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Home > Matrix & Vector calculators > Null Space example (Nullity of a matrix)
24. Null Space example ( Enter your problem )
( Enter your problem )
  1. Example `[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`
  2. Example `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]`
  3. Example `[[3,-1,-1],[2,-2,1]]`
  4. Example `[[-2,2,6,0],[0,6,7,5],[1,5,4,5]]`
 		Other related methods
  1. Transforming matrix to Row Echelon Form
  2. Transforming matrix to Reduced Row Echelon Form
  3. Rank of matrix
  4. Characteristic polynomial of matrix
  5. Eigenvalues
  6. Eigenvectors (Eigenspace)
  7. Triangular Matrix
  8. LU decomposition using Gauss Elimination method of matrix
  9. LU decomposition using Doolittle's method of matrix
  10. LU decomposition using Crout's method of matrix
  11. Diagonal Matrix
  12. Cholesky Decomposition
  13. QR Decomposition (Gram Schmidt Method)
  14. QR Decomposition (Householder Method)
  15. LQ Decomposition
  16. Pivots
  17. Singular Value Decomposition (SVD)
  18. Moore-Penrose Pseudoinverse
  19. Power Method for dominant eigenvalue
  20. determinants using Sarrus Rule
  21. determinants using properties of determinants
  22. Row Space
  23. Column Space
  24. Null Space
3. Example `[[3,-1,-1],[2,-2,1]]`
(Previous example)

4. Example `[[-2,2,6,0],[0,6,7,5],[1,5,4,5]]`


Find Null Space ...
`[[-2,2,6,0],[0,6,7,5],[1,5,4,5]]`

Solution:
`-2``2``6``0`
`0``6``7``5`
`1``5``4``5`


Now, reduce the matrix to reduced row echelon form
`R_1 larr R_1-:-2`

 = 
`1``-1``-3``0`
`0``6``7``5`
`1``5``4``5`


`R_3 larr R_3- R_1`

 = 
`1``-1``-3``0`
`0``6``7``5`
`0``6``7``5`


`R_2 larr R_2-:6`

 = 
`1``-1``-3``0`
`0``1``7/6``5/6`
`0``6``7``5`


`R_1 larr R_1+ R_2`

 = 
`1``0``-11/6``5/6`
`0``1``7/6``5/6`
`0``6``7``5`


`R_3 larr R_3-6xx R_2`

 = 
`1``0``-11/6``5/6`
`0``1``7/6``5/6`
`0``0``0``0`


The rank of a matrix is the number of non all-zeros rows
`:. Rank = 2`

Null Space :
Now, solve the matrix equation
`1``0``-11/6``5/6`
`0``1``7/6``5/6`
`0``0``0``0`
 
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`0`
`0`
`0`


`x_1-11/6x_3+5/6x_4=0`

`x_2+7/6x_3+5/6x_4=0`

Add equation for each free variable
`x_1-11/6x_3+5/6x_4=0`

`x_2+7/6x_3+5/6x_4=0`

`x_3=x_3`

`x_4=x_4`

Solve for each variable in terms of the free variables
`x_1=11/6x_3-5/6x_4`

`x_2=-7/6x_3-5/6x_4`

`x_3=x_3`

`x_4=x_4`

Convert this into vectors
`x_1`
`x_2`
`x_3`
`x_4`
 = 
`11/6x_3-5/6x_4`
`-7/6x_3-5/6x_4`
`x_3`
`x_4`
 = `[[11/6],[-7/6],[1],[0]]``x_3``+``[[-5/6],[-5/6],[0],[1]]``x_4`


Thus, the basis for the null space is
`[[11/6],[-7/6],[1],[0]],[[-5/6],[-5/6],[0],[1]]`

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


3. Example `[[3,-1,-1],[2,-2,1]]`
(Previous example)


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