Question

Solve the following equations by reduclion method 

x+3y+3z= 16 ,  x+4y+4z=21 , x+3y+4z = 19 

Answer 1

Matrix equation is 

`[(1,3,3),(1,4,4),(1,3,4)][(x),(y),(z)] = [(16),(21),(19)]`

R₂ → R₂ - R₃

`[(1,3,3),(0,1,0),(1,3,4)] [(x),(y),(z)] = [(16),(2),(19)]`

R₃ → R₃ - R₁

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

`[(x+3y+3z),(0+y+0),(0+0+z)] = [(16),(2),(3)]`

∴ x+3y+3z = 16

y = 2

z = 3

∴ x + 3(2) + 3(3) = 16

∴ x + 6 + 9 = 16

∴ x = 16 - 15 = 1

∴ x = 1 , y = 2 , z = 3