3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
Solve Equations 2x+3y-z=5,3x+2y+z=10,x-5y+3z=0 using LU decomposition using Gauss Elimination method
Solution: Total Equations are `3`
`2x+3y-z=5 -> (1)`
`3x+2y+z=10 -> (2)`
`x-5y+3z=0 -> (3)`
Now converting given equations into matrix form `[[2,3,-1],[3,2,1],[1,-5,3]] [[x],[y],[z]]=[[5],[10],[0]]`
Now, A = `[[2,3,-1],[3,2,1],[1,-5,3]]`, X = `[[x],[y],[z]]` and B = `[[5],[10],[0]]`
`LU` decomposition : If we have a square matrix A, then an upper triangular matrix U can be obtained without pivoting under Gaussian Elimination method, and there exists lower triangular matrix L such that A=LU.
Here `A` | = | | `2` | `3` | `-1` | | | `3` | `2` | `1` | | | `1` | `-5` | `3` | |
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Using Gaussian Elimination method `R_2 larr R_2-``(3/2)``xx R_1` `[:.L_(2,1)=color{blue}{3/2}]`
= | | `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `1` | `-5` | `3` | |
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`R_3 larr R_3-``(1/2)``xx R_1` `[:.L_(3,1)=color{blue}{1/2}]`
= | | `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `0` | `-13/2` | `7/2` | |
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`R_3 larr R_3-``(13/5)``xx R_2` `[:.L_(3,2)=color{blue}{13/5}]`
= | | `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `0` | `0` | `-3` | |
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`:.U` | = | | `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `0` | `0` | `-3` | |
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`L` is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal.
`:.L` | = | | `1` | `0` | `0` | | | `color{blue}{3/2}` | `1` | `0` | | | `color{blue}{1/2}` | `color{blue}{13/5}` | `1` | |
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`:.` LU decomposition for A is
`A` | = | | `2` | `3` | `-1` | | | `3` | `2` | `1` | | | `1` | `-5` | `3` | |
| = | | `1` | `0` | `0` | | | `3/2` | `1` | `0` | | | `1/2` | `13/5` | `1` | |
| `xx` | | `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `0` | `0` | `-3` | |
| = | `LU` |
Now, `Ax=B`, and `A=LU => LUx=B`
let `Ux=y`, then `Ly=B =>`
| `1` | `0` | `0` | | | `3/2` | `1` | `0` | | | `1/2` | `13/5` | `1` | |
| `xx` | | = | |
| | | `` | `y_1` | | | | | `=` | `5` | `` | | | | `` | `3/2y_1` | `+` | `y_2` | | | `=` | `10` | `` | | | | `` | `1/2y_1` | `+` | `13/5y_2` | `+` | `y_3` | `=` | `0` | `` |
Now use forward substitution method From (1) `y_1=5`
From (2) `3/2y_1+y_2=10`
`=>(3(5))/(2)+y_2=10`
`=>15/2+y_2=10`
`=>y_2=10-15/2`
`=>y_2=5/2`
From (3) `1/2y_1+13/5y_2+y_3=0`
`=>((5))/(2)+(13(5/2))/(5)+y_3=0`
`=>9+y_3=0`
`=>y_3=0-9`
`=>y_3=-9`
Now, `Ux=y`
| `2` | `3` | `-1` | | | `0` | `-5/2` | `5/2` | | | `0` | `0` | `-3` | |
| `xx` | | = | |
| | | `` | `2x` | `+` | `3y` | `-` | `z` | `=` | `5` | `` | | | | | | `-` | `5/2y` | `+` | `5/2z` | `=` | `5/2` | `` | | | | | | | | `-` | `3z` | `=` | `-9` | `` |
Now use back substitution method From (3) `-3z=-9`
`=>z=(-9)/(-3)=3`
From (2) `-5/2y+5/2z=5/2`
`=>-(5y)/(2)+(5(3))/(2)=5/2`
`=>-(5y)/(2)+15/2=5/2`
`=>-(5y)/(2)=5/2-15/2`
`=>-(5y)/(2)=-5`
`=>y=-5xx-2/5=2`
From (1) `2x+3y-z=5`
`=>2x+3(2)-(3)=5`
`=>2x+3=5`
`=>2x=5-3`
`=>2x=2`
`=>x=(2)/(2)=1`
Solution by LU decomposition method is `x=1,y=2 and z=3`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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