MATLAB: Doesn’t A*P produce the right answer

matrix manipulationmatrix multiplication

Doing a simple matrix multiplication in MatLab seems to be giving me the incorrect answer. If

A = [ 1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16 ];

and

P = [ 0 0 0 1; 1 0 0 0; 0 0 1 0; 0 1 0 0 ];

then

A*P = [ 4 1 3 2; 8 5 7 6; 12 9 11 10; 16 13 15 14 ];

However, when I do A*P in MatLab, I get the following answer:

A*P = [ 2 4 3 1; 6 8 7 5; 10 12 11 9; 14 16 15 13 ];

Can someone please explain to me what is happening?

Best Answer

MATLAB is producing the correct answer. Your understanding of matrix multiplication is not correct. The operation that gives the result you were expecting is actually

A*P.'

Also, you have a typo in your post. I assume the P you meant is:

P = [ 0 0 0 1; 1 0 0 0; 0 0 1 0; 0 1 0 0 ];

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MATLAB: Matrix differential equations, correct answer

Hi Sara,

this plot

figure(1)
tvec = 0:.01:6;
reliability_d = double(subs(reliability,t,tvec));
rel_d = double(subs(rel,t,tvec));
plot(tvec,reliability_d,tvec,rel_d)
grid on

shows that betwen the two versions, the first one has issues. As you can see, it's even increasing in the second half of the plot.

The second version 'rel' is correct, as verified here using an eigenvalue calculation. (In the last line of your code I upped the significant figures with

rel=vpa(rel,5)

). Due to the form of D, with all zeros in the lower left corner, time derivates of variables 1-12 do not depend on variables 13-28. Therefore the problem can be reduced to the upper left 12x12 corner of D.

tvec = 0:.01:6;
rel_d = double(subs(rel,t,tvec));         % your second result
% reduce problem to 12x12
D12 = D(1:12,1:12);
phi0 = C(1:12);                           % initial condition
[V lambda]= eig(D12);
phi0V = phi0*V;
inVSum = sum(inv(V),2);
cof = phi0V.*inVSum';                     % coefficients of exp(lambda*t) terms
[tvecx lambdax] = meshgrid(tvec,diag(lambda));  % time across,lambda down
phimat = exp(lambdax.*tvecx);
relnew = cof*phimat;
figure(2)
plot(tvec,relnew,tvec,rel_d +.01)        % add .01 for visual purposes
grid on
rel
[cof;diag(lambda)']                     % coefficients, lambdas

The last line shows that the solution has three exponentials of the form exp(lambda*t) with nonzero coefficients. If you scroll through the somewhat inconvenient rel line, you will find the same ones.

Best Answer

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