Rank of a Matrix: Solved Example Problems

Example 1.6

Find the rank of the matrix A= 

Solution :

The order of A is 3 × 3.

∴ ρ(A) ≤ 3.

Let us transform the matrix A to an echelon form by using elementary transformations.

The number of non zero rows is 2

∴Rank of A is 2.

ρ (A) = 2.

Note

A row having atleast one non -zero element is called as non-zero row.

Example 1.7

Find the rank of the matrix A= 

Solution:

The order of A is 3 × 4.

∴ ρ (A)≤3.

Let us transform the matrix A to an echelon form

The number of non zero rows is 3. ∴ ρ(A) =3.

Example 1.8

Find the rank of the matrix A= 

Solution:

The order of A is 3 × 4.

∴ ρ(A) ≤ 3.

Let us transform the matrix A to an echelon form

The number of non zero rows is 3.

∴ ρ (A) =3.

Testing the consistency of non homogeneous linear equations (two and three variables) by rank method : Solved Example Problems

Example 1.9

Show that the equations x + y = 5, 2x + y = 8 are consistent and solve them.

Solution:

The matrix equation corresponding to the given system is

AX=B

Number of non-zero rows is 2.  

ρ (A )= ρ ([ A, B]) = 2 = Number of unknowns.  

The given system is consistent and has unique solution. 

Now, the given system is transformed into

x + y = 5

y = 2

∴ (1) ⇒ x + 2 = 5

x = 3

Solution is x = 3, y = 2

Example 1.10

Show that the equations 2x + y = 5, 4x + 2 y = 10 are consistent and solve them.

Solution:

The matrix equation corresponding to the system is

ρ ( A ) = ρ ([ A, B]) = 1 < number of unknowns

∴ The given system is consistent and has infinitely many solutions.

Now, the given system is transformed into the matrix equation.

Let us take y = k, k ∈R

⇒ 2x + k = 5

x = 1/2 ( 5 − k)

x = 1/2 ( 5 − k) , y = k for all k ∈R

Thus by giving different values for k, we get different solution. Hence the system has infinite number of solutions.

Example 1.11

Show that the equations 3x − 2 y = 6, 6x − 4 y = 10 are inconsistent.

Solution:

The matrix equation corresponding to the given system is

∴The given system is inconsistent and has no solution.

Example 1.12

Show that the equations 2x + y + z = 5, x + y + z = 4, x − y + 2z = 1 are consistent and hence solve them.

Solution:

The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.

ρ( A ) = ρ( [A, B] )= 3 = Number of unknowns .

The given system is consistent and has unique solution.

To find the solution, let us rewrite the above echelon form into the matrix form.

x + y + z = 4 (1)

 y + z = 3 (2)

3z = 3 (3)

(3)⇒ z = 1

(2)⇒ y = 3 − z = 2

(1) ⇒ x = 4 − y − z

x=1

∴ x = 1, y = 2, z = 1

Example 1.13

Show that the equations x + y + z = 6, x + 2 y + 3z = 14, x + 4 y + 7z = 30 are consistent and solve them.

Solution:

The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has two non-zero rows.

∴ ρ ( [A, B] ) = 2, ρ ( A) = 2

ρ ( A ) = ρ ( [A, B] ) = 2 < Number of unknowns.

The given system is consistent and has infinitely many solutions.

The given system is equivalent to the matrix equation,

x + y + z = 6  (1)

y + 2z = 8 (2)

(2)⇒ y = 8 − 2z,

(1)⇒ x = 6 − y − z = 6 − (8 − 2 z) − z = z – 2

Let us take z = k, k ∈R , we get x = k − 2, y = 8 − 2k , Thus by giving different values for k we get different solutions. Hence the given system has infinitely many solutions.

Example 1.14

Show that the equations x − 4 y + 7z = 14, 3x + 8 y − 2z = 13, 7x − 8 y + 26z = 5 are inconsistent.

Solution:

The matrix equation corresponding to the given system is

The last equivalent matrix is in the echelon form. [A, B] has 3 non-zero rows and [A] has 2 non-zero rows.

The system is inconsistent and has no solution.

Example 1.15

Find k, if the equations x + 2 y − 3z = −2, 3x − y − 2z = 1, 2x + 3y − 5z = k are consistent.

Solution:

The matrix equation corresponding to the given system is

For the equations to be consistent, ρ ( [A, B] ) = ρ ( A)= 2

∴21 + 7k = 0

7k = −21 .

k = −3

Example 1.16

Find k, if the equations x + y + z = 7, x + 2 y + 3z = 18, y + kz = 6 are inconsistent.

Solution:

The matrix equation corresponding to the given system is

For the equations to be inconsistent

ρ ( [A, B] ) ≠ ρ ( A)

It is possible if k − 2 = 0 .

K=2

Example 1.17

Investigate for what values of ‘a’ and ‘b’ the following system of equations

x + y + z = 6, x + 2 y + 3z = 10, x + 2 y + az = b have 

(i) no solution  (ii) a unique solution (iii) an infinite number of solutions.

Solution:

The matrix equation corresponding to the given system is

Case (i) For no solution:

The system possesses no solution only when ρ ( A )≠ ρ ([ A, B]) which is possible only when a − 3 = 0 and b − 10 ≠ 0

Hence for a = 3, b ≠ 10 , the system possesses no solution.

Case (ii) For a unique solution:

The system possesses a unique solution only when ρ ( A ) = ρ ([ A, B]) = number of unknowns.

i.e when ρ ( A ) = ρ ([ A, B]) = 3

Which is possible only when a − 3 ≠ 0 and b may be any real number as we can observe .

Hence for a ≠ 3 and b ∈R , the system possesses a unique solution.

Case (iii) For an infinite number of solutions:

The system possesses an infinite number of solutions only when

ρ ( A )= ρ ([ A, B]) < number of unknowns

i,e when ρ ( A)= ρ ([ A, B])= 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b − 10 = 0

Hence for a = 3, b =10, the system possesses infinite number of solutions.

Example 1.18

The total number of units produced (P) is a linear function of amount of over times in labour (in hours) (l), amount of additional machine time (m) and fixed finishing time (a)

i.e, P = a + bl + cm

From the data given below, find the values of constants a, b and c

Estimate the production when overtime in labour is 50 hrs and additional machine time is 15 hrs.

Solution:

We have, P = a + bl + cm

Putting above values we have

6,950 = a + 40b + 10c

6,725 = a + 35b + 9c

7,100 = a + 40b + 12c

The Matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation

∴ The production equation is P = 5000 + 30l + 75m

 Pat l = 50, m=15         = 5000 + 30(50) + 75(15)

=7625 units.

∴The production = 7,625 units.