2. UNIT-V- Solution of System of Linear Equations
using Matrix
οΆ Equivalent System of Linear Equations:
Two linear systems using the same set of variables are equivalent if each of
the equations in the second system can be derived algebraically from the
equations in the first system, and vice-versa.
OR
Two systems are equivalent if either both are inconsistent or each equation of
any of them is a linear combination of the equations of the other one.
OR
Two linear systems are equivalent if and only if they have the same solution
set.
οΆ Example-1.Prove that following system of linear equations (1) is
equivalent to (2).
(1) π + ππ = βππ , ππ β π = π
(2) ππ + ππ = βπ, βππ + ππ = βππ
Solution:
Here we have
ο π₯ + 4π¦ = β10 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(1)
3π₯ β π¦ = 9 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(2)
ο Multiply equation (2) by 4 and add in to (1) we get
π₯ + 4π¦ = β10
3. 12π₯ β 4π¦ = 36
_______________________
13π₯ = 26
β΄ π₯ = 2
ο Put π₯ = 2 in equation (1) we get 2 + 4π¦ = β10
β΄ 4π¦ = β12 βΉ π¦ = β3
ο Hence we get solution set (π₯, π¦) = (2, β3)for system of linear equations
(1). Now we have to check for System (2) .
ο Here we have,
4π₯ + 3π¦ = β1β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (1)
β2π₯ + 5π¦ = β19 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... (2)
ο Multiply equation (2) by 2 and add in to (1) we get
4π₯ + 3π¦ = β1
β4π₯ + 10π¦ = β38
_______________________
13π¦ = β39
β΄ π¦ = β3
ο Now put π¦ = β3in equation (1) we get 4π₯ β 9 = β1
βΉ 4π₯ = 8
βΉ π₯ = 2
ο Hence we get solution set (2,-3) for system of linear equations (2)
ο β΄ Both systems have same solution therefore we can say that system (1)
is equivalent to system (2).
οΆ Homogeneous Systemof linear Equations:
4. A system of linear equations is homogeneous if all of the constant terms are
zero:
A homogeneous system is equivalent to a matrix equation of the form AX=0
Where A is an m Γ n matrix, x is a column vector with n entries, and 0 is the
zero vectors with m entries.
For a system of homogeneous linear equations π΄π = 0.
1) π = 0 is always a solution. This solution in which each unknown has
the value zero is called the Null solution or the trivial solution. Thus a
homogeneous system is always consistent. (i.e it has solution)
2) A system of homogeneous linear equations has either the trivial
solution or an infinite number of solutions.
3) If π ( π΄) = number of unknowns, the system has only the trivial
solution.
4) If π (π΄) < number of unknowns, the system has an infinite number of
non βtrivial solutions.
5. οΆ Example-1. Determine βbβ such that the system of homogeneous
equations
ππ + π + ππ = π
π + π + ππ = π
ππ + ππ + ππ = π
Has (i) trivial solution
(ii) Non-trivial solution.
Find the NonβTrivial solution using matrix method.
Solution:
ο Here we have,
2π₯ + π¦ + 2π§ = 0
π₯ + π¦ + 3π§ = 0
4π₯ + 3π¦ + ππ§ = 0
A systemof homogeneous linear equations
AX=0
Always has a solution
If R(A) equal to n
Unique or trivial solution
If R(A) less than n
Infiniteno. of non trivial
solution
6. ο (i) For trivial Solution: We know that π₯ = 0, π¦ = 0 πππ π§ = 0. So, b
can have any value.
ο (ii) For non trivial Solution: The given equations are written in the
matrix form as
ο [
2 1 2
1 1 3
4 3 π
][
π₯
π¦
π§
] = [
0
0
0
]
ο [
1 1 3
2 1 2
4 3 π
][
π₯
π¦
π§
] = [
0
0
0
] π 1 β π 2
ο [
1 1 3
0 β1 β4
0 β1 π β 12
][
π₯
π¦
π§
] = [
0
0
0
] π 2 β π 2 β 2π 1,π 3 β π 3 β 4π 1
ο [
1 1 3
0 β1 β4
0 0 π β 8
][
π₯
π¦
π§
] = [
0
0
0
] π 3 β π 3 β π 2
ο For non trivial solution Infinite solutions =π ( π΄) = 2 < Number of
unknowns
π β 8 = 0 β΄ π = 8
οΆ Example-2. Solve the homogeneous linearsystem of equations:
π π + ππ π + π π = π
π π + ππ π + ππ π = π
βππ π β ππ π β π π = π
ππ π β ππ π + π π + π π = π
π π β ππ π β π π + π π = π
Solution:
ο We have given system of equation
π₯1 + 3π₯2 + π₯4 = 0
π₯1 + 4π₯2 + 2π₯3 = 0
β2π₯2 β 2π₯3 β π₯4 = 0
8. ο ~
[
1 3 0
0 1 2
0
0
0
0
0
0
1
0
0
1
β1
β3/2
1
0 ]
[
π₯1
π₯2
π₯3
π₯4
] =
[
0
0
0
0
0]
π 4 (
37
2
) + π 5
β΄ π₯1 + 3π₯2 + π₯4 = 0 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.(1)
π₯2 + 2π₯3 β π₯4 = 0 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (2)
π₯3 β
3
2
π₯4 = 0 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ (3)
π₯4 = 0 ... β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. (4)
ο Since π₯4 = 0 from equation (3) we get π₯3 = 0.
ο Since π₯3 = 0, π₯4 = 0 from equation (2) we get π₯2 = 0
ο Since π₯2 = 0, π₯4 = 0 from equation (1) we get π₯1 = 0.
ο Hence π₯1 = 0, π₯2 = 0, π₯3 = 0, π₯4 = 0
ο i.e. system has only a trivial solution.
οΆ Non Homogeneous Systemof Linear equations and its Solution:
ο The vector equation is equivalent to a matrix equation of the form
ο
ο Where A is an mΓn matrix, x is a column vector with n entries, and b is
a column vector with m entries.
9. ο The above system of equations AX=b is known as non homogeneous
system of equations.
ο Here C=[A/B] is Augmented matrix
οΆ Example-1. Show that the non Homogeneous systemof linear equation
are not consistant.
ππ + ππ = βππ
ππ + πππ β ππ = βπ
ππ β πππ = βπ
Solution:
ο In the matrix form π΄π = π΅ we can write
A systemof non-homogeneous linear equations
AX=B
if R(A)=R(C)
solution exists
systemis consistant
if R(A)=R(C)=n
systemhas unique
solution
if R(A)=R(C) less than n
InfiniteSolution
if R(A)# R(C)
solution does not exist
systemis inconsistant
10. [
2 6 0
6 20 β6
0 6 β18
] [
π₯
π¦
π§
] = [
β11
β3
β1
]
ο ~ [
2 6 0
0 2 β6
0 6 β18
] [
π₯
π¦
π§
] = [
β11
30
β1
] (β3) π 1 + π 2
ο ~ [
2 6 0
0 2 β6
0 0 0
] [
π₯
π¦
π§
] = [
β11
30
β91
] (β3) π 2 + π 3
ο β΄ Rank of C = 3 and Rank of A=2
ο Here, ( Rank of C = 3) β (Rank of A=2)
ο β΄ System has no solution that means system is inconsistent.
οΆ Example-2. Testfor consistencythe following system of equations and if
consistentthen solve them
π π + ππ π β π π = π
ππ π β π π + ππ π = π
ππ π β ππ π + ππ π = π
π π β π π + π π = βπ
Solution:
ο We have given system of solution
π₯1 + 2π₯2 β π₯3 = 3
3π₯1 β π₯2 + 2π₯3 = 1
2π₯1 β 2π₯2 + 3π₯3 = 2
π₯1 β π₯2 + π₯3 = β1
ο The System of equation can be written in matrix form as
[
1 2 β1
3 β1 2
2
1
β2
β1
3
1
][
π₯1
π₯2
π₯3
] = [
3
1
2
β1
]