1. The system of equations X + Y + Z = 6, 2X + 3Y - Z = 5, and 7X - 8Y + 26Z = 31 is inconsistent. The ranks of the coefficient matrix A and the extended matrix A:B are not equal, with the rank of A:B being 3 and the rank of A being 2. 2. For a system of three equations with three variables, the ranks must be equal for the system to be consistent. 3. An inconsistent system has no solution or has multiple solutions, meaning the equations are dependent and do not represent a unique set of values for the variables.

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.