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Topic: Error : Inner matrix dimensions must agree When computing the three dimensional integral.
Replies: 7   Last Post: Jun 24, 2013 11:14 AM

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 Steven Lord Posts: 18,038 Registered: 12/7/04
Re: Error : Inner matrix dimensions must agree When computing the three dimensional integral.
Posted: Jun 20, 2013 10:51 AM

"someone" <someone@somewhere.net> wrote in message
news:kpt46r\$qf3\$1@newscl01ah.mathworks.com...
> "yolanda" wrote in message <kpskm1\$d1b\$1@newscl01ah.mathworks.com>...
>> Error : Inner matrix dimensions must agree When computing the three
>> dimensional integral. The code is as follows:
>>
>> m1 = 20;
>> v1 = 60;
>> m2 = 20;
>> v2 = 60; m3 = 20;
>> v3 = 60; cov = [ 60,0,0;
>> 0,60,0;
>> 0,0,60]
>> cov1=inv(cov)
>> f = @(z,y,x) exp(
>> ([(x-m1),(y-m2),(z-m3)]*(cov1)*[(x-m1);(y-m2);(z-m3)])./(-2) )

"The first input, fun, is a function handle. fun(x,y,z) must accept a vector
x and scalars y and z, and return a vector of values of the integrand."

When your function is called with a vector z and scalar x and y, the
expressions (x-m1) and (y-m2) will be scalars but the expression (z-m3) will
be a vector. One of the concatenation steps will not work:

[(x-m1),(y-m2),(z-m3)] % Doesn't work if z is a column vector
[(x-m1);(y-m2);(z-m3)] % Doesn't work if z is a row vector

I believe z will be a row vector, so the latter step will not work.

--
Steve Lord
slord@mathworks.com
http://www.mathworks.com

Date Subject Author
6/19/13 yolanda
6/19/13 Curious
6/19/13 yolanda
6/19/13 yolanda
6/19/13 yolanda
6/20/13 Steven Lord
6/20/13 yolanda
6/24/13 Steven Lord