|
|
Home > Matrix & Vector calculators > LU decomposition using Crout's method of Matrix example
|
|
10. LU decomposition using Crout's method of matrix example
( Enter your problem )
|
- 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]]`
|
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
|
|
1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
1. Find LU decomposition using Crout's method of Matrix ... `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
Solution: Crout's method for LU decomposition Let `A=LU`
| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `l_(11)` | `0` | `0` | | | `l_(21)` | `l_(22)` | `0` | | | `l_(31)` | `l_(32)` | `l_(33)` | |
| `xx` | | `1` | `u_(12)` | `u_(13)` | | | `0` | `1` | `u_(23)` | | | `0` | `0` | `1` | |
|
| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `l_(11)` | `l_(11)u_(12)` | `l_(11)u_(13)` | | | `l_(21)` | `l_(21)u_(12) + l_(22)` | `l_(21)u_(13) + l_(22)u_(23)` | | | `l_(31)` | `l_(31)u_(12) + l_(32)` | `l_(31)u_(13) + l_(32)u_(23) + l_(33)` | |
|
This implies `l_(11)=8`
`l_(11)u_(12)=-6=>8xxu_(12)=-6=>u_(12)=-3/4`
`l_(11)u_(13)=2=>8xxu_(13)=2=>u_(13)=1/4`
`l_(21)=-6`
`l_(21)u_(12) + l_(22)=7=>(-6)xx(-3/4) + l_(22)=7=>l_(22)=5/2`
`l_(21)u_(13) + l_(22)u_(23)=-4=>(-6)xx1/4 + 5/2xxu_(23)=-4=>u_(23)=-1`
`l_(31)=2`
`l_(31)u_(12) + l_(32)=-4=>2xx(-3/4) + l_(32)=-4=>l_(32)=-5/2`
`l_(31)u_(13) + l_(32)u_(23) + l_(33)=3=>2xx1/4 + (-5/2)xx(-1) + l_(33)=3=>l_(33)=0`
`:.A=L xx U=LU`
| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `8` | `0` | `0` | | | `-6` | `5/2` | `0` | | | `2` | `-5/2` | `0` | |
| `xx` | | `1` | `-3/4` | `1/4` | | | `0` | `1` | `-1` | | | `0` | `0` | `1` | |
| = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
|
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
|
|
|
|
Share this solution or page with your friends.
|
|
|
|