|
|
|
|
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
|
|
|
|
Null Space calculator
|
 1. `[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`
2. `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]`
3. `[[3,-1,-1],[2,-2,1]]`
4. `[[-2,2,6,0],[0,6,7,5],[1,5,4,5]]`
5. `[[1,4,8,-1],[2,4,8,0],[3,2,4,1]]`
6. `[[-3,6,-1,1,-7],[1,-2,2,3,-1],[2,-4,5,8,-4]]`
|
Example1. Find Null 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` Null Space : Now, solve the matrix equation | `1` | `0` | `2` | `0` | `1` | | | `0` | `1` | `1` | `0` | `1` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `0` | `0` | |
| | | `x_1` | | | `x_2` | | | `x_3` | | | `x_4` | | | `x_5` | |
| = | |
`x_1+2x_3+x_5=0` `x_2+x_3+x_5=0` `x_4-x_5=0` Add equation for each free variable `x_1+2x_3+x_5=0` `x_2+x_3+x_5=0` `x_3=x_3` `x_4-x_5=0` `x_5=x_5` Solve for each variable in terms of the free variables `x_1=-2x_3-x_5` `x_2=-x_3-x_5` `x_3=x_3` `x_4=x_5` `x_5=x_5` Convert this into vectors | `x_1` | | | `x_2` | | | `x_3` | | | `x_4` | | | `x_5` | |
| = | | `-2x_3-x_5` | | | `-x_3-x_5` | | | `x_3` | | | `x_5` | | | `x_5` | |
| = | `[[-2],[-1],[1],[0],[0]]` | `x_3` | `+` | `[[-1],[-1],[0],[1],[1]]` | `x_5` |
Thus, the basis for the null space is `[[-2],[-1],[1],[0],[0]],[[-1],[-1],[0],[1],[1]]` |
|
|
|
|
|
