
Rank of a Matrix
The rank of a matrix A , denoted r=rank(A) , is the number of leading 1s in the RREF of the matrix.
Example
Given [A ∣ b ]=100010120001,
- r(A)= ________
- r([A ∣ b ])= ________
Full rank
A matrix has full rank if every column or every row of the RREF of the matrix contains a leading 1.
Note: rank(Am×n)≤min(m,n)
Example
Find the maximum rank of the matrix A=[a11a21a12a22a13a23] .
A has size 2×3 so rank(A)≤2.
In this case, if A has full rank, each row must contain a leading 1 (not every column, since the rank can be at most 2).
Types of Solutions
A system of linear equations may be:
- Inconsistent (no solution)⟺rank(A)<rank([ A ∣ b ])
- Consistent (one solution, or infinitely many solutions) ⟺rank(A)=rank([ A ∣ b ])
No Solutions
In the RREF of [A ∣ b ], there is a leading entry in the augmented column: [00⋯01]
Exactly One Solution
There is a unique solution ⟺rank(A)=rank([ A ∣ b ])=n.
In the RREF of [A ∣ b ]:
- There is no leading entry in the augmented column
- every column of the coefficient matrix (left) has a leading 1
Infinitely Many Solutions
There are infinitely many solutions ⟺rank(A)=rank([ A ∣ b ])<n.
In the RREF of [A ∣ b ]:
- There is no leading entry in the augmented column
- There is at least one column without a leading 1
This results in an k-parameter family of solutions where k=n−rank(A).

Example: Rank of a Matrix
- Find the ranks of the following matrices and state the number of solutions.
- 120010120
120010120R2 → R2−2R1100010100
- rank(A)=2 which is full rank because A has only 2 columns.
- rank([A ∣ b ])=2 (there is no leading 1 in the augmented column)
- n=2 because this represents a system of equations with 2 unknowns (2 columns)
Since rank(A)=rank([A ∣ b ])=n, this system has a unique solution.
- 1000102−10310
This matrix is already in RREF.
- rank(A)=2 which is not full rank.
- rank([A ∣ b ])=2
- n=3 because this represents a system of equations with 3 unknowns (3 columns)
Since rank(A)=rank([A ∣ b ])<n, this system has infinitely many solutions.
In particular, since n−rank(A) = 3−2 = 1, there is a 1-parameter family of solutions.
- Suppose A is a 3×5 matrix with rank(A)=2 and rank([A ∣ b ])=3. How many solutions are there for the linear system with coefficient matrix A and constant vector b?
- rank(A)=2
- rank([A ∣ b ])=3
- n=5 (read from the size of Am×n=A3×5)
Since rank(A)<rank([A ∣ b ]), this system has no solutions.
That's because, in RREF, the augmented column must contain a leading 1: [00…∣ 1].
This means that the sum of 0 times every variable equals 1, i.e. 0=1, which is impossible.