|
|
|
|
|
|
|
|
|
|
|
Solution
|
Solution provided by AtoZmath.com
|
|
|
|
LU decomposition using Crout's method of Matrix calculator
|
 1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
2. `[[3,2,4],[2,0,2],[4,2,3]]`
3. `[[1,1,1],[-1,-3,-3],[2,4,4]]`
4. `[[1,2,3],[0,1,0],[2,3,1]]`
5. `[[1,2,1],[6,-1,0],[-1,-2,-1]]`
6. `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
7. `[[2,3],[4,10]]`
8. `[[5,1],[4,2]]`
9. `[[6,3],[4,5]]`
|
Example1. 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)=-0.75` `l_(11)u_(13)=2=>8xxu_(13)=2=>u_(13)=0.25` `l_(21)=-6` `l_(21)u_(12) + l_(22)=7=>(-6)xx(-0.75) + l_(22)=7=>l_(22)=2.5` `l_(21)u_(13) + l_(22)u_(23)=-4=>(-6)xx0.25 + 2.5xxu_(23)=-4=>u_(23)=-1` `l_(31)=2` `l_(31)u_(12) + l_(32)=-4=>2xx(-0.75) + l_(32)=-4=>l_(32)=-2.5` `l_(31)u_(13) + l_(32)u_(23) + l_(33)=3=>2xx0.25 + (-2.5)xx(-1) + l_(33)=3=>l_(33)=0` Now checking `A=LU` ? | `LU` | = | | `8` | `0` | `0` | | | `-6` | `2.5` | `0` | | | `2` | `-2.5` | `0` | |
| `xx` | | `1` | `-0.75` | `0.25` | | | `0` | `1` | `-1` | | | `0` | `0` | `1` | |
| = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
|
| And `A` | = | | `8` | `-6` | `2` | | | `-6` | `7` | `-4` | | | `2` | `-4` | `3` | |
|
Solution is possible. |
|
|
|
|
|
