|
|
|
|
Code is changed on 22.07.2025, Now it also works for Complex Number.
For wrong or incomplete solution, please submit the feedback form.
So, I will try my best to improve it soon.
|
|
|
|
|
|
|
|
|
Solution
|
Solution provided by AtoZmath.com
|
|
|
|
Row Space calculator
|
 1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
2. `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
3. `[[3,2,4],[2,0,2],[4,2,3]]`
4. `[[1,1,1],[-1,-3,-3],[2,4,4]]`
5. `[[2,3],[4,10]]`
6. `[[5,1],[4,2]]`
|
Example1. Find Row Space ...
`[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`Solution: | `1` | `-2` | `0` | `3` | `-4` | | | `3` | `2` | `8` | `1` | `4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
Now, reduce the matrix to reduced row echelon form interchanging rows `R_1 harr R_2` | = | | `3` | `2` | `8` | `1` | `4` | | | `1` | `-2` | `0` | `3` | `-4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
|
`R_1 larr R_1-:3` | = | | `1` | `2/3` | `8/3` | `1/3` | `4/3` | | | `1` | `-2` | `0` | `3` | `-4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
|
`R_2 larr R_2- R_1` | = | | `1` | `2/3` | `8/3` | `1/3` | `4/3` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
|
`R_3 larr R_3-2xx R_1` | = | | `1` | `2/3` | `8/3` | `1/3` | `4/3` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `-1` | `2` | `0` | `4` | `-3` | |
|
`R_4 larr R_4+ R_1` | = | | `1` | `2/3` | `8/3` | `1/3` | `4/3` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
|
`R_2 larr R_2xx(-3/8)` | = | | `1` | `2/3` | `8/3` | `1/3` | `4/3` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
|
`R_1 larr R_1-2/3xx R_2` | = | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
|
`R_3 larr R_3-5/3xx R_2` | = | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
|
`R_4 larr R_4-8/3xx R_2` | = | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `0` | `0` | `7` | `-7` | |
|
interchanging rows `R_3 harr R_4` | = | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `7` | `-7` | | | `0` | `0` | `0` | `3` | `-3` | |
|
`R_3 larr R_3-:7` | = | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `3` | `-3` | |
|
`R_1 larr R_1- R_3` | = | | `1` | `0` | `2` | `0` | `1` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `3` | `-3` | |
|
`R_2 larr R_2+ R_3` | = | | `1` | `0` | `2` | `0` | `1` | | | `0` | `1` | `1` | `0` | `1` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `3` | `-3` | |
|
`R_4 larr R_4-3xx R_3` | = | | `1` | `0` | `2` | `0` | `1` | | | `0` | `1` | `1` | `0` | `1` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `0` | `0` | |
|
The rank of a matrix is the number of non all-zeros rows `:. Rank = 3` Row Space : The nonzero rows in the reduced row-echelon form are a basis for the row space of the matrix `[[1,0,2,0,1]],` `[[0,1,1,0,1]],` `[[0,0,0,1,-1]]` |
|
|
|
|
|
