Advertisement Remove all ads

Solve the Following Equations by the Method of Reduction 2x-y + z=1,  x + 2y +3z = 8, 3x + y-4z=1. - Mathematics and Statistics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

Solve the following equations by the method of reduction :

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

Advertisement Remove all ads

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

Concept: Elementary Transformations
  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×