Solve Equations 2x+3y-z=5,3x+2y+z=10,x-5y+3z=0 using LU decomposition using Gauss Elimination method

Solution:
Total Equations are `3`

`2x+3y-z=5 -> (1)`

`3x+2y+z=10 -> (2)`

`x-5y+3z=0 -> (3)`

Now converting given equations into matrix form
`[[2,3,-1],[3,2,1],[1,-5,3]] [[x],[y],[z]]=[[5],[10],[0]]`

Now, A = `[[2,3,-1],[3,2,1],[1,-5,3]]`, X = `[[x],[y],[z]]` and B = `[[5],[10],[0]]`

Here `A`

=

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

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

=	`2` `3` `-1` `0` `-5/2` `5/2` `1` `-5` `3`

`R_3 larr R_3-``(1/2)``xx R_1` `[:.L_(3,1)=color{blue}{1/2}]`

=	`2` `3` `-1` `0` `-5/2` `5/2` `0` `-13/2` `7/2`

`R_3 larr R_3-``(13/5)``xx R_2` `[:.L_(3,2)=color{blue}{13/5}]`

=	`2` `3` `-1` `0` `-5/2` `5/2` `0` `0` `-3`

`:.U`

=

`2` `3` `-1` `0` `-5/2` `5/2` `0` `0` `-3`

`L` is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal.

`:.L`

=

`1` `0` `0` `color{blue}{3/2}` `1` `0` `color{blue}{1/2}` `color{blue}{13/5}` `1`

`:.` LU decomposition for A is

`A`

=

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

=

`1` `0` `0` `3/2` `1` `0` `1/2` `13/5` `1`

`xx`

`2` `3` `-1` `0` `-5/2` `5/2` `0` `0` `-3`

=

`LU`

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

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

`1` `0` `0` `3/2` `1` `0` `1/2` `13/5` `1`

`xx`

`y_1` `y_2` `y_3`

=

`5` `10` `0`

			``	`y_1`					`=`	`5`	``
			``	`3/2y_1`	`+`	`y_2`			`=`	`10`	``
			``	`1/2y_1`	`+`	`13/5y_2`	`+`	`y_3`	`=`	`0`	``

Now use forward substitution method
From (1)
`y_1=5`

From (2)
`3/2y_1+y_2=10`

`=>(3(5))/(2)+y_2=10`

`=>15/2+y_2=10`

`=>y_2=10-15/2`

`=>y_2=5/2`

From (3)
`1/2y_1+13/5y_2+y_3=0`

`=>((5))/(2)+(13(5/2))/(5)+y_3=0`

`=>9+y_3=0`

`=>y_3=0-9`

`=>y_3=-9`

Now, `Ux=y`

`2` `3` `-1` `0` `-5/2` `5/2` `0` `0` `-3`

`xx`

`x` `y` `z`

=

`5` `5/2` `-9`

			``	`2x`	`+`	`3y`	`-`	`z`	`=`	`5`	``
					`-`	`5/2y`	`+`	`5/2z`	`=`	`5/2`	``
							`-`	`3z`	`=`	`-9`	``

Now use back substitution method
From (3)
`-3z=-9`

`=>z=(-9)/(-3)=3`

From (2)
`-5/2y+5/2z=5/2`

`=>-(5y)/(2)+(5(3))/(2)=5/2`

`=>-(5y)/(2)+15/2=5/2`

`=>-(5y)/(2)=5/2-15/2`

`=>-(5y)/(2)=-5`

`=>y=-5xx-2/5=2`

From (1)
`2x+3y-z=5`

`=>2x+3(2)-(3)=5`

`=>2x+3=5`

`=>2x=5-3`

`=>2x=2`

`=>x=(2)/(2)=1`

Solution by LU decomposition method is
`x=1,y=2 and z=3`

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