1. Solve Equations 2x+5y=21,x+2y=8 using Crout's 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]]`

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

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

=

`l_(11)` `0` `l_(21)` `l_(22)`

`xx`

`1` `u_(12)` `0` `1`

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

=

`l_(11)` `l_(11)u_(12)` `l_(21)` `l_(21)u_(12) + l_(22)`

This implies
`l_(11)=2`

`l_(11)u_(12)=5=>2xxu_(12)=5=>u_(12)=5/2`



`l_(21)=1`

`l_(21)u_(12) + l_(22)=2=>1xx5/2 + l_(22)=2=>l_(22)=-1/2`



`:.A=L xx U=LU`

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

=

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

`xx`

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

=

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

---

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

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

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

`xx`

`y_1` `y_2`

=

`21` `8`

			``	`2y_1`			`=`	`21`	``
			``	`y_1`	`-`	`1/2y_2`	`=`	`8`	``

Now use forward substitution method
From (1)
`2y_1=21`

`=>2y_1=21`

`=>y_1=(21)/(2)=21/2`

From (2)
`y_1-1/2y_2=8`

`=>(21/2)-(y_2)/(2)=8`

`=>21/2-(y_2)/(2)=8`

`=>-(y_2)/(2)=8-21/2`

`=>-(y_2)/(2)=-5/2`

`=>y_2=-5/2xx-2=5`

Now, `Ux=y`

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

`xx`

`x` `y`

=

`21/2` `5`

			``	`x`	`+`	`5/2y`	`=`	`21/2`	``
					``	`y`	`=`	`5`	``

Now use back substitution method
From (2)
`y=5`

From (1)
`x+5/2y=21/2`

`=>x+(5(5))/(2)=21/2`

`=>x+25/2=21/2`

`=>x=21/2-25/2`

`=>x=-2`

Solution by Crout's method is
`x=-2 and y=5`