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Home > Matrix & Vector calculators > Solving systems of linear equations using LU decomposition using Doolittle's method example
9. LU decomposition using Doolittle's method example ( Enter your problem )
( Enter your problem )
  1. Example `2x+5y=21,x+2y=8`
  2. Example `2x+5y=16,3x+y=11`
  3. Example `2x+3y-z=5,3x+2y+z=10,x-5y+3z=0`
  4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`
 		Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method
2. Example `2x+5y=16,3x+y=11`
(Previous example)		4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`
(Next example)

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 Doolittle's 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]]`

Doolittle's method for LU decomposition
Let `A=LU`

`2``3``-1`
`3``2``1`
`1``-5``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)`


`2``3``-1`
`3``2``1`
`1``-5``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)`


This implies
`u_(11)=2`

`u_(12)=3`

`u_(13)=-1`



`l_(21)u_(11)=3=>l_(21)xx2=3=>l_(21)=3/2`

`l_(21)u_(12) + u_(22)=2=>3/2xx3 + u_(22)=2=>u_(22)=-5/2`

`l_(21)u_(13) + u_(23)=1=>3/2xx(-1) + u_(23)=1=>u_(23)=5/2`



`l_(31)u_(11)=1=>l_(31)xx2=1=>l_(31)=1/2`

`l_(31)u_(12) + l_(32)u_(22)=-5=>1/2xx3 + l_(32)xx(-5/2)=-5=>l_(32)=13/5`

`l_(31)u_(13) + l_(32)u_(23) + u_(33)=3=>1/2xx(-1) + 13/5xx5/2 + u_(33)=3=>u_(33)=-3`



`:.A=L xx U=LU`

`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`
 = 
`2``3``-1`
`3``2``1`
`1``-5``3`



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`
`y_2`
`y_3`
 = 
`5`
`10`
`0`


```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` 
`x`
`y`
`z`
 = 
`5`
`5/2`
`-9`


```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 Doolittle's 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 Submit Here


2. Example `2x+5y=16,3x+y=11`
(Previous example)		4. Example `x+y+z=3,2x-y-z=3,x-y+z=9`
(Next example)


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