Solve Equations 2x+5y=21,x+2y=8 using LU decomposition using Gauss Elimination 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]]`
`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.
Using Gaussian Elimination method
`R_2 larr R_2-``(1/2)``xx R_1` `[:.L_(2,1)=color{blue}{1/2}]`
`L` is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal.
| `:.L` | = | | `1` | `0` | | | `color{blue}{1/2}` | `1` | |
|
`:.` LU decomposition for A is
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 LU decomposition method is
`x=-2 and y=5`
This material is intended as a summary. Use your textbook for detail explanation.
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