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

**Testing the consistency of non homogeneous
linear equations (two and three variables) by rank method.**

Consider the equations *A X*= *B* in ‘*n*’
unknowns.

(i) If *ρ* ([*A*, *B*] )
= *ρ* ( *A*) , then the equations
are consistent.

(ii) If *ρ*[([*A*, *B*] )
= *ρ* ( *A* )= *n* , then the
equations are consistent and have unique solution.

(iii) If *ρ* ([ *A*, *B*] )
= *ρ* ( *A* ) < *n* , then the
equations are consistent and have infinitely many solutions.

(iv) If *ρ* ([*A*, *B* ])
≠ *ρ* ( *A*) then the equations are
inconsistent and has no solution.

**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}
*

=7625 units.

∴The production = 7,625
units.

Tags : with Solved Example Problems , 12th Business Maths and Statistics : Chapter 1 : Applications of Matrices and Determinants

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12th Business Maths and Statistics : Chapter 1 : Applications of Matrices and Determinants : Testing the consistency of non homogeneous linear equations (two and three variables) by rank method | with Solved Example Problems

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