Solving linear system - but using only parts of the Matrix

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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

Accepted Answer

Star Strider
Star Strider on 3 May 2021
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
.

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