Question

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

(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5 

(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4 

(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4

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Thanks for answering

Answer 1

(i) Here the number of unknowns = 3.

The matrix form of the system is AX = B where

(i.e) AX = B 

The augmented matrix (A, B) is

Applying Gaussian elimination method on [A,B] we get

The above matrix is in echelon form also ρ(A, B) 

= ρ(A) = 3 = number of unknowns

The system of equations is consistent with a unique solution. To find the solution. Now writing the equivalent equations we get

x – y + 2z = 2 

3y = 3 ⇒ y = 1 

7z = -7 ⇒ z = 1 

Substituting z = y = 1 in (1) we get x – 1 + 2 = 2 ⇒ x = 1 

⇒ x = y = z = 1

(ii) Here the number of unknowns is 3. 

The matrix form of the given system of equations is:

AX = B

(i.e) Now the augmented matrix [A, B] is

The above matrix is in echelon form also 

ρ(A, B) = ρ(A) = 2 < number of unknowns 

The system of equations is consistent with the infinite number of solutions. 

To find the solution: 

Now writing the equivalent equations we get

(iii) Here the number of unknowns is 3. 

The matrix form of the given equation is

AX = B

The above matrix is in echelon form. 

Here ρ(A, B) = 3; ρ(A) = 2 

So ρ(A, B) ≠ ρ(A) 

The system of equations is inconsistent with no solution. 

(iv) Here the number of unknowns is 3. 

The matrix form of the given equation is

AX = B

The augmented matrix [A, B] is

The above matrix is in echelon form also ρ(A, B) 

= ρ(A) = 1 < number of unknowns 

The system of equations is consistent with the infinite number of solutions. 

To find the Solution 

Now writing the equivalent equations we get