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From: Matt J on 26 Feb 2010 10:55
"Pascal Schulthess" <news(a)pascalschulthess.de> wrote in message <hm81f1$sp5$1(a)fred.mathworks.com>...

> this works great, although I hoped to complete this task a little faster ;) (it takes about 50 minutes)
>
> Another idea I had was to benefit from the form of of A since there's identity matrix in there. Perhaps something like
>
> inv(A) = inv([ I ; A0 ]) = [ I , inv(A0) ]
====================

If you don't mind some numerical risks, you could resort to the normal equation. Namely A\b is equivalent to

(A.'*A)\(A.'*b)

which in your case is

(I+A0.'*A0)\(b+A0.'*b)

Note that (I+A0.'*A0) has reduced dimensions 9303x9303, so you've saved yourself something perhaps.
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