Solving linear system - but using only parts of the Matrix

Hi everyone,

I'm trying to solve a linear system of equations but want to use only selectet rows for that operation.

For example:

D=
0	0	0	0	0.345265357599750	0.457309520629080	0.548726230994530	0	0	0	0
0	0	0	0	0.200398189744228	0.345265357599750	0.457309520629080	0	0	0	0
0	0	0	0	-0.00530900651840216	0.200398189744228	0.345265357599750	0	0	0	0
0	0	0	0	-0.369187964058863	-0.00530900651840216	0.200398189744228	0	0	0	0
0	0	0	0	-1.19314718055995	-0.369187964058863	-0.00530900651840216	0	0	0	0
0	0	0	0	-0.369187964058863	-1.19314718055995	-0.369187964058863	0	0	0	0
0	0	0	0	-0.00530900651840216	-0.369187964058863	-1.19314718055995	0	0	0	0
0	0	0	0	0.200398189744228	-0.00530900651840216	-0.369187964058863	0	0	0	0
0	0	0	0	0.345265357599750	0.200398189744228	-0.00530900651840216	0	0	0	0
0	0	0	0	0.457309520629080	0.345265357599750	0.200398189744228	0	0	0	0
0	0	0	0	0.548726230994530	0.457309520629080	0.345265357599750	0	0	0	0

U=
0
0
0
0
1.14881496080949e-07
3.47991801628408e-07
1.14881496080949e-07
0
0
0
0

I want to perform the operation

P=D\U

But only with the 3 rows that contain a result in the variable U.

P should be

P=
0
0
0
0
-7.42753614188724e-09
-2.87062227792614e-07
-7.42753614188723e-09
0
0
0
0

Can somebody help me, how to program this?

Very best

Christian

Answer 1

[Star Strider]

The lsqr function is appropriate here —

D = [...
0	0	0	0	0.345265357599750	0.457309520629080	0.548726230994530	0	0	0	0
0	0	0	0	0.200398189744228	0.345265357599750	0.457309520629080	0	0	0	0
0	0	0	0	-0.00530900651840216	0.200398189744228	0.345265357599750	0	0	0	0
0	0	0	0	-0.369187964058863	-0.00530900651840216	0.200398189744228	0	0	0	0
0	0	0	0	-1.19314718055995	-0.369187964058863	-0.00530900651840216	0	0	0	0
0	0	0	0	-0.369187964058863	-1.19314718055995	-0.369187964058863	0	0	0	0
0	0	0	0	-0.00530900651840216	-0.369187964058863	-1.19314718055995	0	0	0	0
0	0	0	0	0.200398189744228	-0.00530900651840216	-0.369187964058863	0	0	0	0
0	0	0	0	0.345265357599750	0.200398189744228	-0.00530900651840216	0	0	0	0
0	0	0	0	0.457309520629080	0.345265357599750	0.200398189744228	0	0	0	0
0	0	0	0	0.548726230994530	0.457309520629080	0.345265357599750	0	0	0	0];

U = [...
0
0
0
0
1.14881496080949e-07
3.47991801628408e-07
1.14881496080949e-07
0
0
0
0];

ix = U~=0;

format long g
P = lsqr(D(ix,:),U(ix))

lsqr converged at iteration 2 to a solution with relative residual 1e-15.

P = 11×1

0 0 0 0 -7.42753614188729e-09 -2.87062227792613e-07 -7.42753614188729e-09 0 0 0

format short

.

[Christian S.]

It works perfectly! Thank you very much Star Strider!

Very best

Christian

[Star Strider]

As always, my pleasure!