1. Solve Equations 2x+5y=21,x+2y=8 using Doolittle's method
Solution:
Total Equations are `2`
`2x+5y=21 -> (1)`
`x+2y=8 -> (2)`
Now converting given equations into matrix form
`[[2,5],[1,2]] [[x],[y]]=[[21],[8]]`
Now, A = `[[2,5],[1,2]]`, X = `[[x],[y]]` and B = `[[21],[8]]`
Doolittle's method for LU decomposition
Let `A=LU`
| = | | `xx` | | `u_(11)` | `u_(12)` | | | `0` | `u_(22)` | |
|
| = | | `u_(11)` | `u_(12)` | | | `l_(21)u_(11)` | `l_(21)u_(12) + u_(22)` | |
|
This implies
`u_(11)=2`
`u_(12)=5`
`l_(21)u_(11)=1=>l_(21)xx2=1=>l_(21)=1/2`
`l_(21)u_(12) + u_(22)=2=>1/2xx5 + u_(22)=2=>u_(22)=-1/2`
`:.A=L xx U=LU`
Now, `Ax=B`, and `A=LU => LUx=B`
let `Ux=y`, then `Ly=B =>`
| | | `` | `y_1` | | | `=` | `21` | `` |
| | | `` | `1/2y_1` | `+` | `y_2` | `=` | `8` | `` |
Now use forward substitution method
From (1)
`y_1=21`
From (2)
`1/2y_1+y_2=8`
`=>((21))/(2)+y_2=8`
`=>21/2+y_2=8`
`=>y_2=8-21/2`
`=>y_2=-5/2`
Now, `Ux=y`
| | | `` | `2x` | `+` | `5y` | `=` | `21` | `` |
| | | | | `-` | `1/2y` | `=` | `-5/2` | `` |
Now use back substitution method
From (2)
`-1/2y=-5/2`
`=>y=-5/2xx-2=5`
From (1)
`2x+5y=21`
`=>2x+5(5)=21`
`=>2x+25=21`
`=>2x=21-25`
`=>2x=-4`
`=>x=(-4)/(2)=-2`
Solution by Doolittle's method is
`x=-2 and y=5`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then