Business Maths and Statistics : Applications of Matrices and Determinants: Rank of a Matrix: Solved Example Problems with Answers, Solution and Explanation

Example 1.1

Find the rank of the matrix

*Solution:*

Let A=

Order of *A* is 2 × 2 ∴ *ρ*(*A*)≤ 2

Consider the second order minor

There is a minor of order 2, which is not zero. ∴*ρ* (*A*) = 2

Example 1.2

Find the rank of the matrix

*Solution:*

Let A=

Order of *A* is 2 × 2 ∴*ρ*(*A*)≤ 2

Consider the second order minor

Since the second order minor vanishes, *ρ*(*A*) ≠ 2

Consider a first order minor |−5| ≠ 0

There is a minor of order 1, which is not zero

∴* ρ *(*A*)* *=* *1

Example 1.3

Find the rank of the matrix

*Solution:*

Let A=

Order Of A is 3x3

∴* ρ *(*A*)* *≤* *3

Consider the third order minor = 6 ≠ 0

There is a minor of order 3, which is not zero

∴*ρ *(*A*)* *=* *3.

Example 1.4

Find the rank of the matrix

*Solution:*

Let A=

Order Of A is 3x3

∴* ρ *(*A*)* *≤* *3

Consider the third order minor

Since the third order minor vanishes, therefore *ρ*(*A*) ≠ 3

Consider a second order minor

There is a minor of order 2, which is not zero.

∴* ρ*(*A*)* *=* *2.

Example 1.5

Find the rank of the matrix

*Solution:*

Let *A* =

Order of *A* is 3 × 4

∴* ρ*(*A*)≤* *3.

Consider the third order minors

Since all third order minors vanishes, *ρ*(*A*) ≠ 3.

Now, let us consider the second order minors,

Consider one of the second order minors

There is a minor of order 2 which is not zero.

∴ρ* *(*A*)* *=* *2.

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.

Example 1.9

Show that the equations *x* + *y* = 5, 2*x* + *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 2*x* + *y* = 5, 4*x* + 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*

⇒ 2*x* + *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 3*x* − 2 *y* = 6, 6*x* − 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 2*x* + *y* + *z* = 5, *x* + *y* + *z* = 4, *x* − *y* + 2*z* = 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* + 3*z* = 14, *x* + 4 *y* + 7*z* = 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* *−* *2*z*,

(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 − 2*k* , 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* + 7*z* = 14, 3*x* + 8 *y* − 2*z* = 13, 7*x* − 8 *y* + 26*z* = 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* − 3*z* = −2, 3*x* − *y* − 2*z* = 1, 2*x* + 3*y* − 5*z* = *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 + 7*k* = 0

7*k* = −21 .

*k *= −3

Example 1.16

Find *k,* if the equations *x* + *y* + *z* = 7, *x* + 2 *y* + 3*z* = 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* + 40*b* + 10*c*

6,725 = *a* + 35*b* + 9*c*

7,100 = *a* + 40*b* + 12*c*

The Matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation

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

*P*at* l *= 50,* m=*15 = 5000 + 30(50) + 75(15)

=7625 units.

∴The production = 7,625 units.

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12th Business Maths and Statistics : Chapter 1 : Applications of Matrices and Determinants : Rank of a Matrix: Solved Example Problems |

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