Solving systems of linear equations using LU decomposition using Doolittle's method calculator

SolutionHelp

$2x+y+z=5,3x+5y+2z=15,2x+y+4z=8$

$2x+5y=16,3x+y=11$

$2x+5y=21,x+2y=8$

$2x+y=8,x+2y=1$

$2x+3y-z=5,3x+2y+z=10,x-5y+3z=0$

$x+y+z=3,2x-y-z=3,x-y+z=9$

$x+y+z=7,x+2y+2z=13,x+3y+z=13$

$2x-y+3z=1,-3x+4y-5z=0,x+3y-6z=0$

Example

1. Solve Equations 2x+5y=21,x+2y=8 using Doolittle'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 =

$[[21],[8]]$

Doolittle's method for LU decomposition
Let

$A=LU$

	$2$	$5$
	$1$	$2$

	$1$	$0$
	$l_(21)$	$1$

$xx$

	$u_(11)$	$u_(12)$
	$0$	$u_(22)$

	$2$	$5$
	$1$	$2$

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

This implies

$u_(11)=2$

$u_(12)=5$

$l_(21)u_(11)=1=>l_(21)xx2=1=>l_(21)=1/2$

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

$:.A=L xx U=LU$

	$2$	$5$
	$1$	$2$

	$1$	$0$
	$1/2$	$1$

$xx$

	$2$	$5$
	$0$	$-1/2$

	$2$	$5$
	$1$	$2$

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 Doolittle's method is

$x=-2 and y=5$