Question

Solve the following equations by the method of reduction :

2x-y + z=1,  x + 2y +3z = 8, 3x + y-4z=1.

Answer 1

2x-y + z=1

x + 2y +3z = 8

3x + y-4z=1

`[[2,-1,1],[1,2,3],[3,1,-4]][[x],[y],[z]]=[[1],[8],[1]]`

R₃ → R₃ – 3R₂

`[[2,-1,1],[1,2,3],[0,-5,-13]][[x],[y],[z]]=[[1],[8],[-23]]`

R₂ → R₂ – 1/2 R₁

`[[2,-1,1],[0,5/2,5/2],[0,-5,-13]][[x],[y],[z]]=[[1],[15/2],[-23]]`

R₂ → 2/5 R₂

`[[2,-1,1],[0,1,1],[0,-5,-13]][[x],[y],[z]]=[[1],[3],[-23]]`

R₃ → R₃ + 5R₂

`[[2,-1,1],[0,1,1],[0,0,-8]][[x],[y],[z]]=[[1],[3],[-8]]`

R₁→ R₁ + R₂ ;

R₃→ R₃ x  -(1/8 )

`[[2,0,2],[0,1,1],[0,0,1]][[x],[y],[z]]=[[4],[3],[1]]`

R₁→ R₁ x (1/2)

R₂→ R₂ – R₃

`[[1,0,1],[0,1,0],[0,0,1]][[x],[y],[z]]=[[2],[2],[1]]`

R₁ → R₁ – R₃

`[[1,0,0],[0,1,0],[0,0,1]][[x],[y],[z]]=[[1],[2],[1]]`

` [[x],[y],[z]]=[[1],[2],[1]]`

x=1, y=2, z=1