Question

Solve the following equations by reduction method : 

x + 2y + z = 8 

2x+ 3y - z = 11 

3x - y - 2z = 5

Answer 1

Given :

x + 2y + z = 8 

2x+ 3y - z = 11 

3x - y - 2z = 5

Its matrix form is 

`[(1,2,1),(2,3,-1),(3,-1,-2)] [(x),(y),(z)] = [(8),(11),(5)]`

R₂ → R₂ - 2R₁ , R₃ → R₃ - 3R₁

`[(1,2,1),(0,-1,-3),(0,-7,-5)] [(x),(y),(z)] = [(8),(-5),(-19)]`

R₃ → R₃ - 7 R₂

`[(1,2,1),(0,-1,-3),(0,0,16)] [(x),(y),(z)] = [(8),(-5),(16)]`

Now write the equation in the original form 

x + 2y + z = 8 ... (i) 

- y- 3z = -5 ... (ii) 

I6z  = 16   .....(iii)

From (iii).  z = 1 

Putting z = 1 in equation (ii) 

We get y = 2 

Putting z = 1, y = 2 in equation (i) 

we get x = 3 

∴ Solution is x = 3, y = 2, z = 1.