2. Gauss-Jordan Elimination method example
( Enter your problem )
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- Example `[[3,1,1],[-1,2,1],[1,1,1]]`
- Example `[[2,3,1],[0,5,6],[1,1,2]]`
- Example `[[2,3],[4,10]]`
- Example `[[5,1],[4,2]]`
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Other related methods
- Adjoint method
- Gauss-Jordan Elimination method
- Cayley Hamilton method
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1. Example `[[3,1,1],[-1,2,1],[1,1,1]]` (Previous example) | 3. Example `[[2,3],[4,10]]` (Next example) |
2. Example `[[2,3,1],[0,5,6],[1,1,2]]`
2. Find Inverse of matrix using Gauss-Jordan Elimination method `A=[[2,3,1],[0,5,6],[1,1,2]]`
Solution: Given matrix is
| `2` | `3` | `1` | | | `0` | `5` | `6` | | | `1` | `1` | `2` | |
Now finding inverse of the given matrix
| `2` | `3` | `1` | | `1` | `0` | `0` | | | `0` | `5` | `6` | | `0` | `1` | `0` | | | `1` | `1` | `2` | | `0` | `0` | `1` | |
`R_1 larr R_1-:2`
= | | `1` `1=2-:2` `R_1 larr R_1-:2` | `3/2` `3/2=3-:2` `R_1 larr R_1-:2` | `1/2` `1/2=1-:2` `R_1 larr R_1-:2` | | `1/2` `1/2=1-:2` `R_1 larr R_1-:2` | `0` `0=0-:2` `R_1 larr R_1-:2` | `0` `0=0-:2` `R_1 larr R_1-:2` | | | `0` | `5` | `6` | | `0` | `1` | `0` | | | `1` | `1` | `2` | | `0` | `0` | `1` | |
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`R_3 larr R_3- R_1`
= | | `1` | `3/2` | `1/2` | | `1/2` | `0` | `0` | | | `0` | `5` | `6` | | `0` | `1` | `0` | | | `0` `0=1-1` `R_3 larr R_3- R_1` | `-1/2` `-1/2=1-3/2` `R_3 larr R_3- R_1` | `3/2` `3/2=2-1/2` `R_3 larr R_3- R_1` | | `-1/2` `-1/2=0-1/2` `R_3 larr R_3- R_1` | `0` `0=0-0` `R_3 larr R_3- R_1` | `1` `1=1-0` `R_3 larr R_3- R_1` | |
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`R_2 larr R_2-:5`
= | | `1` | `3/2` | `1/2` | | `1/2` | `0` | `0` | | | `0` `0=0-:5` `R_2 larr R_2-:5` | `1` `1=5-:5` `R_2 larr R_2-:5` | `6/5` `6/5=6-:5` `R_2 larr R_2-:5` | | `0` `0=0-:5` `R_2 larr R_2-:5` | `1/5` `1/5=1-:5` `R_2 larr R_2-:5` | `0` `0=0-:5` `R_2 larr R_2-:5` | | | `0` | `-1/2` | `3/2` | | `-1/2` | `0` | `1` | |
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`R_1 larr R_1-3/2xx R_2`
= | | `1` `1=1-3/2xx0` `R_1 larr R_1-3/2xx R_2` | `0` `0=3/2-3/2xx1` `R_1 larr R_1-3/2xx R_2` | `-13/10` `-13/10=1/2-3/2xx6/5` `R_1 larr R_1-3/2xx R_2` | | `1/2` `1/2=1/2-3/2xx0` `R_1 larr R_1-3/2xx R_2` | `-3/10` `-3/10=0-3/2xx1/5` `R_1 larr R_1-3/2xx R_2` | `0` `0=0-3/2xx0` `R_1 larr R_1-3/2xx R_2` | | | `0` | `1` | `6/5` | | `0` | `1/5` | `0` | | | `0` | `-1/2` | `3/2` | | `-1/2` | `0` | `1` | |
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`R_3 larr R_3+1/2xx R_2`
= | | `1` | `0` | `-13/10` | | `1/2` | `-3/10` | `0` | | | `0` | `1` | `6/5` | | `0` | `1/5` | `0` | | | `0` `0=0+1/2xx0` `R_3 larr R_3+1/2xx R_2` | `0` `0=-1/2+1/2xx1` `R_3 larr R_3+1/2xx R_2` | `21/10` `21/10=3/2+1/2xx6/5` `R_3 larr R_3+1/2xx R_2` | | `-1/2` `-1/2=-1/2+1/2xx0` `R_3 larr R_3+1/2xx R_2` | `1/10` `1/10=0+1/2xx1/5` `R_3 larr R_3+1/2xx R_2` | `1` `1=1+1/2xx0` `R_3 larr R_3+1/2xx R_2` | |
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`R_3 larr R_3xx10/21`
= | | `1` | `0` | `-13/10` | | `1/2` | `-3/10` | `0` | | | `0` | `1` | `6/5` | | `0` | `1/5` | `0` | | | `0` `0=0xx10/21` `R_3 larr R_3xx10/21` | `0` `0=0xx10/21` `R_3 larr R_3xx10/21` | `1` `1=21/10xx10/21` `R_3 larr R_3xx10/21` | | `-5/21` `-5/21=-1/2xx10/21` `R_3 larr R_3xx10/21` | `1/21` `1/21=1/10xx10/21` `R_3 larr R_3xx10/21` | `10/21` `10/21=1xx10/21` `R_3 larr R_3xx10/21` | |
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`R_1 larr R_1+13/10xx R_3`
= | | `1` `1=1+13/10xx0` `R_1 larr R_1+13/10xx R_3` | `0` `0=0+13/10xx0` `R_1 larr R_1+13/10xx R_3` | `0` `0=-13/10+13/10xx1` `R_1 larr R_1+13/10xx R_3` | | `4/21` `4/21=1/2+13/10xx-5/21` `R_1 larr R_1+13/10xx R_3` | `-5/21` `-5/21=-3/10+13/10xx1/21` `R_1 larr R_1+13/10xx R_3` | `13/21` `13/21=0+13/10xx10/21` `R_1 larr R_1+13/10xx R_3` | | | `0` | `1` | `6/5` | | `0` | `1/5` | `0` | | | `0` | `0` | `1` | | `-5/21` | `1/21` | `10/21` | |
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`R_2 larr R_2-6/5xx R_3`
= | | `1` | `0` | `0` | | `4/21` | `-5/21` | `13/21` | | | `0` `0=0-6/5xx0` `R_2 larr R_2-6/5xx R_3` | `1` `1=1-6/5xx0` `R_2 larr R_2-6/5xx R_3` | `0` `0=6/5-6/5xx1` `R_2 larr R_2-6/5xx R_3` | | `2/7` `2/7=0-6/5xx-5/21` `R_2 larr R_2-6/5xx R_3` | `1/7` `1/7=1/5-6/5xx1/21` `R_2 larr R_2-6/5xx R_3` | `-4/7` `-4/7=0-6/5xx10/21` `R_2 larr R_2-6/5xx R_3` | | | `0` | `0` | `1` | | `-5/21` | `1/21` | `10/21` | |
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Solution By Gauss-Jordan Elimination method. `[[4/21,-5/21,13/21],[2/7,1/7,-4/7],[-5/21,1/21,10/21]]`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
1. Example `[[3,1,1],[-1,2,1],[1,1,1]]` (Previous example) | 3. Example `[[2,3],[4,10]]` (Next example) |
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