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]]`

Here `A`

=

`2` `5` `3` `1`

Using Gaussian Elimination method
`R_2 larr R_2-``(3/2)``xx R_1` `[:.L_(2,1)=color{blue}{3/2}]`

=	`2` `5` `0` `-13/2`

`:.U`

=

`2` `5` `0` `-13/2`

`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

`A`

=

`2` `5` `3` `1`

=

`1` `0` `3/2` `1`

`xx`

`2` `5` `0` `-13/2`

=

`LU`

Now, `Ax=B`, and `A=LU => LUx=B`

let `Ux=y`, then `Ly=B =>`

`1` `0` `3/2` `1`

`xx`

`y_1` `y_2`

=

`16` `11`

			``	`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`

`2` `5` `0` `-13/2`

`xx`

`x` `y`

=

`16` `-13`

			``	`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`

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