Home > Matrix & Vector calculators > Row Space example

26. Row Space example ( 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. Determinant by gaussian elimination
  21. Expanding determinant along row / column
  22. Determinants using montante (bareiss algorithm)
  23. Leibniz formula for determinant
  24. determinants using Sarrus Rule
  25. determinants using properties of determinants
  26. Row Space
  27. Column Space
  28. Null Space

25. determinants using properties of determinants
(Previous method)
2. Example `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]`
(Next example)

1. Example `[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`





1. 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]]`


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



25. determinants using properties of determinants
(Previous method)
2. Example `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]`
(Next example)





Share this solution or page with your friends.
 
 
Copyright © 2025. All rights reserved. Terms, Privacy
 
 

.