Write down the given system of equations in the form of a matrix equation AX = B.
Step 1 :
Find the augmented matrix [A, B] of the system of equations.
Step 2 :
Find the rank of A and rank of [A, B] by applying only elementary row operations.
Note :
Column operations should not be applied.
Step 3 :
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 there are n unknowns in the system AX = B
ρ(A) = ρ([A| B]) < n
then the system is consistent and has infinitely many solutions and these solutions.
Case 3 :
If ρ(A) ≠ ρ([A| B])
then the system AX = B is inconsistent and has no solution.
Problem :
Test for consistency and if possible, solve the following systems of equations by rank method.
(i) x − y + 2z = 2, 2x + y + 4z = 7, 4x − y + z = 4
Solution :
ρ (A) = ρ ([A| B]) = 3
Since ρ (A) and ρ ([A| B]) are equal, it has unique solution.
x - y + 2z = 2 ------(1)
3y = 3
-7z = -7
z = 1
y = 1
By applying the values of y and z in (1), we get
x - 1 + 2(1) = 2
x - 1 + 2 = 2
x = 1
Hence the solution is (1, 1, 1)
(ii) 3x + y + z = 2, x − 3y + 2z =1, 7x − y + 4z = 5
Solution :
ρ (A) = ρ ([A| B]) = 2
Since ρ (A) and ρ ([A| B]) are equal and it is less than 3, it has infinitely many solution.
3x + y + z = 2 ------(1)
10y - 5z = -1
z = t
10y = 5z
10y = 5t
y = t/2
By applying the values of y and z, we get
3x + (t/2) + t = 2
3x = 2 - t - (t/2)
3x = 2 - (3t/2)
3x = (4 - 3t)/2
x = (4 - 3t) /6
Hence the solutions are ((4 - 3t)/6, t/2, t) , where t ∈ R.
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