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Home > Matrix & Vector calculators > LU decomposition using Doolittle's method of Matrix example
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9. LU decomposition using Doolittle's 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 Doolittle's method of Matrix ... `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
Solution: Doolittle's method for LU decomposition Let `A=LU`
| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `1` | `0` | `0` | | | `l_(21)` | `1` | `0` | | | `l_(31)` | `l_(32)` | `1` | |
| `xx` | | `u_(11)` | `u_(12)` | `u_(13)` | | | `0` | `u_(22)` | `u_(23)` | | | `0` | `0` | `u_(33)` | |
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| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `u_(11)` | `u_(12)` | `u_(13)` | | | `l_(21)u_(11)` | `l_(21)u_(12) + u_(22)` | `l_(21)u_(13) + u_(23)` | | | `l_(31)u_(11)` | `l_(31)u_(12) + l_(32)u_(22)` | `l_(31)u_(13) + l_(32)u_(23) + u_(33)` | |
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This implies `u_(11)=8`
`u_(12)=-6`
`u_(13)=2`
`l_(21)u_(11)=-6=>l_(21)xx8=-6=>l_(21)=-3/4`
`l_(21)u_(12) + u_(22)=7=>(-3/4)xx(-6) + u_(22)=7=>u_(22)=5/2`
`l_(21)u_(13) + u_(23)=-4=>(-3/4)xx2 + u_(23)=-4=>u_(23)=-5/2`
`l_(31)u_(11)=2=>l_(31)xx8=2=>l_(31)=1/4`
`l_(31)u_(12) + l_(32)u_(22)=-4=>1/4xx(-6) + l_(32)xx5/2=-4=>l_(32)=-1`
`l_(31)u_(13) + l_(32)u_(23) + u_(33)=3=>1/4xx2 + (-1)xx(-5/2) + u_(33)=3=>u_(33)=0`
`:.A=L xx U=LU`
| `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
| = | | `1` | `0` | `0` | | | `-3/4` | `1` | `0` | | | `1/4` | `-1` | `1` | |
| `xx` | | `8` | `-6` | `2` | | | `0` | `5/2` | `-5/2` | | | `0` | `0` | `0` | |
| = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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