3d inverse
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From: Andy on 15 Jul 2010 15:36 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i1nlvr$bus$1(a)fred.mathworks.com>... > "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <i1nleg$6di$1(a)fred.mathworks.com>... > > "Andy " <dcg48(a)cornell.edu> wrote in message <i1nk5r$dla$1(a)fred.mathworks.com>... > > > > > > It would be clearest to post for us the formula for T(i) in terms of A(i,j,k) and B > > > I'm not sure the formula for T(i). T as a whole is the summation of A(:,j,k)*B(j,k) for all 15 columns (j) and 7 rows (k). > > ===== > > > > In that case, the first solution I gave you should have worked. I suggest you check the dimensions of A and T and post your full error messages. > ================= > > Note, for example, how the following does not result in an error. If all the matrices have the dimensions you claim, there should be no problem. > > >> A=rand(64^3,15,7); T=rand(64^3,1); > >> B=reshape(A,64^3,[])\T; whos B > > Name Size Bytes Class Attributes > > B 105x1 840 double Matt, Thanks that worked perfectly. I found my typo when I had previously tried it too. Since B is now 105x1, how can I tell if it indexed it first by row then column, or vice versa?
From: Matt J on 15 Jul 2010 17:36
"Andy " <dcg48(a)cornell.edu> wrote in message <i1nnv4$l2g$1(a)fred.mathworks.com>... > Matt, > Thanks that worked perfectly. I found my typo when I had previously tried it too. Since B is now 105x1, how can I tell if it indexed it first by row then column, or vice versa? =============== If the original 64^3x15x7 array A was indexed A(i,j,k) then B15x7=reshape(B,15,7) will have rows correspondingly indexed by j and columns indexed by k. |