Test of consistency

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 1 st 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 x AX = B   A:B =   Rank 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 = Here A = , X = and B = A:B = Rank of A A = = Two rows are independent  

7. Solution - Continue Rank of A:B = = 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 = = = = = We can get unit square matrix of A  

9. Solution - Continue Also we can write A:B = as x = 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 = AX = B A:B =  

12. Solution - Continue A:B = =  

13. Solution - Continue = = 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. Sol: = A= X= B= A:B =  

15. Solution - Continue = 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 Sol: =  

17. Solution - continue =  

18. Solution - Continue 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.