Solve Equations 2x+5y=16,3x+y=11 using Crout's 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]]`

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

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

=

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

`xx`

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

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

=

`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)=3`

`l_(21)u_(12) + l_(22)=1=>3xx5/2 + l_(22)=1=>l_(22)=-13/2`



`:.A=L xx U=LU`

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

=

`2` `0` `3` `-13/2`

`xx`

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

=

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

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

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

`2` `0` `3` `-13/2`

`xx`

`y_1` `y_2`

=

`16` `11`

			``	`2y_1`			`=`	`16`	``
			``	`3y_1`	`-`	`13/2y_2`	`=`	`11`	``

Now use forward substitution method
From (1)
`2y_1=16`

`=>2y_1=16`

`=>y_1=(16)/(2)=8`

From (2)
`3y_1-13/2y_2=11`

`=>3(8)-(13y_2)/(2)=11`

`=>24-(13y_2)/(2)=11`

`=>-(13y_2)/(2)=11-24`

`=>-(13y_2)/(2)=-13`

`=>y_2=-13xx-2/13=2`

Now, `Ux=y`

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

`xx`

`x` `y`

=

`8` `2`

			``	`x`	`+`	`5/2y`	`=`	`8`	``
					``	`y`	`=`	`2`	``

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

From (1)
`x+5/2y=8`

`=>x+(5(2))/(2)=8`

`=>x+5=8`

`=>x=8-5`

`=>x=3`

Solution by Crout's method is
`x=3 and y=2`

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