Solving 7x7 Matrix Returns Very Long Fractions

8 Ansichten (letzte 30 Tage)
balsip
balsip am 12 Okt. 2016
Kommentiert: Steven Lord am 13 Okt. 2016
Hello, I'm solving a 7x7 matrix. After inputting the A matrix and B vector and solving, the returned solutions for X are very, very long fractions. How can I specify that I would like these returned in rounded numbers (in scientific notation, for example), not in fractions? Code is below, and thank you.
Cx = 6 + 0.0806;
Cy = 6.8359;
syms w1 w2 w3 w4 w5 w6 w7;
eqn1 = + ((Cx)*w1) - (4*w2) + (1*w3) + (0*w4) + (0*w5) + (0*w6) - (4*w7) == (Cy)*(2*.125);
eqn2 = - (4*w1) + ((Cx)*w2) - (4*w3) + (1*w4) + (0*w5) + (0*w6) + (0*w7) == (Cy)*sin(2*pi*.250);
eqn3 = + (1*w1) - (4*w2) + ((Cx)*w3) - (4*w4) + (1*w5) + (0*w6) + (0*w7) == (Cy)*sin(2*pi*.375);
eqn4 = + (0*w1) + (1*w2) - (4*w3) + ((Cx)*w4) - (4*w5) + (1*w6) + (0*w7) == (Cy)*sin(2*pi*.500);
eqn5 = + (0*w1) + (0*w2) + (1*w3) - (4*w4) + ((Cx)*w5) - (4*w6) + (1*w7) == (Cy)*sin(2*pi*.625);
eqn6 = + (0*w1) + (0*w2) + (0*w3) + (1*w4) - (4*w5) + ((Cx)*w6) - (4*w7) == (Cy)*sin(2*pi*.750);
eqn7 = - (4*w1) + (0*w2) + (0*w3) + (0*w4) + (1*w5) - (4*w6) + ((Cx)*w7) == (Cy)*sin(2*pi*.875);
[A,B] = equationsToMatrix([eqn1, eqn2, eqn3, eqn4, eqn5, eqn6, eqn7], [w1, w2, w3, w4, w5, w6, w7]);
X = linsolve(A,B)
This returns the following (only w1 solution showing due to length):
X = 2182838934989084222859735148265186349320750668882016039009/735711874176482141060943596914818753670316261979795226624

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 12 Okt. 2016
Use vpa() or double() on the answer
  4 Kommentare
balsip
balsip am 13 Okt. 2016
Bearbeitet: balsip am 13 Okt. 2016
That was entirely unintended. It seems like the symbolic variables are the default output of the linsolve() function.
Steven Lord
Steven Lord am 13 Okt. 2016
Only when you pass symbolic variables into linsolve.
A = gallery('minij', 7);
b = sum(A, 2);
xd = linsolve(A, b); % double
xs = linsolve(sym(A), b); % sym
whos xd xs

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by