Create a free account to get started

Unlock full course notes by signing up

  • Video lessons taught by top educators
  • Practice questions with guided solutions
  • Written review notes for reference
  • Expert answers anytime from course tutors

Rank of a Matrix

The rank of a matrix AA , denoted r=rank(A)r = {\rm rank}(A) , is the number of leading 1s in the RREF of the matrix.

Example

Given [A ∣ b⃗ ]=[101001200001]\left[A \ |\ \vec{b}\ \right] = \left[\begin{array}{ccc|c} 1&0&1&0\\ 0&1&2&0\\ 0&0&0&1 \end{array}\right],

  • r(A)=r(A)= ________
  • r([A ∣ b⃗ ])=r([A \ |\ \vec{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){\rm rank}(A_{m \times n}) \le \min(m,n)

Example

Find the maximum rank of the matrix A=[a11a12a13a21a22a23]A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ \end{bmatrix} .

AA has size 2×32 \times 3 so rank(A)≤2{\rm rank}(A) \le 2.

In this case, if AA 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 ∣ b⃗ ])\iff {\rm rank}(A) < {\rm rank}([\ A\ |\ \vec b\ ])
  • Consistent (one solution, or infinitely many solutions)   ⟺  rank(A)=rank([ A ∣ b⃗ ])\iff {\rm rank}(A) \bm= {\rm rank}([\ A\ |\ \vec b\ ])

No Solutions

In the RREF of [A ∣ b⃗ ][A \ |\ \vec b\ ], there is a leading entry in the augmented column: [00⋯01]\left[\begin{array}{cccc|c} 0&0&\cdots&0&1 \end{array}\right]

Exactly One Solution

There is a unique solution   ⟺  rank(A)=rank([ A ∣ b⃗ ])=n\iff {\rm rank}(A) \bm= {\rm rank}([\ A\ |\ \vec b\ ]) \colorTwo{\bm = n}.

In the RREF of [A ∣ b⃗ ][A \ |\ \vec 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 ∣ b⃗ ])<n\iff {\rm rank}(A) \bm= {\rm rank}([\ A\ |\ \vec b\ ]) \colorTwo{\bm < n}.

In the RREF of [A ∣ b⃗ ][A \ |\ \vec 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\colorFour{\bm k}-parameter family of solutions where k=n−rank(A)k=n-{\rm rank}(A).

Create a free account to get started

Unlock full course notes by signing up

  • Video lessons taught by top educators
  • Practice questions with guided solutions
  • Written review notes for reference
  • Expert answers anytime from course tutors

Example: Rank of a Matrix

  1. Find the ranks of the following matrices and state the number of solutions.
  2. [102100∣120]\left[\begin{array}{l} 1& 0 \\ 2& 1 \\ 0 & 0 \\ \end{array}\right.\left|\begin{array}{l} 1\\2\\0\end{array}\right]

[101212000]→R2  →  R2−2R1[101010000]\left[\begin{array}{cc|c} 1&0&1\\ 2&1&2\\ 0&0&0 \end{array}\right] \xrightarrow{R_2 \ \ \rightarrow \ \ R_2-2R_1} \left[\begin{array}{cc|c} 1&0&1\\ 0&1&0\\ 0&0&0 \end{array}\right]

  • rank(A)=2{\rm rank}(A)=2 which is full rank because AA has only 2 columns.
  • rank([A ∣ b⃗ ])=2{\rm rank}([A \ |\ \vec{b}\ ])=2 (there is no leading 1 in the augmented column)
  • n=2n=2 because this represents a system of equations with 2 unknowns (2 columns)


Since rank(A)=rank([A ∣ b⃗ ])=n{\rm rank}(A)={\rm rank}([A \ |\ \vec{b}\ ])=n, this system has a unique solution.

  1. [10201−1000∣310]\left[\begin{array}{l} 1 & 0 & 2 \\ 0 &1 & -1 \\ 0 & 0 & 0 \\ \end{array}\right.\left|\begin{array}{l} 3\\1\\0\end{array}\right]

This matrix is already in RREF.

  • rank(A)=2{\rm rank}(A)=2 which is not full rank.
  • rank([A ∣ b⃗ ])=2{\rm rank}([A\ |\ \vec{b}\ ])=2
  • n=3n=3 because this represents a system of equations with 3 unknowns (3 columns)

Since rank(A)=rank([A ∣ b⃗ ])<n{\rm rank}(A)={\rm rank}([A\ |\ \vec{b}\ ])<n, this system has infinitely many solutions.

In particular, since n−rank(A) = 3−2 = 1n-{\rm rank}(A) \ =\ 3-2 \ =\ 1, there is a 1-parameter family of solutions.


  1. Suppose AA is a 3×53\times 5 matrix with rank(A)=2{\rm rank}(A)=2 and rank([A ∣ b⃗ ])=3{\rm rank}([A\ |\ \vec{b}\ ])=3. How many solutions are there for the linear system with coefficient matrix AA and constant vector b⃗\vec b?
  • rank(A)=2{\rm rank}(A)=2
  • rank([A ∣ b⃗ ])=3{\rm rank}([A\ |\ \vec{b}\ ])=3
  • n=5n=5 (read from the size of Am×n=A3×5A_{m \times \colorTwo{n}} = A_{3 \times \colorTwo{5}})

Since rank(A)<rank([A ∣ b⃗ ]){\rm rank}(A)<{\rm rank}([A\ |\ \vec{b}\ ]), this system has no solutions.

That's because, in RREF, the augmented column must contain a leading 1: [00…∣  1]\begin{bmatrix} 0& 0& \dots &|\ \ 1\\ \end{bmatrix} .

This means that the sum of 0 times every variable equals 1, i.e. 0=10=1, which is impossible.

In the reduced row echelon form of the coefficient matrix AA that corresponds to a system of linear equations with 6 equations in 9 unknowns, the maximum number of leading 1s is .


Create a free account to get started

Unlock full course notes by signing up

  • Video lessons taught by top educators
  • Practice questions with guided solutions
  • Written review notes for reference
  • Expert answers anytime from course tutors

Determine the number of solutions for the system of linear equations Ax⃗=b⃗A\vec{x}=\vec{b} with 5 equations and 5 unknowns, given:

The rank of AA is 5 and the rank of [A ∣ b⃗ ][A\ |\ \vec{b}\ ] is 5.

Choose an answer

Create a free account to get started

Unlock full course notes by signing up

  • Video lessons taught by top educators
  • Practice questions with guided solutions
  • Written review notes for reference
  • Expert answers anytime from course tutors

Extra Practice