Examples on Methods to Find Rank of a Matrix

Example 1: Find the rank of the matrix A = [Tex]\begin {bmatrix} 4 & 5\\ 0 & 2 \end {bmatrix}[/Tex] using minor method

Solution:

First, find the determinant of the give 2 × 2 matrix.

|A| = [Tex]\begin {vmatrix} 4 & 5\\ 0 & 2 \end {vmatrix}[/Tex]

|A| = (4 × 2) – (0 × 5)

|A| = 8 – 0

|A| = 8

Since, determinant of A is not equal to zero.

Rank of A = Order of matrix A

R(A) = 2

Example 2: Find the rank of the matrix B = [Tex]\begin {bmatrix} 3 & 6\\ 1 & 2 \end {bmatrix}[/Tex] using Echelon method

Solution:

First, find the Echelon form of matrix B by using row operation.

B = [Tex]\begin {bmatrix} 3 & 6\\ 1 & 2 \end {bmatrix}[/Tex]

R₁ → R₁ / 3

B = [Tex]\begin {bmatrix} 1 & 2\\ 1 & 2 \end {bmatrix}[/Tex]

R₂ → R₂ – R₁

B = [Tex]\begin {bmatrix} 1 & 2\\ 0 & 0 \end {bmatrix}[/Tex]

The above matrix represents the Echelon form of matrix B.

Now, find the number of non-zero rows i.e., = 2

Rank of matrix B = R(B) = 2

Example 3: Find the rank of the matrix X = [Tex]\begin {bmatrix} 1 & 6 & 4\\ 1 & 3 & 2\\ 1 & 12& 8 \end {bmatrix}[/Tex] using minor method

Solution:

First find the determinant of the matrix.

|X| = [Tex]\begin {vmatrix} 1 & 6 & 4\\ 1 & 3 & 2\\ 1 & 12& 8 \end {vmatrix}[/Tex]

|X| = 1 [(3 × 8) – (12 × 2)] – 6[(8 × 1) – (2× 0)] + 4[(12× 1) – (3× 0)]

|X| = [24 – 24] – 6× 8 + 4×12

|X| = 0 – 48 + 48 = 0

Since, |X| = 0 then we find at least one non-zero 2 × 2 minor.

Minor of A = [Tex]\begin {vmatrix} 1 & 6\\ 1 & 3 \end {vmatrix}[/Tex]

|Minor of A| = 3 – 6 = -3 ≠ 0

Therefore, rank of matrix X = 2 (order of minor)

Example 4: Find the rank of the matrix D = [Tex] \begin {bmatrix} 2 & 4 & 8\\ 1 & 0 & 0\\ 0 & 2& 3 \end {bmatrix} [/Tex] using Echelon method

Solution:

First convert matrix D into its Echelon form using elementary row operations.

D = [Tex] \begin {bmatrix} 2 & 4 & 8\\ 1 & 0 & 0\\ 0 & 2& 3 \end {bmatrix} [/Tex]

R₁ → R₁ /2

D = [Tex] \begin {bmatrix} 1 & 2 & 4\\ 1 & 0 & 0\\ 0 & 2& 3 \end {bmatrix} [/Tex]

R₂ → R₂ – R₁

D = [Tex] \begin {bmatrix} 1 & 2 & 4\\ 0 & -2 & -4\\ 0 & 2& 3 \end {bmatrix} [/Tex]

R₁ → R₁ + R₂, and R₃ →R₃ + R₂

D = [Tex] \begin {bmatrix} 1 & 0 & 0\\ 0 & -2 & -4\\ 0 & 0& -1 \end {bmatrix} [/Tex]

R₂ → R₂ / (-2), R₃ → R₃/ (-1)

D = [Tex] \begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0& 1 \end {bmatrix} [/Tex]

The above matrix is Echelon form of matrix D.

Now, find the number of non-zero rows i.e., = 3

Rank of matrix D = 3

Example 5: Find the rank of the matrix G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 2 & 1 & 0&-1\\ -1 & -6 & -3 &0\\ 0 & 4 & 5 & 6 \end {bmatrix} [/Tex] using the normal form method.

Solution:

First, find the normal form of matrix G using the elementary row and column operations.

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 2 & 1 & 0&-1\\ -1 & -6 & -3 &0\\ 0 & 4 & 5 & 6 \end {bmatrix} [/Tex]

R₂ → R₂ – 2R₁ and R₃→ R₃ + R₁

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 0 & -3 & -2&-1\\ 0 & -4 & -2 &0\\ 0 & 4 & 5 & 6 \end {bmatrix} [/Tex]

R₂ → R₂ / (-3)

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 0 & 1 & 2/3&1/3\\ 0 & -4 & -2 &0\\ 0 & 4 & 5 & 6 \end {bmatrix} [/Tex]

R₃ → R₃ + 4R₂ and R₄→ R₄ – 4R₂

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 0 & 1 & 2/3&1/3\\ 0 & 0 & 2/3 &4/3\\ 0 & 0 & -2/3 & 14/3 \end {bmatrix} [/Tex]

R₄ → R₄ + R₃

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 0 & 1 & 2/3&1/3\\ 0 & 0 & 2/3 &4/3\\ 0 & 0 & 0 & 6 \end {bmatrix} [/Tex]

R₃ → (3/2)R₃ and R₄ → R₄ /6

G = [Tex] \begin {bmatrix} 1 & 2 & 1&0\\ 0 & 1 & 2/3&1/3\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1 \end {bmatrix} [/Tex]

C₂ → C₂ – 2C₁ and C₃ → C₃ – C₁

G = [Tex] \begin {bmatrix} 1 & 0 & 0&0\\ 0 & 1 & 2/3&1/3\\ 0 & 0 & 1 & 2\\ 0 & 0 & 0 & 1 \end {bmatrix} [/Tex]

C₃ → C₃ – (2/3)C₂ and C₄ → C₄ – (1/3)C₂

G = [Tex] \begin {bmatrix} 1 & 0 & 0&0\\ 0 & 1 & 0&0\\ 0 & 0 & 0 & 2\\ 0 & 0 & 0 & 1 \end {bmatrix} [/Tex]

C₃ ↔ C₄

G = [Tex] \begin {bmatrix} 1 & 0 & 0&0\\ 0 & 1 & 0&0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 1 & 0 \end {bmatrix} [/Tex]

R₃ → R₃ / 2 and R₄ → R₄ – R₃

G = [Tex] \begin {bmatrix} 1 & 0 & 0&0\\ 0 & 1 & 0&0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end {bmatrix} [/Tex]

The above matrix is the normal form of matrix G i.e., [Tex]\begin {bmatrix} I_3 & 0\\ 0 & 0 \end {bmatrix}[/Tex]

So, the rank of matrix G = R(G) = 3

To find the rank of a matrix find the highest order of the non-zero minor within the matrix. Rank of a matrix in the number that represents the number of non-zeros rows or columns in the matrix. If the rank of the matrix is r then the matrix contains at least one minor with order r and the minors with order greater than r is zero. The second method to find the rank of matrix is by converting it into Echelon form.

In this article we will discuss methods to find rank of a matrix in depth along with the rank definition, methods to find rank of a matrix i.e., by minors and by Echelon form. Also, we will discuss the properties of rank and solve some examples including both the methods. Let’s start our learning on the topic “Methods to Find Rank of a Matrix. “