Inaccuracy in solving simultaneous equations using matrix
古いコメントを表示
So i was solving a system of n linear equations. My coefficient matrix is a tridiagonal one.
However as i was decreasing the values inside matrix or increasing n the error was increasing rapidly.
clear
n=1000;
B=(1:n);
B=B';
A=full(gallery('tridiag',n,0.341,0.232,0.741));
x=A\B;
c=A*x-B;
error=0;
for i=1:n
error=error+abs(c(i,1));
end
error
%error = 2.174626266011847e+155
Here the system is in the form Ax=B
ideally c should contain only zero.
Can anyone a suggest a method so that i can decrease the net error.
NOTE: I also tried the Thomas Algorithm even that gave an error of similar order .
3 件のコメント
Walter Roberson
2020 年 6 月 3 日
>> rcond(A)
ans =
5.96371748872238e-170
Much too small for reliable numeric solution. rank(A) says 999 rather than 1000.
Switching to symbolic toolbox is able to get full rank and exact solution.
Susmit Kumar Mishra
2020 年 6 月 3 日
Walter Roberson
2020 年 6 月 3 日
編集済み: Walter Roberson
2020 年 6 月 3 日
n = 1000;
B = (1:n).';
A = sym( full(gallery('tridiag',n,0.341,0.232,0.741)) ); %11 seconds
x = A\B; %not fast!! 43 seconds
c = A*x-B;
error = sum(abs(c));
disp(error)
採用された回答
その他の回答 (0 件)
カテゴリ
ヘルプ センター および File Exchange で Linear Algebra についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!