II B. Com

1. Test of
Consistency (or)
Consistency of
Linear Equations
Dr. R. MUTHUKRISHNAVENI
SAIVA BHANU KSHATRIYA
COLLEGE, ARUPPUKOTTAI

2. Consistency of linear equations
– A system of linear equation is said to be consistent. It refers to the
linear equations have only one solution
– Ex. 1 – The system of linear equations
– X + Y =3
– 2X – Y = 3
– Has a solution X = 2 and Y = 1.
– Hence the system of linear equation is consistent
X + Y =3
2X - Y= 3
3X = 6
X=6/3=2
X=2
X+Y=3
2+Y=3
Y=3-2
Y=1

3. Inconsistency of Linear equations
– A system of linear equation is said to be inconsistent. It refers to the
linear equations have more than one solution
– Ex. 2 The system of linear equation
– X +Y = 3
– 3X + 3Y = 9
– Has more than one solution. The equation is a multiple of 1st
equation. From the equation, X has the value of 1 or 2 like wise Y has
the value of 2 or 1. That means X = 1; Y = 2 or X= 2; Y =1

4. How to find
consistency/inconsistency
– Step 1: Convert the equations in to matrix form(AX = B)
– Step 2: club the first matrix and answer matrix (A:B)
– Step 3. Find the rank of A and A:B with help of Gauss elimination
method(only row elimination that is row rank(row echelon)
– Step 4: If the Rank of A = Rank of A:B then the equations are
consistent
– Step 5: If the rank of a matrix(A:B) = the number of variables then the
equations have unique solution for variable
– Step 6: Repeated the row elimination process find the variable value
X + 2Y = 5
4X – Y = 2
1 2
4 −1
x
𝑋
𝑌
=
5
2
AX = B
A:B =
1 2
4 −1
|
5
2Rank of A =
2(Two rows
are
independent)
Rank of A:B =
2(Two rows are
independent)
2=2(X&Y)

5. Illustration 1
– Test the consistency of the system of linear equation and also find the
value of X and Y
– X + 2Y = 5
– 4X – Y = 2

6. Solution
– X + 2Y = 5; 4X – Y =2
– Convert into Matrix form
–
1 2
4 −1
𝑥
𝑋
𝑌
=
5
2
Here A =
1 2
4 −1
, X =
𝑋
𝑌
and B =
5
2
– A:B =
1 2
4 −1
5
2
– Rank of A
– A =
1 2
4 −1
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 4𝑅1
≅
1 2
4 − 4 −1 − 8
=
1 2
0 −9
– Two rows are independent

7. Solution - Continue
– Rank of A:B =
1 2
4 −1
5
2
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 4𝑅1
≅
1 2
4 − 4 −1 − 8
5
2 − 20
– =
1 2
0 −9
5
−18
Two rows are independent Rank of A:B = 2
– Therefore Rank of A = Rank of A:B = Number of variables(X and Y)
– 2 = 2 = 2 Hence the system linear equations is consistent and have a
unique solution

8. Solution - Continue
– Repeated the Row elimination method
– A:B =
1 2
4 −1
5
2
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 4𝑅1
≅
1 2
4 − 4 −1 − 8
5
2 − 20
– =
1 2
0 −9
5
−18
𝑅1 → 𝑅1
𝑅2 → 𝑅2 /−9
≅
1 2
0/−9 −9/−9
5
−18/−9
=
1 2
0 1
5
2
– =
1 2
0 1
5
2
𝑅1 → 𝑅1 − 2𝑅2
𝑅2 → 𝑅2
≅
1 − 0 2 − 2
0 1
5 − 4
2
=
1 0
0 1
1
2
– We can get unit square matrix of A

9. Solution - Continue
– Also we can write A:B =
1 0
0 1
1
2
as
–
1 0
0 1
x
𝑋
𝑌
=
1
2
– 1X +0Y =1 → X = 1
– 0X+1Y = 2 → Y= 2
– Hence it is a unique solution of the system.

10. Illustration 2
– Test the consistency of the system of linear equation and also find the
value of X , Y and Z
– X + Y + Z =9
– 2X + 5Y + 7Z = 52
– 2X + Y – Z = 0

11. Solution
– X + Y + Z =9
– 2X + 5Y + 7Z = 52
– 2X + Y – Z = 0
–
1 1 1
2 5 7
2 1 −1
𝑋
𝑌
𝑍
=
9
52
0
– AX = B
– A:B =
1 1 1
2 5 7
2 1 −1
9
52
0

12. Solution - Continue
– A:B =
1 1 1
2 5 7
2 1 −1
9
52
0
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 2𝑅1
𝑅3 → 𝑅 − 2𝑅1
≅
1 1 1
2 − 2 5 − 2 7 − 2
2 − 2 1 − 2 −1 − 2
9
52 − 18
0 − 18
– =
1 1 1
0 3 5
0 −1 −3
9
34
−18
𝑅1 → 𝑅1
𝑅2 → 𝑅2
𝑅3 → −3𝑅3 − 𝑅2
≅
1 1 1
0 3 5
0 3 − 3 9 − 5
9
34
54 − 34

13. Solution - Continue
–
1 1 1
0 3 5
0 3 − 3 9 − 5
9
34
54 − 34
=
1 1 1
0 3 5
0 0 4
9
34
20
–
1 1 1
0 3 5
0 0 4
𝑋
𝑌
𝑍
=
9
34
20
𝑋 + 𝑌 + 𝑍 = 9
3𝑌 + 5𝑍 = 34
4𝑍 = 20
– Z = 20/4 =5
– 3Y+ 5(5) = 34; 3Y = 34 – 25; 3Y = 9; Y = 9/3 = 3
– X+3+5 = 9; X = 9 – 3 – 5; X = 1
– X = 1; Y = 3; Z = 5
Three rows of A and A:B
are independent
Rank of both matrices = 3
Number of variables = 3
(X,Y&Z)

14. Illustration 3
– Test the consistency of the system of equations.
–
𝑋 + 𝑌 + 𝑍 = 6
2𝑋 + 3𝑌 − 𝑍 = 5
– Sol:
–
1 1 1
2 3 −1
𝑋
𝑌
𝑍
=
6
5
⇒ 𝐴𝑋 = 𝐵; A=
1 1 1
2 3 −1
X=
𝑋
𝑌
𝑍
B=
6
5
– A:B =
1 1 1
2 3 −1
6
5
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 2𝑅1
≅
1 1 1
2 − 2 3 − 2 −1 − 2
6
5 − 12

15. Solution - Continue
–
1 1 1
2 − 2 3 − 2 −1 − 2
6
5 − 12
=
1 1 1
0 1 −3
6
−7
– Rank of A = Rank of A:B = 2 (Two rows independent)
– But Number of variables = 3
– Therefore the system of linear equation is consistent and the system
has infinite number of solution

16. Illustration 4
– Test the consistency of the system of equations
–
𝑋 − 4𝑌 + 7𝑍 = 8
3𝑋 + 8𝑌 − 2𝑍 = 6
7𝑋 − 8𝑌 + 26𝑍 = 31
– Sol:
–
1 −4 7
3 8 −2
7 −8 26
𝑋
𝑌
𝑍
=
8
6
31

17. Solution - continue
–
1 −4 7
3 8 −2
7 −8 26
8
6
31
𝑅1 → 𝑅1
𝑅2 → 𝑅2 − 3𝑅1
𝑅3 → 𝑅3 − 7𝑅1
– ≅
1 −4 7
3 − 3 8 + 12 −2 − 21
7 − 7 −8 + 28 26 − 49
8
6 − 24
31 − 56
=
1 −4 7
0 20 −23
0 20 −23
8
−18
−25
𝑅1 → 𝑅1
𝑅2 → 𝑅2
𝑅3 → 𝑅3 − 𝑅2
≅
1 −4 7
0 20 −23
0 20 − 20 −23 + 23
8
−18
−25 + 18

18. Solution - Continue
–
1 −4 7
0 20 −23
0 0 0
8
−18
−7
– Rank of A = 2 (Two rows are independent) and Rank of A:B = 3 (Three
rows are independent) are not equal.
– The system of equations are inconsistent.