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Click here to Find the value of h,k for which the system of equations has a Unique or Infinite or no solution calculator
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Solution
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Solution provided by AtoZmath.com
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Solving systems of linear equations using Gauss-Jordan Elimination method calculator
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 1. `2x+y+z=5,3x+5y+2z=15,2x+y+4z=8`
2. `2x+5y=16,3x+y=11`
3. `2x+5y=21,x+2y=8`
4. `2x+y=8,x+2y=1`
5. `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
6. `x+y+z=3,2x-y-z=3,x-y+z=9`
7. `x+y+z=7,x+2y+2z=13,x+3y+z=13`
8. `2x-y+3z=1,-3x+4y-5z=0,x+3y-6z=0`
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Example1. Solve Equations 2x+5y=21,x+2y=8 using Gauss-Jordan Elimination methodSolution:Total Equations are `2` `2x+5y=21 -> (1)` `x+2y=8 -> (2)` Converting given equations into matrix form `R_1 larr R_1-:2` | = | | `1` `1=2-:2`
`R_1 larr R_1-:2` | `5/2` `5/2=5-:2`
`R_1 larr R_1-:2` | | `21/2` `21/2=21-:2`
`R_1 larr R_1-:2` | | | `1` | `2` | | `8` | |
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`R_2 larr R_2- R_1` | = | | `1` | `5/2` | | `21/2` | | | `0` `0=1-1`
`R_2 larr R_2- R_1` | `-1/2` `-1/2=2-5/2`
`R_2 larr R_2- R_1` | | `-5/2` `-5/2=8-21/2`
`R_2 larr R_2- R_1` | |
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`R_2 larr R_2xx-2` | = | | `1` | `5/2` | | `21/2` | | | `0` `0=0xx-2`
`R_2 larr R_2xx-2` | `1` `1=-1/2xx-2`
`R_2 larr R_2xx-2` | | `5` `5=-5/2xx-2`
`R_2 larr R_2xx-2` | |
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`R_1 larr R_1-5/2xx R_2` | = | | `1` `1=1-5/2xx0`
`R_1 larr R_1-5/2xx R_2` | `0` `0=5/2-5/2xx1`
`R_1 larr R_1-5/2xx R_2` | | `-2` `-2=21/2-5/2xx5`
`R_1 larr R_1-5/2xx R_2` | | | `0` | `1` | | `5` | |
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`i.e.` `x=-2` `y=5` Solution By Gauss jordan elimination method. `x = -2` `y = 5`
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