Examples of Rank of a Matrix

Example 1: Find the rank of matrix [Tex]\bold{\begin{bmatrix} -1 & -2 & -3\\ -4 & -5 & -6 \\ -7 & -8 & -7 \end{bmatrix}} [/Tex] using minor method.

Solution:

Given [Tex]A = \begin{bmatrix} -1 & -2 & -3\\ -4 & -5 & -6 \\ -7 & -8 & -7 \end{bmatrix} [/Tex]

Step 1: Calculate the determinant of A

det(A) = -1 (35 – 48) + 2 (28 – 42) – 3 (32 – 35)

det(A) = 13 – 28 – 9 = -24

As det(A) ≠ 0, ρ(A) = order of A = 3

Example 2. Find the rank of matrix [Tex]\bold{\begin{bmatrix} 2 & 4 & 6\\ 8 & 10 & 12 \\ 14 & 16 & 0 \end{bmatrix}} [/Tex] using minor method.

Solution:

Given [Tex]A = \begin{bmatrix} 2 & 4 & 6\\ 8 & 10 & 12 \\ 14 & 16 & 0 \end{bmatrix} [/Tex]

Step 1: Calculate the determinant of A

det(A) = 2(0-192) – 4(0-168) + 6(128-140)

det(A) = -384 + 672 – 72 = 216

As det(A) ≠ 0, ρ(A) = order of A = 3

Example 3. Find the rank of matrix [Tex]\bold{\begin{bmatrix} -1 & -2 & -3\\ -4 & -5 & -6 \\ -7 & -8 & -9 \end{bmatrix}} [/Tex] using Echelon form method.

Solution:

Given [Tex]A = \begin{bmatrix} -1 & -2 & -3\\ -4 & -5 & -6 \\ -7 & -8 & -9 \end{bmatrix} [/Tex]

Step 1: Convert A to echelon form

Apply R₂ = R₂ – 4R₁

Apply R₃ = R₃ – 7R₁

[Tex]A = \begin{bmatrix} -1 & -2 & -3\\ 0 & 3 & 6 \\ 0 & 6 & 12 \end{bmatrix} [/Tex]

Apply R₃ = R₃ – 2R₂

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

As matrix A is now in lower triangular form, it is in Echelon Form.

Step 2: Number of non-zero rows in A = 2. Thus ρ(A) = 2

Example 4. Find the rank of matrix [Tex]\bold{\begin{bmatrix} 2 & 4 & 6\\ 8 & 10 & 12 \\ 14 & 16 & 18 \end{bmatrix}} [/Tex] using Echelon form method.

Solution:

Given [Tex]A = \begin{bmatrix} 2 & 4 & 6\\ 8 & 10 & 12 \\ 14 & 16 & 18 \end{bmatrix} [/Tex]

Step 1: Convert A to echelon form

Apply R₂ = R₂ – 4R₁

Apply R₃ = R₃ – 7R₁

[Tex]A = \begin{bmatrix} 2 & 4 & 6\\ 0 & -6 & -12 \\ 0 & -12 & -24 \end{bmatrix} [/Tex]

Apply R₃ = R₃ – 2R₂

[Tex]A = \begin{bmatrix} 2 & 4 & 6\\ 0 & -6 & -12 \\ 0 & 0 & 0 \end{bmatrix} [/Tex]

As matrix A is now in lower triangular form, it is in Echelon Form.

Step 2: Number of non-zero rows in A = 2. Thus ρ(A) = 2

Example 5. Find the rank of matrix [Tex]\bold{\begin{bmatrix} 2 & 4 & 2 & 4\\ 2 & 6 & 4 & 4 \\ 4 & 8 & 6 & 8 \\ 6 & 14 & 8 & 12 \end{bmatrix}} [/Tex] using normal form method.

Solution:

Given [Tex]A = \begin{bmatrix} 2 & 4 & 2 & 4\\ 2 & 6 & 4 & 4 \\ 4 & 8 & 6 & 8 \\ 6 & 14 & 8 & 12 \end{bmatrix} [/Tex]

Apply R₂ = R₂ – R₁ , R₃ = R₃ – 2R₁ and R₄ = R₄ – 3R₁

[Tex]A = \begin{bmatrix} 2 & 4 & 2 & 4\\ 0 & 2 & 2 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 2 & 2 & 0 \end{bmatrix} [/Tex]

Apply R₁ = R₁ – 2R₂ and R4 = R₄ – R₂

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

Apply R₁ = R₁ + R₃ and R₂ = R₂ – R₃

[Tex]A = \begin{bmatrix} 2 & 0 & 0 & 4\\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} [/Tex]

Apply C₄ → C₄ – 2C₁

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

Apply R₁ = R₁/2, R₂ = R₂/2, R₃ = R₃/2

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

Thus A can be written as [Tex]\begin{bmatrix} I_3 & 0\\ 0 & 0\\ \end{bmatrix} [/Tex]

Thus, ρ(A) = 3

Rank of a Matrix is defined as the dimension of the vector space formed by its columns. Rank of a Matrix is a very important concept in the field of Linear Algebra, as it helps us to know if we can find a solution to the system of equations or not. Rank of a matrix also helps us know the dimensionality of its vector space.

This article explores, the concept of the Rank of a Matrix in detail including its definition, how to calculate the rank of the matrix as well as a nullity and its relation with rank. We will also learn how to solve some problems based on the rank of a matrix. So, let’s start with the definition of the rank of the matrix first.