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)
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM