1. Solve Equations 2x+5y=21,x+2y=8 using Crout'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]]`
Crout's method for LU decomposition
Let `A=LU`
| = | | `l_(11)` | `0` | | | `l_(21)` | `l_(22)` | |
| `xx` | |
| = | | `l_(11)` | `l_(11)u_(12)` | | | `l_(21)` | `l_(21)u_(12) + l_(22)` | |
|
This implies
`l_(11)=2`
`l_(11)u_(12)=5=>2xxu_(12)=5=>u_(12)=5/2`
`l_(21)=1`
`l_(21)u_(12) + l_(22)=2=>1xx5/2 + l_(22)=2=>l_(22)=-1/2`
`:.A=L xx U=LU`
Now, `Ax=B`, and `A=LU => LUx=B`
let `Ux=y`, then `Ly=B =>`
| | | `` | `2y_1` | | | `=` | `21` | `` |
| | | `` | `y_1` | `-` | `1/2y_2` | `=` | `8` | `` |
Now use forward substitution method
From (1)
`2y_1=21`
`=>2y_1=21`
`=>y_1=(21)/(2)=21/2`
From (2)
`y_1-1/2y_2=8`
`=>(21/2)-(y_2)/(2)=8`
`=>21/2-(y_2)/(2)=8`
`=>-(y_2)/(2)=8-21/2`
`=>-(y_2)/(2)=-5/2`
`=>y_2=-5/2xx-2=5`
Now, `Ux=y`
| | | `` | `x` | `+` | `5/2y` | `=` | `21/2` | `` |
| | | | | `` | `y` | `=` | `5` | `` |
Now use back substitution method
From (2)
`y=5`
From (1)
`x+5/2y=21/2`
`=>x+(5(5))/(2)=21/2`
`=>x+25/2=21/2`
`=>x=21/2-25/2`
`=>x=-2`
Solution by Crout's method is
`x=-2 and y=5`
This material is intended as a summary. Use your textbook for detail explanation.
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