Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Gauss Elimination Back Substitution method
Solution:
Total Equations are `3`
`x+y+z=3 -> (1)`
`2x-y-z=3 -> (2)`
`x-y+z=9 -> (3)`
Converting given equations into matrix form
| `1` | `1` | `1` | | `3` | |
| `2` | `-1` | `-1` | | `3` | |
| `1` | `-1` | `1` | | `9` | |
`R_2 larr R_2-2xx R_1`
| = | | `1` | `1` | `1` | | `3` | | | `0` | `-3` | `-3` | | `-3` | | | `1` | `-1` | `1` | | `9` | |
|
`R_3 larr R_3- R_1`
| = | | `1` | `1` | `1` | | `3` | | | `0` | `-3` | `-3` | | `-3` | | | `0` | `-2` | `0` | | `6` | |
|
`R_3 larr R_3-0.6667xx R_2`
| = | | `1` | `1` | `1` | | `3` | | | `0` | `-3` | `-3` | | `-3` | | | `0` | `0` | `2` | | `8` | |
|
`i.e.`
`x+y+z=3 ->(1)`
`-3y-3z=-3 ->(2)`
`2z=8 ->(3)`
Now use back substitution method
From (3)
`2z=8`
`=>z=(8)/(2)=4`
From (2)
`-3y-3z=-3`
`=>-3y-3(4)=-3`
`=>-3y-12=-3`
`=>-3y=-3+12=9`
`=>y=(9)/(-3)=-3`
From (1)
`x+y+z=3`
`=>1x+1(-3)+1(4)=3`
`=>1x+1=3`
`=>1x=3-1=2`
Solution using back substitution method.
`x=2,y=-3 and z=4`
This material is intended as a summary. Use your textbook for detail explanation.
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