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How to retrieve an unknown value from a matrix

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Scott Sanders
Scott Sanders il 31 Ott 2018
Commentato: Scott Sanders il 31 Ott 2018
Point of this task is to retrieve a value of "w" from the matrix "A", where we know that the determinant of this matrix is equal to zero. There will be more than one value of "w" but how do I make the "w" value my output in such a way?
m=9000000;
Ig=1350000000;
D=2.5;
t=0.02;
I=pi()*(D^4-(D-2*t)^4)/64;
l1=16;
l2=14;
l3=17;
l4=13;
k=1000000;
A=[((4*k)-((w^2)*m)) 0 (2*k*(l2-l1)); ...
0 ((4*k)-((w^2)*m)) (2*k*(l4-l3)); ...
(2*k*(l2-l1)) (2*k*(l4-l3)) ((2*k*(l1^2+l2^2-l3^2-l4^2)-(w^2*Ig)))];
det(A)=0;
  2 Commenti
the cyclist
the cyclist il 31 Ott 2018
Do you mean that you want to solve for the value of w that will make the determinant of A equal to zero?

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Risposte (2)

YT
YT il 31 Ott 2018
Modificato: YT il 31 Ott 2018
Instead of
det(A) = 0; %which should've created an error for you
You could use solve like this to find w
syms w
... %your other lines
solve(det(A) == 0,w)
More info on solve can be found here
  1 Commento
Scott Sanders
Scott Sanders il 31 Ott 2018
When I run this it does not produce any values for w. Just outputs w as a symbol.

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the cyclist
the cyclist il 31 Ott 2018
Assuming you want to solve for w, as I mentioned in my comment, then this will do it:
m=9000000;
Ig=1350000000;
D=2.5;
t=0.02;
I=pi()*(D^4-(D-2*t)^4)/64;
l1=16;
l2=14;
l3=17;
l4=13;
k=1000000;
detA = @(w) det([((4*k)-((w.^2)*m)) 0 (2*k*(l2-l1));
0 ((4*k)-((w.^2)*m)) (2*k*(l4-l3));
(2*k*(l2-l1)) (2*k*(l4-l3)) ((2*k*(l1^2+l2^2-l3^2-l4^2)-(w.^2*Ig)))]);
w_critical = fzero(detA,0.6);
figure
hold on
for w = 0.66:0.001:0.7
h = plot(w,detA(w),'o');
set(h,'Color','k')
end
A couple things to note:
First, I defined detA as a function of w.
Second, I had to choose a very good initial guess for w_critical, because your function varies over a huge range, so fzero will fail if you are not close. I found a good value by plotting your function, and zooming in on one part near zero. Here is the final plot I used:
  1 Commento
Scott Sanders
Scott Sanders il 31 Ott 2018
This isn't quite what I was looking for but think it is because I have not established the problem enough yet so will refine my input before requesting any further help. Thank you for your help

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