# CONSISTENCY OF SYSTEM OF LINEAR EQUATIONS BY RANK METHOD

A system of linear equations, written in the form of AX = B, is consistent if and only if the rank of the coefficient of matrix is equal to the rank of the augmented matrix, that is

ρ(A)  =  ρ([A|B])

We have the following cases :

Case 1 :

If there are n unknowns in the system of equations and

ρ(A)  =  ρ([A|B]) = n

then the system AX = B, is consistent and has a unique solution.

Case 2 :

if ρ(A)  =  ρ([A|B]) < n

then ρ(A)  =  ρ([A|B]) = 2 or 1. Then, the system will have infinitely many solution.

Case 3 :

if ρ(A)   ρ([A|B])

Then, the system is inconsistent and it has no solution.

Test for consistency and if possible, solve the following systems of equations by rank method.

Problem 1 :

x - y + 2z = 2, 2x + y + 4z = 7, 4z - y + z = 4

Solution :

ρ(A)  =  ρ([A|B]) = 3

The system is consistent and it has unique solution.

x - y + 2z = 2 -----(1)

3y = 3 ----(2)

-7z = -7 ------(3)

z = 1

y = 3/3

y = 1

Applying the value of y and z in (1), we get

x - 1 + 2(1) = 2

x = 1

So, the solution is (1, 1, 1).

Problem 2 :

3x + y + z = 2, x - 3y + 2z = 1 and 7x - y + 4z = 5

Solution :

ρ(A)  =  ρ([A|B]) = 2

The system is consistent and it has infinitely many solution.

x - 3y + 2z = 1 ----(1)

10y - 5z = -1 -----(2)

Let z = t

10y = -1 + 5t

y = (5t - 1)/10

Applying the value of y and z in (1), we get

So, the solution is

Problem 3 :

2x + 2y + z = 5, x - y + z = 1 and 3x + y + 2z = 4

Solution :

Here ρ(A)  ρ([A|B]), so the given system of equations is having no solution.

Problem 4 :

2x - y + z = 2, 6x - 3y + 3z = 6 and 4x - 2y + 2z = 4

Solution :

ρ(A)  =  ρ([A|B]) = 1

The system is consistent and it has infinitely many solution.

6x - 3y + 3z = 6

y = s and z = t

6x - 3s + 3t = 6

6x = 6 + 3s - 3t

x = (6 + 3s - 3t)/6

x = (2 + s - t)/3

So, the solution is

(2 + s - t)/3, s, t)

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