form

A pivot is the first nonzero entry of a nonzero row in echelon form. As rows proceed downward, pivot positions move strictly right and zero rows come last; entries below each pivot are zero. Reduced row-echelon form further makes each pivot one and its column zero elsewhere. Echelon form permits back substitution; reduced form makes parameterization especially direct.

Worked unique solution: x+2y=5 and 3x+5y=13 give [[1,2|5],[3,5|13]]. R₂←R₂−3R₁ gives [0,−1|−2], so y=2. The first equation then gives x=1. Check both originals: 1+4=5 and 3+10=13. Row reduction is a method; substitution into the original equations is an independent check.

3. Rank and consistency

Rank(A) is the number of pivots in any echelon form of A. It equals both the maximum number of independent rows and the maximum number of independent columns. Thus 0≤rank(A)≤min(m,n). Nonzero rows must be counted after reduction, not in the original matrix: proportional nonzero rows are dependent.

The system is consistent exactly when rank(A)=rank([A|b]). A row [0 … 0 | c] with c≠0 states 0=c and proves inconsistency; the augmented column has created an extra pivot. If consistent with rank r, there are n−r free variables. If r=n, the solution is unique. If r<n, infinitely many real solutions exist. The number of equations alone cannot distinguish these cases.

Example: x+y=1 and 2x+2y=3 reduce to [0,0|1], so there is no solution. Changing the second right side to 2 produces a zero row instead and infinitely many solutions (x,y)=(1−t,t). Both systems have the same singular coefficient matrix; the right-hand side decides consistency.

4. Parameterizing all solutions and nullity

For x+2y−z=3 and y+z=1, choose the nonpivot variable z=t. Then y=1−t and x=1+3t. Every solution is

(x,y,z)ᵀ=(1,1,0)ᵀ+t(3,−1,1)ᵀ, t∈R.

The first vector is one particular solution xₚ; the direction vector solves the homogeneous system Ax=0. In general, when Ax=b is consistent, every solution is xₚ+z with z in the nullspace ker(A)={z:Az=0}. Subtract two particular solutions to see why their difference is homogeneous.

A homogeneous system always has the zero solution. It has a nonzero solution exactly when rank(A)<n. The nullity (dimension of ker(A)) is n−rank(A), the rank–nullity relation. For a 4×6 matrix of rank 3, the nullspace has dimension 3, not 4−3. A consistent nonhomogeneous solution set is generally an affine translate, not a vector space: if b≠0 it does not contain the zero vector.

5. Parameters require case splits

For x+y=1 and 2x+ty=s, elimination gives (t−2)y=s−2. If t≠2, y=(s−2)/(t−2) and x=1−y: unique. If t=2 and s=2, the equations are dependent and there is one free parameter. If t=2 and s≠2, the system is inconsistent. Dividing by t−2 before considering zero would discard both exceptional cases.

For square n×n A, the following are equivalent: det(A)≠0; rank(A)=n; Ax=0 has only the zero solution; A is invertible; and Ax=b has exactly one solution for every b. For a rectangular matrix, full column rank guarantees uniqueness when a solution exists, but not consistency for every b. Full row rank guarantees consistency for every b, but may leave free variables. Example: a 2×3 matrix of rank 2 has a solution for every two-entry b and at least one free variable.

Solving checklist and pitfalls

Write the unknown order, reduce [A|b], and identify contradictions before counting free variables. Count pivots in A, not automatically in the augmented matrix. Express every nonpivot variable independently, then back substitute. If a pivot depends on a parameter, branch before division. Test a proposed general solution in every original equation. Do not use determinant tests on rectangular matrices or equate rank with the number of solutions.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

LS01. What is the rank of [[1,2,3],[2,4,6],[0,1,1]]?

A. 1 B. 2 C. 3 D. 0

Answer: B

A: The first and third rows are independent.

B: Row 2 is twice row 1; rows 1 and 3 give two pivots.

C: One row is dependent, so rank cannot be 3.

D: The matrix has nonzero rows.

LS02. A system in three unknowns has rank(A)=rank([A|b])=2 over R. How many solutions does it have?

A. None / कोई नहीं B. Exactly one / ठीक एक C. Exactly two / ठीक दो D. Infinitely many / अनंत

Answer: D

A: Equal ranks establish consistency.

B: There is one free variable, so uniqueness fails.

C: Rank is not a count of solutions.

D: There are 3−2=1 free parameters, giving infinitely many real solutions.

LS03. The reduced augmented matrix contains [0 0 0 | 1]. What follows?

A. x₃=1 B. The system has no solution / कोई हल नहीं C. There is a free variable / एक स्वतंत्र चर है D. The system has exactly one solution / ठीक एक हल है

Answer: B

A: The coefficient of x₃ is zero, not one.

B: This row states 0=1, an inconsistency.

C: Other rows may have free variables, but this contradiction already rules out every solution.

D: A contradictory row prevents any solution.

LS04. A is 4×6 with rank 3. What is the dimension of {x:Ax=0}?

A. 1 B. 3 C. 4 D. 6

Answer: B

A: Use the number of unknowns 6, not rows 4.

B: Nullity is number of columns minus rank: 6−3=3.

C: The number of equations is not nullity.

D: Three pivot constraints reduce the six-dimensional domain.

LS05. For x+y=1 and 2x+ty=3, which statement is correct?

A. Unique for every t / हर t पर अद्वितीय B. Infinite at t=2 / t=2 पर अनंत C. No solution at t=2; unique otherwise / t=2 पर कोई हल नहीं; अन्यथा अद्वितीय D. No solution for t≠2 / t≠2 पर कोई हल नहीं

Answer: C

A: At t=2, subtracting twice the first equation yields 0=1.

B: At t=2 the right sides are incompatible, not redundant.

C: Elimination gives (t−2)y=1; it is inconsistent at 2 and solvable uniquely otherwise.

D: When t−2≠0, y=1/(t−2), then x=1−y.

LS06. Which is the full solution of x+2y−z=3 and y+z=1?

A. (x,y,z)=(1+3t,1−t,t), t∈R B. (x,y,z)=(1−3t,1−t,t), t∈R C. (x,y,z)=(3,0,0) D. No solution / कोई हल नहीं

Answer: A

A: Set z=t; then y=1−t and x=3−2(1−t)+t=1+3t.

B: Substitution into the first equation gives 3−6t, not always 3.

C: The second equation would be 0=1.

D: Choosing t=0 already gives the valid solution (1,1,0).

Source and scope

Supports the official System of linear equations and Matrices headings through rank, consistency and complete solution sets. Six original single-correct questions test foundations, not exhaustive mastery.

Official GATE 2026 CS syllabus (verified 2026-10-14).