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Code is changed on 22.07.2025, Now it also works for Complex Number.
For wrong or incomplete solution, please submit the feedback form.
So, I will try my best to improve it soon.
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Solution
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Solution provided by AtoZmath.com
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LU decomposition using Doolittle's method of Matrix calculator
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 1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
2. `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
3. `[[3,2,4],[2,0,2],[4,2,3]]`
4. `[[1,1,1],[-1,-3,-3],[2,4,4]]`
5. `[[2,3],[4,10]]`
6. `[[5,1],[4,2]]`
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Example1. 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)=-0.75` `l_(21)u_(12) + u_(22)=7=>(-0.75)xx(-6) + u_(22)=7=>u_(22)=2.5` `l_(21)u_(13) + u_(23)=-4=>(-0.75)xx2 + u_(23)=-4=>u_(23)=-2.5` `l_(31)u_(11)=2=>l_(31)xx8=2=>l_(31)=0.25` `l_(31)u_(12) + l_(32)u_(22)=-4=>0.25xx(-6) + l_(32)xx2.5=-4=>l_(32)=-1` `l_(31)u_(13) + l_(32)u_(23) + u_(33)=3=>0.25xx2 + (-1)xx(-2.5) + u_(33)=3=>u_(33)=0` Now checking `A=LU` ? | `LU` | = | | `1` | `0` | `0` | | | `-0.75` | `1` | `0` | | | `0.25` | `-1` | `1` | |
| `xx` | | `8` | `-6` | `2` | | | `0` | `2.5` | `-2.5` | | | `0` | `0` | `0` | |
| = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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| And `A` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
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Solution is possible. |
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