inv Matrix must be square

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Naema
Naema 2014-6-20
评论： Andrew 2014-6-21
how to invert alpha in the following relation.so if alpha values are from 3 to 11 with 50 steps (size of alpha is "1x50"), how can do the following:
r=log(0.001)./(-inv(alpha));
I got the following error: Error using ==> inv Matrix must be square.
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Matt J
Matt J 2014-6-20
It is not clear what operation you're trying to achieve. Since alpha is not square, inv(alpha) is indeed undefined.

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回答（1 个）

Andrew
Andrew 2014-6-20
编辑：Andrew 2014-6-20
Naema, what do you mean by inverse? If you want the element wise inverse such as
1/alpha_ij
then you would use
r=log(0.001)./(-(alpha.^(-1));
If you want alpha flipped, as in you defined alpha as linspace(3,11,50) but actually want linspace(11,3,50) then you can use the fliplr function like this:
r=log(0.001)./(-fliplr(alpha))
If you want the matrix inverse of alpha, well that's impossible to actually find (you can do some funny things with pseudo inverses but they're not real inverses).
Hope one of these is what you are looking for.
Andrew
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Naema
Naema 2014-6-20
by the way, the plot command can be: plot(h_Au,l10,'r','linewidth',2); or: plot(h_Au,r10,'r','linewidth',2); please let me know what happens with you asaaaap! thanks, Naema
Andrew
Andrew 2014-6-21
Hey, so the issue is in the r10 line.
you need to add dots to make it
r10=(1-exp(-fliplr(alpha)*(2*pi())/1310e-9).*l10).*a;
However you still have a problem because the data you have is generating INF values from this code because the exponents are too large. Unfortunately that's not something I can help you with. You probably have a typo in your code or have misinterpreted an equation, so you'll have to check that out.
Sorry for the late response. I got sidetracked by somehting.
Andrew

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