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Home > Matrix & Vector calculators > LU Decomposition using Gauss Elimination method of Matrix example
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8. LU decomposition using Gauss Elimination method of matrix example
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- Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
- Example `[[3,2,4],[2,0,2],[4,2,3]]`
- Example `[[1,1,1],[-1,-3,-3],[2,4,4]]`
- Example `[[2,3],[4,10]]`
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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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1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
1. Find LU Decomposition using Gauss Elimination method of Matrix ... `[[3,2,4],[2,0,2],[4,2,3]]`
Solution:
`LU` decomposition : If we have a square matrix A, then an upper triangular matrix U can be obtained without pivoting under Gaussian Elimination method, and there exists lower triangular matrix L such that A=LU.
Here `A` | = | | `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `3` | |
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Using Gaussian Elimination method `R_2 larr R_2-``(2/3)``xx R_1` `[:.L_(2,1)=color{blue}{2/3}]`
= | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `4` | `2` | `3` | |
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`R_3 larr R_3-``(4/3)``xx R_1` `[:.L_(3,1)=color{blue}{4/3}]`
= | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` | `-2/3` | `-7/3` | |
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`R_3 larr R_3-``(1/2)``xx R_2` `[:.L_(3,2)=color{blue}{1/2}]`
= | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` | `0` | `-2` | |
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`:.U` | = | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` | `0` | `-2` | |
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`L` is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal.
`:.L` | = | | `1` | `0` | `0` | | | `color{blue}{2/3}` | `1` | `0` | | | `color{blue}{4/3}` | `color{blue}{1/2}` | `1` | |
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`:.` LU decomposition for A is
`A` | = | | `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `3` | |
| = | | `1` | `0` | `0` | | | `2/3` | `1` | `0` | | | `4/3` | `1/2` | `1` | |
| `xx` | | `3` | `2` | `4` | | | `0` | `-4/3` | `-2/3` | | | `0` | `0` | `-2` | |
| = | `LU` |
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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