Solve Equations 2x+5y=16,3x+y=11 using LU decomposition using Gauss Elimination method
Solution:
Total Equations are `2`
`2x+5y=16 -> (1)`
`3x+y=11 -> (2)`
Now converting given equations into matrix form
`[[2,5],[3,1]] [[x],[y]]=[[16],[11]]`
Now, A = `[[2,5],[3,1]]`, X = `[[x],[y]]` and B = `[[16],[11]]`
`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-``(3/2)``xx R_1` `[:.L_(2,1)=color{blue}{3/2}]`
`L` is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal.
`:.L` | = | | `1` | `0` | | | `color{blue}{3/2}` | `1` | |
|
`:.` LU decomposition for A is
Now, `Ax=B`, and `A=LU => LUx=B`
let `Ux=y`, then `Ly=B =>`
| | | `` | `y_1` | | | `=` | `16` | `` |
| | | `` | `3/2y_1` | `+` | `y_2` | `=` | `11` | `` |
Now use forward substitution method
From (1)
`y_1=16`
From (2)
`3/2y_1+y_2=11`
`=>(3(16))/(2)+y_2=11`
`=>24+y_2=11`
`=>y_2=11-24`
`=>y_2=-13`
Now, `Ux=y`
| | | `` | `2x` | `+` | `5y` | `=` | `16` | `` |
| | | | | `-` | `13/2y` | `=` | `-13` | `` |
Now use back substitution method
From (2)
`-13/2y=-13`
`=>y=-13xx-2/13=2`
From (1)
`2x+5y=16`
`=>2x+5(2)=16`
`=>2x+10=16`
`=>2x=16-10`
`=>2x=6`
`=>x=(6)/(2)=3`
Solution by LU decomposition method is
`x=3 and y=2`
This material is intended as a summary. Use your textbook for detail explanation.
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