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Home > Matrix & Vector calculators > LU decomposition using Crout's method of Matrix example
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10. LU decomposition using Crout'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]]` (Previous example) | 3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]` (Next example) |
2. Example `[[3,2,4],[2,0,2],[4,2,3]]`
Find LU decomposition using Crout's method of Matrix ... `[[3,2,4],[2,0,2],[4,2,3]]`
Solution: Crout's method for LU decomposition Let `A=LU`
| `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `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` | |
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| `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `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)` | |
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This implies `l_(11)=3`
`l_(11)u_(12)=2=>3xxu_(12)=2=>u_(12)=2/3`
`l_(11)u_(13)=4=>3xxu_(13)=4=>u_(13)=4/3`
`l_(21)=2`
`l_(21)u_(12) + l_(22)=0=>2xx2/3 + l_(22)=0=>l_(22)=-4/3`
`l_(21)u_(13) + l_(22)u_(23)=2=>2xx4/3 + (-4/3)xxu_(23)=2=>u_(23)=1/2`
`l_(31)=4`
`l_(31)u_(12) + l_(32)=2=>4xx2/3 + l_(32)=2=>l_(32)=-2/3`
`l_(31)u_(13) + l_(32)u_(23) + l_(33)=3=>4xx4/3 + (-2/3)xx1/2 + l_(33)=3=>l_(33)=-2`
`:.A=L xx U=LU`
| `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `3` | |
| = | | `3` | `0` | `0` | | | `2` | `-4/3` | `0` | | | `4` | `-2/3` | `-2` | |
| `xx` | | `1` | `2/3` | `4/3` | | | `0` | `1` | `1/2` | | | `0` | `0` | `1` | |
| = | | `3` | `2` | `4` | | | `2` | `0` | `2` | | | `4` | `2` | `3` | |
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This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
1. Example `[[8,-6,2],[-6,7,-4],[2,-4,3]]` (Previous example) | 3. Example `[[1,1,1],[-1,-3,-3],[2,4,4]]` (Next example) |
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