1. 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]]`

Here `A`

=

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

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

=	`2` `5` `0` `-1/2`

`:.U`

=

`2` `5` `0` `-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

`A`

=

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

=

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

`xx`

`2` `5` `0` `-1/2`

=

`LU`

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

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

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

`xx`

`y_1` `y_2`

=

`21` `8`

			``	`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`

`2` `5` `0` `-1/2`

`xx`

`x` `y`

=

`21` `-5/2`

			``	`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`