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Home > Matrix & Vector calculators > Row Space example
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22. Row Space example
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- Example `[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`
- Example `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]`
- Example `[[3,-1,-1],[2,-2,1]]`
- Example `[[-2,2,6,0],[0,6,7,5],[1,5,4,5]]`
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Other related methods
- Transforming matrix to Row Echelon Form
- Transforming matrix to Reduced Row Echelon Form
- Rank of matrix
- Characteristic polynomial of matrix
- Eigenvalues
- Eigenvectors (Eigenspace)
- Triangular Matrix
- LU decomposition using Gauss Elimination method of matrix
- LU decomposition using Doolittle's method of matrix
- LU decomposition using Crout's method of matrix
- Diagonal Matrix
- Cholesky Decomposition
- QR Decomposition (Gram Schmidt Method)
- QR Decomposition (Householder Method)
- LQ Decomposition
- Pivots
- Singular Value Decomposition (SVD)
- Moore-Penrose Pseudoinverse
- Power Method for dominant eigenvalue
- determinants using Sarrus Rule
- determinants using properties of determinants
- Row Space
- Column Space
- Null Space
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21. 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 `R_2 larr R_2-3xx R_1`
= | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_3 larr R_3-2xx R_1`
= | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `7` | `7` | `-4` | `11` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_4 larr R_4+ R_1`
= | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `8` | `8` | `-8` | `16` | | | `0` | `7` | `7` | `-4` | `11` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_2 larr R_2-:8`
= | | `1` | `-2` | `0` | `3` | `-4` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `7` | `7` | `-4` | `11` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_1 larr R_1+2xx R_2`
= | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `7` | `7` | `-4` | `11` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_3 larr R_3-7xx R_2`
= | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `0` | `0` | `7` | `-7` | |
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`R_3 larr R_3-:3`
= | | `1` | `0` | `2` | `1` | `0` | | | `0` | `1` | `1` | `-1` | `2` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `7` | `-7` | |
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`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` | `7` | `-7` | |
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`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` | `7` | `-7` | |
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`R_4 larr R_4-7xx R_3`
= | | `1` | `0` | `2` | `0` | `1` | | | `0` | `1` | `1` | `0` | `1` | | | `0` | `0` | `0` | `1` | `-1` | | | `0` | `0` | `0` | `0` | `0` | |
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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
21. determinants using properties of determinants (Previous method) | 2. Example `[[1,2,3,2],[3,0,1,8],[2,-2,-2,6]]` (Next example) |
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