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I'm currently having trouble getting my head around solving this matrix (eigenvalue problem?) transcendental equation.

I have a matrix 4x4 where, $\det(M)=0$. And where ij of the matrix is a function of 4 variables, 2 of them are known where 2 are unknown, $\alpha_r$ (alphar) and $\epsilon$ (epsilon, the strain). I've set strain to a reasonable value (0.05) which leaves me one variable, alphar, to solve for. The matrix I have is below,

matrix

and the Mathematica code is (with a=alphar)

M = {{6.08259*10^22 Cosh[0.76724 a], 2.98856*10^23 Cosh[3.53664 a], -1.60097*10^21 Cosh[0.191442 a], -6.58085*10^21 Cosh[0.823236 a]},       {-7.24194*10^22 Sinh[0.76724 a], 1.63044*10^24 Sinh[3.53664 a], -1.89318*10^21 Sinh[0.191442 a], 3.40266*10^22 Sinh[0.823236 a]},       {2.21319*10^11 Sinh[0.76724 a], 2.21319*10^11 Sinh[3.53664 a], 1.46171*10^11 Sinh[0.191442 a], 1.46171*10^11 Sinh[0.823236 a]},       {-3.57353*10^11 Cosh[0.76724 a], 1.01793*10^12 Cosh[3.53664 a], 5.59395*10^10 Cosh[0.191442 a], -1.44394*10^11 Cosh[0.823236 a]}} 

I need to find the root so that $\det(M)=0$. I've tried using

FindRoot[Det[M]==0, {a,0}] 

But I don't get a reasonable value (depending on initial guess ranging from 0 to anything between 5 for example) I get 0 or almost 0, I should hopefully get a value between 0.1 and 0.8 depending on initial strain value.

From similar papers on the topic it seems I should have one transcendental equation in terms of the two variables, $\alpha_r$ and $\epsilon$. Would I be right in assuming this would be the characteristic polynomial of the matrix?

Sorry for the vague post, I tried getting in as much info as possible without blabbering on!

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    That is an exceedingly complicated transcendental equation that you have there... and the numbers are quite large, too. You might want to consider recasting your data in different units...2011-12-04
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    What does your matrix look like before picking a value for strain?2011-12-04
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    dunks: If you plot the determinant, you see that 0 is clearly the only solution, which agrees with the result of FindRoot. Maybe you have an error in your matrix? (Or maybe I had a copying error - I added the Mma code for the matrix to your post, can you check it?)2011-12-04
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    In fact, the determinant is the sum of terms like $-k_1 \cosh(k_2 \alpha_r)\cosh(k_3 \alpha_r)\sinh(k_4 \alpha_r)\sinh(k_5 \alpha_r)$ (where each $k_i>0$ is different), so it is clearly negative for all real $\alpha_r \neq 0$. And is only zero for $\alpha_r = 0$.2011-12-04
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    Thanks for the replies. J.M, tell me about it. I'm trying to keep the numbers as they are due to Er and Em being Young's Modulus of the material.2011-12-04
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    Aleks Vlasev, the matrix before setting a value for strain I'm afraid is too large to show you with pasting the Mathematica code.2011-12-04
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    @Simon , thanks for the help. I checked your mma code and all seems to be in order, I've double checked my original code and it should be fine as well. The value of alphar for Det[matrix]==0 should not be a negative number, with a strain range from 0.01 to 0.1 alphar should be from 0.1 to around 0.8. I've included a link to a pastebin of my mma code if you would be so kind to have a look? pastebin.com/raw.php?i=C3mPV4G32011-12-04
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    @Simon http://pastebin.com/raw.php?i=MfXNRkpc updated mma code.2011-12-04
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    @dunks: Not looking good. At the end of [your code](http://pastebin.com/MfXNRkpc) (removing the `MatrixForm`), try `dbqs = Det[10^-15 bqs];Plot[Evaluate[Table[Tooltip[dbqs, e11] , {e11, .01, .2, .01}]], {ar, 0, .01}]` You'll see that the `Det[]` always seems to be negative. In general, the expression is a bit messy, but it might be possible to prove this. Since you say that the Det should not be zero, I guess you've made a mistake somewhere.2011-12-05

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