Solve Equations x+y+z=3,2x-y-z=3,x-y+z=9 using Crout's method

Solution:
Total Equations are `3`

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

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

`x-y+z=9 -> (3)`

Now converting given equations into matrix form
`[[1,1,1],[2,-1,-1],[1,-1,1]] [[x],[y],[z]]=[[3],[3],[9]]`

Now, A = `[[1,1,1],[2,-1,-1],[1,-1,1]]`, X = `[[x],[y],[z]]` and B = `[[3],[3],[9]]`

Crout's method for LU decomposition
Let `A=LU`

`1` `1` `1` `2` `-1` `-1` `1` `-1` `1`

=

`l_(11)` `0` `0` `l_(21)` `l_(22)` `0` `l_(31)` `l_(32)` `l_(33)`

`xx`

`1` `u_(12)` `u_(13)` `0` `1` `u_(23)` `0` `0` `1`

`1` `1` `1` `2` `-1` `-1` `1` `-1` `1`

=

`l_(11)` `l_(11)u_(12)` `l_(11)u_(13)` `l_(21)` `l_(21)u_(12) + l_(22)` `l_(21)u_(13) + l_(22)u_(23)` `l_(31)` `l_(31)u_(12) + l_(32)` `l_(31)u_(13) + l_(32)u_(23) + l_(33)`

This implies
`l_(11)=1`

`l_(11)u_(12)=1=>1xxu_(12)=1=>u_(12)=1`

`l_(11)u_(13)=1=>1xxu_(13)=1=>u_(13)=1`



`l_(21)=2`

`l_(21)u_(12) + l_(22)=-1=>2xx1 + l_(22)=-1=>l_(22)=-3`

`l_(21)u_(13) + l_(22)u_(23)=-1=>2xx1 + (-3)xxu_(23)=-1=>u_(23)=1`



`l_(31)=1`

`l_(31)u_(12) + l_(32)=-1=>1xx1 + l_(32)=-1=>l_(32)=-2`

`l_(31)u_(13) + l_(32)u_(23) + l_(33)=1=>1xx1 + (-2)xx1 + l_(33)=1=>l_(33)=2`



`:.A=L xx U=LU`

`1` `1` `1` `2` `-1` `-1` `1` `-1` `1`

=

`1` `0` `0` `2` `-3` `0` `1` `-2` `2`

`xx`

`1` `1` `1` `0` `1` `1` `0` `0` `1`

=

`1` `1` `1` `2` `-1` `-1` `1` `-1` `1`

---

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

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

`1` `0` `0` `2` `-3` `0` `1` `-2` `2`

`xx`

`y_1` `y_2` `y_3`

=

`3` `3` `9`

			``	`y_1`					`=`	`3`	``
			``	`2y_1`	`-`	`3y_2`			`=`	`3`	``
			``	`y_1`	`-`	`2y_2`	`+`	`2y_3`	`=`	`9`	``

Now use forward substitution method
From (1)
`y_1=3`

From (2)
`2y_1-3y_2=3`

`=>2(3)-3y_2=3`

`=>6-3y_2=3`

`=>-3y_2=3-6`

`=>-3y_2=-3`

`=>y_2=(-3)/(-3)=1`

From (3)
`y_1-2y_2+2y_3=9`

`=>(3)-2(1)+2y_3=9`

`=>1+2y_3=9`

`=>2y_3=9-1`

`=>2y_3=8`

`=>y_3=(8)/(2)=4`

Now, `Ux=y`

`1` `1` `1` `0` `1` `1` `0` `0` `1`

`xx`

`x` `y` `z`

=

`3` `1` `4`

			``	`x`	`+`	`y`	`+`	`z`	`=`	`3`	``
					``	`y`	`+`	`z`	`=`	`1`	``
							``	`z`	`=`	`4`	``

Now use back substitution method
From (3)
`z=4`

From (2)
`y+z=1`

`=>y+(4)=1`

`=>y+4=1`

`=>y=1-4`

`=>y=-3`

From (1)
`x+y+z=3`

`=>x+(-3)+(4)=3`

`=>x+1=3`

`=>x=3-1`

`=>x=2`

Solution by Crout's method is
`x=2,y=-3 and z=4`

---