Question

Solve the following systems of linear equations by Gaussian elimination method:

(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1 

(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2

Answer 1

(i) 2x – 2y + 3z = 2, x + 2y – z = 3 and 3x – y + 2z = 1 

The matrix form of the above equations is

(i.e) AX = B 

The augment matrix (A, B) is

The above matrix is in echelon form. 

Now writing the equivalent equations

Substituting z = 4 in (2) we get 

-6y + 20 = -4 

⇒ -6y = -4 – 20 = -24 

⇒ y = 4 

Substituting z = 4 and y = 4 in (1) we get 

x + 8 – 4 = 3 

⇒ x + 4 = 3 

⇒ x = 3 – 4 = -1 

So, x = -1; y = 4; z = 4 

(ii) 2x + 4y + 6z = 22 …… (1) 

3x + 8y + 5z = 27 ……. (2)

-x + y + 2z = 2 ……. (3) 

Divide equation (1) by 2 we get 

x + 2y + 3z = 11 ……. (1) 

3x + 8y + 5z = 27 …….. (2) 

-x + y + 2z = 2 ……. (3) 

The matrix form of the above equations is

(i.e) AX = B 

The augment matrix (A, B) is

The above matrix is in echelon form. 

Now writing the equivalent equations

Substituting z = 2 in (2) we get 

y – 4 = -3 

⇒ y = -3 + 4 = 1 

Substituting z = 2, y = 1 in (1) we get 

x + 2(1) + 3(2) = 11 

⇒ x + 2 + 6 = 11

⇒ x + 8 = 11 

⇒ x = 11 – 8 = 3 

x = 3, y = 1, z = 2