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Question
Solve the following equations by the method of reduction :
2x-y + z=1, x + 2y +3z = 8, 3x + y-4z=1.
Solution
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
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