Solve Equations 2x+3y-z=5,3x+2y+z=10,x-5y+3z=0 using Gauss Elimination Back Substitution method
Solution:
Total Equations are `3`
`2x+3y-z=5 -> (1)`
`3x+2y+z=10 -> (2)`
`x-5y+3z=0 -> (3)`
Converting given equations into matrix form
| `2` | `3` | `-1` | | `5` | |
| `3` | `2` | `1` | | `10` | |
| `1` | `-5` | `3` | | `0` | |
`R_2 larr R_2-1.5xx R_1`
| = | | `2` | `3` | `-1` | | `5` | | | `0` | `-2.5` | `2.5` | | `2.5` | | | `1` | `-5` | `3` | | `0` | |
|
`R_3 larr R_3-0.5xx R_1`
| = | | `2` | `3` | `-1` | | `5` | | | `0` | `-2.5` | `2.5` | | `2.5` | | | `0` | `-6.5` | `3.5` | | `-2.5` | |
|
`R_3 larr R_3-2.6xx R_2`
| = | | `2` | `3` | `-1` | | `5` | | | `0` | `-2.5` | `2.5` | | `2.5` | | | `0` | `0` | `-3` | | `-9` | |
|
`i.e.`
`2x+3y-z=5 ->(1)`
`-2.5y+2.5z=2.5 ->(2)`
`-3z=-9 ->(3)`
Now use back substitution method
From (3)
`-3z=-9`
`=>z=(-9)/(-3)=3`
From (2)
`-2.5y+2.5z=2.5`
`=>-2.5y+2.5(3)=2.5`
`=>-2.5y+7.5=2.5`
`=>-2.5y=2.5-7.5=-5`
`=>y=-5xx-2/5=2`
From (1)
`2x+3y-z=5`
`=>2x+3(2)-1(3)=5`
`=>2x+3=5`
`=>2x=5-3=2`
`=>x=(2)/(2)=1`
Solution using back substitution method.
`x=1,y=2 and z=3`
This material is intended as a summary. Use your textbook for detail explanation.
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