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Rank of a Matrix and Some Special Matrices

The maximum number of linearly independent columns (or rows) of a matrix is called the rank of a matrix. The rank of a matrix cannot exceed the number of its rows or columns.Β 

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If we consider a square matrix, the columns (rows) are linearly independent only if the matrix is nonsingular. In other words, the rank of any nonsingular matrix of order m is m. The rank of a matrix A is denoted by ρ(A).

The rank of a null matrix is zero. A null matrix has no non-zero rows or columns. So, there are no independent rows or columns. Hence, the rank of a null matrix is zero.Β 

How to Find the Rank of a Matrix?

To find the rank of a matrix, we will transform the matrix into its echelon form.Β 

Then, determine the rank by the number of non-zero rows.Β 

Consider the following matrix.

\(\begin{array}{l}A= \begin{bmatrix} 2 & 4 &6 \\ 4& 8& 12 \end{bmatrix}\end{array} \)

While observing the rows, we can see that the second row is two times the first row. Here, we have two rows, but it does not count. The rank is considered as 1.Β 

Consider the unit matrix.Β 

\(\begin{array}{l}A = \begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 0 & 0 &1 \end{bmatrix}\end{array} \)

We can see that the rows are independent. Hence, the rank of this matrix is 3.Β 

The rank of a unit matrix of order m is m.Β 

If A matrix is of order mΓ—n, then ρ(A ) ≀ min{m, n } = minimum of m, n.

If A is of order nΓ—n and |A| β‰  0, then the rank of A = n.

If A is of order nΓ—n and |A| = 0, then the rank of A will be less than n.

Rank of a Matrix by Row-Echelon Form

We can transform a given non-zero matrix to a simplified form called a Row-echelon form using the row elementary operations. In this form, we may have rows, all of whose entries are zero. Such rows are called zero rows. A non-zero row is one in which at least one of the elements is not zero.

For example, consider the following matrix.

\(\begin{array}{l}A = \begin{bmatrix} 1 &0 &2 \\ 0& 0 & 1\\ 0 & 0 & 0 \end{bmatrix}\end{array} \)

Here R1 and R2 are non-zero rows.

R3 is a zero row.

A non-zero matrix A is said to be in a row-echelon form if:Β 

(i) All zero rows of A occur below every non-zero row of A.Β 

(ii) The first non-zero element in any row i of A occurs in the jth column of A, and then all other elements in the jth column of A below the first non-zero element of row i are zeros.

(iii) The first non-zero entry in the ith row of A lies to the left of the first non-zero entry in ( i + 1)th row of A.

Note: A non-zero matrix is said to be in a row-echelon form if all zero rows occur as bottom rows of the matrix and if the first non-zero element in any lower row occurs to the right of the first non-zero entry in the higher row.Β 

If a matrix is in row-echelon form, then all elements below the leading diagonal are zeros.

Consider the following matrix.

\(\begin{array}{l}A = \begin{bmatrix} 0 &0 &1 \\ 0& 0 & 5\\ 0 & 0 & 0 \end{bmatrix}\end{array} \)

Check the rows from the last row of the matrix. The third row is a zero row. The first non-zero element in the second row occurs in the third column, and it lies to the right of the first non-zero element in the first row, which occurs in the second column. Hence, matrix A is in row echelon form.

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Rank of a Matrix Solved Examples

Example 1:

Find the rank of matrix A by using the row echelon form.Β 

\(\begin{array}{l}A = \begin{bmatrix} 1 &2 &3 \\ 2& 1 & 4\\ 3 & 0 & 5 \end{bmatrix}\end{array} \)

Solution:

Given,

\(\begin{array}{l}A = \begin{bmatrix} 1 &2 &3 \\ 2& 1 & 4\\ 3 & 0 & 5 \end{bmatrix}\end{array} \)

Now, we apply elementary transformations.

R2 β†’ R2 – 2R1

R3 β†’ R3 – 3R1

We get

\(\begin{array}{l}\begin{bmatrix} 1 &2 &3 \\ 0& -3 & -2\\ 0 & -6 & -4 \end{bmatrix}\end{array} \)

R3 β†’ R3 – 2R2

\(\begin{array}{l}\begin{bmatrix} 1 &2 &3 \\ 0& -3 & -2\\ 0 & 0 & 0 \end{bmatrix}\end{array} \)

The above matrix is in row echelon form.Β 

Number of non-zero rows = 2

Hence, the rank of matrix A = 2

Example 2:

Find the rank of the given matrix.

\(\begin{array}{l}A = \begin{bmatrix} 1 &2 &3 \\ 2& 3 &4\\ 3 & 5 & 7 \end{bmatrix}\end{array} \)

Solution:

Given,

\(\begin{array}{l}A = \begin{bmatrix} 1 &2 &3 \\ 2& 3 &4\\ 3 & 5 & 7 \end{bmatrix}\end{array} \)

Now, we transform matrix A to echelon form by using elementary transformation.

R2 β†’ R2 – 2R1

R3 β†’ R3 – 3R1

\(\begin{array}{l}\begin{bmatrix} 1 &2 &3 \\ 0& -1 &-2\\ 0 & -1 & -2 \end{bmatrix}\end{array} \)

R3 β†’ R3 – R2

\(\begin{array}{l}\begin{bmatrix} 1 &2 &3 \\ 0& -1 &-2\\ 0 & 0 & 0 \end{bmatrix}\end{array} \)

Number of non-zero rows = 2

Hence, the rank of matrix A = 2

Example 3:Β 

Find the rank of the matrix.

\(\begin{array}{l}\begin{bmatrix} 1 &1 &1 \\ 1& 1 &1\\ 1 & 1 & 1 \end{bmatrix}\end{array} \)

Solution:

GivenΒ 

\(\begin{array}{l}\begin{bmatrix} 1 &1 &1 \\ 1& 1 &1\\ 1 & 1 & 1 \end{bmatrix}\end{array} \)

R2 β†’ R2 – R1

R3 β†’ R3 – R1

We get

\(\begin{array}{l}\begin{bmatrix} 1 &1 &1 \\ 0& 0 &0\\ 0 & 0 & 0 \end{bmatrix}\end{array} \)

Here, the number of non-zero rows = 1

Hence, the rank of the matrix = 1

Example 4:

Find the rank of the 2Γ—2 matrix

\(\begin{array}{l}B = \begin{bmatrix} 5 & 6\\ 7& 8 \end{bmatrix}\end{array} \)

Solution:

Given,

\(\begin{array}{l}B = \begin{bmatrix} 5 & 6\\ 7& 8 \end{bmatrix}\end{array} \)

Order of B = 2Γ—2

|B| = 40 – 42 = -2 β‰  0

So, the rank of B = 2

Example 5:

Given,

\(\begin{array}{l}A = \begin{bmatrix} 4& 7\\ 8& 14 \end{bmatrix}\end{array} \)

Find the rank of matrix A.

Solution:

Given,

\(\begin{array}{l}A = \begin{bmatrix} 4& 7\\ 8& 14 \end{bmatrix}\end{array} \)

By observing the rows, we can see that the elements of the second row are twice the elements of the first row.Β 

R1β†’ 2R1 – R2

\(\begin{array}{l}\begin{bmatrix} 0& 0\\ 8& 14 \end{bmatrix}\end{array} \)

Number of non-zero rows = 1

The rank of matrix A = 1.

Example 6:

The rank of the matrix M isΒ 

\(\begin{array}{l}M = \begin{bmatrix} 0 & 1 & 1\\ 1& 0 &1 \\ 1& 1& 0 \end{bmatrix}\end{array} \)

a) 1

b) 2

c) 3

d) 0

Solution:

\(\begin{array}{l}M = \begin{bmatrix} 0 & 1 & 1\\ 1& 0 &1 \\ 1& 1& 0 \end{bmatrix}\end{array} \)

Apply row transformations to make the matrix into echelon form.

Interchange R2 and R1.

\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 1\\ 0& 1 &1 \\ 1& 1& 0 \end{bmatrix}\end{array} \)

R3 β†’ R3 – R1

\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 1\\ 0& 1 &1 \\ 0& 1& -1 \end{bmatrix}\end{array} \)

R3 β†’ R3 – R2

\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 1\\ 0& 1 &1 \\ 0& 0& -2 \end{bmatrix}\end{array} \)

Divide R3 by -2

\(\begin{array}{l}\begin{bmatrix} 1 & 0 & 1\\ 0& 1 &1 \\ 0& 0& 1 \end{bmatrix}\end{array} \)

Since there are three non-zero rows, rank = 3

Hence, option (c) is the answer.

Example 7:

The rank of the following matrix is

\(\begin{array}{l}\begin{bmatrix} 1 & 1 & 0& -2\\ 2& 0& 2 & 2\\ 4& 1 & 3 & 1 \end{bmatrix}\end{array} \)

a) 1

b) 2

c) 3

4) 4

Solution:

Given,

\(\begin{array}{l}\begin{bmatrix} 1 & 1 & 0& -2\\ 2& 0& 2 & 2\\ 4& 1 & 3 & 1 \end{bmatrix}\end{array} \)

We transform the matrix using elementary row operations.

R2 β†’ R2 – 2R1

\(\begin{array}{l}\begin{bmatrix} 1 & 1 & 0& -2\\ 0& -2& 2 & 6\\ 0& -3 & 3 & 9 \end{bmatrix}\end{array} \)

R2 β†’ R2/-2

\(\begin{array}{l}\begin{bmatrix} 1 & 1 & 0& -2\\ 0& 1& -1 &-3\\ 0& -3 & 3 & 9 \end{bmatrix}\end{array} \)

R3 β†’ R3 + 3R2

\(\begin{array}{l}\begin{bmatrix} 1 & 1 & 0& -2\\ 0& 1& -1 &-3\\ 0& 0 & 0 & 0 \end{bmatrix}\end{array} \)

Since the number of non-zero rows is 2, rank = 2

Hence, option (b) is the answer.

Example 8:

\(\begin{array}{l}\text{Let}\ P = \begin{bmatrix} 1 & 1 & -1\\ 2 & -3& 4\\ 3 & -2 & 3 \end{bmatrix}\ \text{and}\ Q = \begin{bmatrix} -1 & -2 & -1\\ 6& 12& 6\\ 5 & 10 & 5 \end{bmatrix}\ \text{be two matrices.}\end{array} \)
Then the rank of P + Q =Β 

a) 1

b) 0

c) 2

d) 3

Solution:

Given,

\(\begin{array}{l}P =\begin{bmatrix} 1 & 1 & -1\\ 2 & -3& 4\\ 3 & -2 & 3 \end{bmatrix}\end{array} \)

\(\begin{array}{l}Q = \begin{bmatrix} -1 & -2 & -1\\ 6& 12& 6\\ 5 & 10 & 5 \end{bmatrix}\end{array} \)

\(\begin{array}{l}P + Q = \begin{bmatrix} 0 & -1 & -2\\ 8& 9& 10\\ 8& 8 & 8 \end{bmatrix}\end{array} \)

Interchange C1 and C2

\(\begin{array}{l}\begin{bmatrix} -1 & 0 & -2\\ 9& 8& 10\\ 8& 8 & 8 \end{bmatrix}\end{array} \)

R2 β†’ R2 + 9R1

R3 β†’ R3 + 8R1

\(\begin{array}{l}\begin{bmatrix} -1 & 0 & -2\\ 0& 8& -8\\ 0& 8 & -8 \end{bmatrix}\end{array} \)

R3 β†’ R3 – R2

\(\begin{array}{l}\begin{bmatrix} -1 & 0 & -2\\ 0& 8& -8\\ 0& 0 & 0 \end{bmatrix}\end{array} \)

R2 β†’ R2/8

\(\begin{array}{l}\begin{bmatrix} -1 & 0 & -2\\ 0& 1& -1\\ 0& 0 & 0 \end{bmatrix}\end{array} \)

Number of non-zero rows = 2

So, the rank = 2

Hence, option (c) is the answer.

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Matrices and Determinants – Important Topics

Matrices and Determinants – Important Questions

Frequently Asked Questions

Q1

What do you mean by the rank of the matrix?

The rank of a matrix is the maximum number of its linearly independent rows (or columns).

Q2

Can the rank of a matrix exceed the number of rows or columns?

The rank of a matrix cannot exceed the number of its rows or columns.

Q3

What do you mean by rank zero matrices?

If all matrix elements become zero, then the matrix is a rank zero matrix.

Q4

How do you find the rank of a matrix?

To find the rank of a matrix, transform the matrix into its echelon form. Then, find the rank by the number of non-zero rows.

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