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# Rank of Matrix

The rank of matrix in simple words may be explained as the number of non-zero rows or columns of a non-zero matrix. We cannot simply find the non-zero rows simply by looking at the given matrix. Hence, to define the rank of matrix more formally, we must know about the minors of a matrix and linearly independent vectors of a matrix.

Minor of a matrix of any order is the determinant of the square sub-matrix of the given matrix. Let A be an m × n, the determinant of any square sub-matrix of A will be a minor of A. So the highest order of any non-singular minor of a matrix is called the rank of matrix.

The rank of matrix is the number of linearly independent vectors of a given matrix. Let us understand more about the rank matrix and its properties.

## What is the Rank of Matrix?

The rank of matrix can be defined in several ways. Let us discuss them in brief:

• Rank of Matrix on the basis of Linear Independent Vectors

The maximum number of linearly independent column or row vectors of matrix is called the rank of matrix. If R1, R2, …., Rm are the row vectors of a matrix A or C1, C2, …, Cn are column vectors of matrix A such that

$$\begin{array}{l}A\begin{bmatrix}R_{1} \\R_{2} \\. \\.\\.\\R_{m}\end{bmatrix}\end{array}$$
or
$$\begin{array}{l}A = \begin{bmatrix} C_{1} & C_{2} & . &. &. &C_{n} \\ \end{bmatrix}\end{array}$$

Then, R1, R2, …., Rm or C1, C2, …, Cn are said to be linearly independent if

A1 R1 + A2 R2 + …..+ Am Rm = 0 ⇔ A1 = A2 = … = Am = 0 where A1, A2, …, Am are scalars

Or k1 C1 + k2 C2 + … + kn Cn = 0 ⇔ k1 = k2 = … = kn = 0 where k1, k2, …, kn are scalars.

• Rank of Matrix on the basis of Minor of Matrix

The highest order of non-zero minor of a matrix is said to be the rank of a matrix. If ‘r’ is the rank of the matrix then atleast one minor of the given matrix is of order r and all other minors of order greater than r is zero.

• Rank of Matrix on the basis on Echelon Form

The number of non-zero rows of a matrix reduced in echelon form is called the rank of the matrix.

A matrix is said to be in echelon form if

i.) there are any zero rows, then they should be placed below the non-zero rows.

ii.) the number of zero in front of any row increases according to the row number.

The non-zero rows of an echelon matrix are that matrix’s linearly independent row vectors.

The rank of a matrix A is denoted by r(A) or ⍴(A).

## How to Find Rank of Matrix?

Let us understand how to find rank of matrix by taking an example. Let A be a matrix of order

4 × 4 such that

$$\begin{array}{l}A=\begin{bmatrix}2 & 3& 0& 1\\1 & 0 & 1 & 2\\-1 & 1 & 1 & -2 \\ 1& 5& 3 & -1 \\\end{bmatrix}\end{array}$$

### Method 1:

Reduce the given matrix A in a row reduced echelon form by applying elementary row operations. Given

$$\begin{array}{l}A=\begin{bmatrix} 2 & 3& 0& 1\\1 & 0 & 1 & 2\\-1 & 1 & 1 & -2 \\ 1& 5& 3 & -1 \\\end{bmatrix}\end{array}$$

Step 1: Applying elementary row operations, exchange R1 with R2, that is R1 ↔ R2

$$\begin{array}{l} A\sim\begin{bmatrix}1 &0 & 1 & 2 \\2 & 3 & 0& 1 \\-1 & 1& 1& -2\\1& 5& 3& -1\\\end{bmatrix}\end{array}$$

Step 2: Applying R2 → R2 + (-2)R1, R3 → R3 + R1 and R4 → R4 + (-1)R1 we get

$$\begin{array}{l}A\sim\begin{bmatrix}1 &0 & 1 & 2 \\0 & 3 & -2& -3 \\0 & 1& 2& 0\\0 & 5& 2& -3\\\end{bmatrix}\end{array}$$

Step 3: Applying R2 ↔ R3, R3 → R3 + (-3)R2 and R4 → R4 + (-5)R2 we get,

$$\begin{array}{l}A\sim\begin{bmatrix}1 &0 & 1 & 2 \\0 & 1 & 2& 0 \\0 & 0& -8& -3\\0 & 0& -8& -3\\\end{bmatrix}\end{array}$$

Step 4: Applying R3 → (-⅛)R3, R4 → R4 + (8)R3

$$\begin{array}{l}A\sim\begin{bmatrix}1 &0 & 1 & 2 \\0 & 1 & 2& 0 \\0 & 0& 1& 3/8\\0 & 0& 0& 0\\\end{bmatrix} \end{array}$$
= B (let)

Therefore, B is the required row reduced Echelon form

Thus, ⍴(A) = 3, since there are 3 non-zero rows.

### Method 2:

By finding non-singular minors of the given matrix A.

Given

$$\begin{array}{l}A=\begin{bmatrix}2 & 3& 0& 1\\1 & 0 & 1 & 2\\-1 & 1 & 1 & -2 \\ 1& 5& 3 & -1 \\\end{bmatrix}\end{array}$$

Clearly, |A| = 0, that is A is singular matrix

But the minor

$$\begin{array}{l}\begin{vmatrix}2 & 3 & 0\\1& 0 & 1 \\-1 & 1 & 1 \\\end{vmatrix} \end{array}$$
= -8 ≠ 0

Hence the highest order of the non-singular minor of A is 3

Therefore, ⍴(A) = 3.

 For matrix of higher order it is suitable to find the rank by reducing the matrix to echelon form than to find it by identifying non-singular minors. To find rank by minors always check first whether the given matrix is non-singular.

## Properties of Rank of Matrix

The following are some important properties of the rank of a matrix.

• Let A be any non-zero matrix of any order and if ⍴(A) < order of A then A is a singular matrix.
• Only the rank of a Null Matrix is zero.
• Rank of an Identity Matrix I is the order of I.
• Rank of matrix Am × n is minimum of m and n.
• If A’ and A* are the transpose and tranjugate of matrix A, then ⍴(A’) = ⍴(A) and

⍴(A*) = ⍴(A).

• Two row equivalent matrices have same rank.
• ⍴(A + B) ≤ ⍴(A) + ⍴(B)
• ⍴(A – B) ≥ ⍴(A) – ⍴(B)
• If A and B are two matrix conformable for matrix multiplication the ⍴(AB) ≤ min{⍴(A), ⍴(B)}.
• If the order of a square matrix A is m, then m – ⍴(A) is the nullity of matrix A.

## Solved Examples on Rank of Matrix

Example 1:

Find the rank of the matrix

$$\begin{array}{l}\begin{bmatrix}1 & 2 & 3 \\2 & 3 & 4 \\3 & 5 & 7 \\\end{bmatrix}\end{array}$$
.

Solution:

Let A =

$$\begin{array}{l}\begin{bmatrix}1 & 2 & 3 \\2 & 3 & 4 \\3 & 5 & 7 \\\end{bmatrix}\end{array}$$

Then |A| = 1( 21 – 20) – 2( 14 – 12) + 3( 10 – 9)

= 1 – 4 + 3 = 0

Thus A is a singular matrix.

But

$$\begin{array}{l}\begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix}\end{array}$$
= 1× 3 – 2 × 2 = 3 – 4 = -1 ≠ 0

Therefore, ⍴(A) = 2.

Example 2:

Are the rows of the matrix

$$\begin{array}{l}\begin{bmatrix} 1& 1& 2 \\ 1& 2& 3 \\ 2& 3& 4 \\\end{bmatrix}\end{array}$$
linearly independent?

Solution:

If order of the matrix = rank of matrix, then the row vectors of the matrix are linearly independent.

Let A =

$$\begin{array}{l}\begin{bmatrix} 1& 1& 2 \\ 1& 2& 3 \\ 2& 3& 4 \\\end{bmatrix}\end{array}$$

Then |A| = 1 × ( 8 – 9) – 1 × (4 – 6) + 2 × (3 – 4) = -1 + 2 – 2 = -1 ≠ 0

Hence, ⍴(A) = 3 = order of A

Therefore, the rows of the given matrix are linearly independent.

## Frequently Asked Questions on Rank of Matrix

Q1

### What is meant by rank of matrix?

The rank of matrix is number of linearly independent row or column vectors of a matrix. The number of linearly independent rows can be easily found by reducing the given matrix in row-reduced echelon form.

Q2

### When is the rank of matrix equal to the order of the matrix?

If the given matrix has linearly independent rows, then the rank of matrix is equal to the order of the matrix.

Q3

### How can the nullity of the matrix be determined by the rank of matrix?

Nullity of the matrix is the difference of order of the matrix and rank of matrix.

Q4

### How do you find the rank of a 3 by 3 matrix?

Firstly, check if the given 3 × 3 is singular or not. If it is singular then the rank of matrix is less than 3 so, check for other non-zero minors of the given matrix. If it is non-singular the 3 is the rank of matrix.

Q5

### How to find the rank of matrix?

The rank of matrix can be determined by reducing the given matrix in row-reduced echelon form, the number of non-zero rows of the echelon form is equal to the rank of matrix. Rank of matrix can also be calculated by finding order of the highest non-singular minor of the given matrix.