Graphing a linear regression line from given points, very simple example, having trouble with matrices dimensions?

I am getting the error Inner matrix dimensions must agree.

Error in ==> spectro at 15
Y = [ones(size(T))  exp(-T)  T.*exp(-T)]*a;

here is the code: %regression

% Enter t=dataC and y=dataABS as columnwise vectors
dataABS=[1.003 1.038 1.079 0.466 0.402 0.469 0.156 0.237 0.188];
dataT=[9.94 9.17 8.34 34.23 39.62 33.93 69.86 58.00 64.85];
dataC=[10 11 10 6 7 6 2 3 2];

% Form the design matrix
X = [ones(size(dataC))  exp(-dataC)  dataC.*exp(-dataC)];

% Calculate model coefficients
a = X\dataC;
T = (0:0.5:10)';
Y = [ones(size(T))  exp(-T)  T.*exp(-T)]*a;
plot(T,Y,'-',dataC,dataABS,'o'), grid on

any suggestions appreciated.

thanks

Answer 1

[Matt Tearle]

% Enter t=dataC and y=dataABS as columnwise vectors

And then you enter them as rows :)

Stick a transpose ( ' ) on those three lines and it works.

ETA: Well, "works" is a subjective term... I suspect the line a = X\dataC; is supposed to be a = X\dataABS;

ETA(2): In response to question below, here's the code using function handles. At the end of this, ymodel is a function of t that you can evaluate ad naseum.

% Enter t=dataC and y=dataABS as columnwise vectors
dataABS=[1.003 1.038 1.079 0.466 0.402 0.469 0.156 0.237 0.188]';
dataT=[9.94 9.17 8.34 34.23 39.62 33.93 69.86 58.00 64.85]';
dataC=[10 11 10 6 7 6 2 3 2]';
% Form the design matrix
dmat = @(t) [ones(size(t))  exp(-t)  t.*exp(-t)];
% Calculate model coefficients
a = dmat(dataC)\dataABS;
T = (0:0.5:10)';
ymodel = @(t) dmat(t)*a;

plot(T,ymodel(T),'-',dataC,dataABS,'o'), grid on

[Adam Quintero]

Excellent! It makes sense now. I am now wondering how to extract the equation of my line of best fit. Is there a function? I'm checking around the help but not having mcuh luck.

[Matt Tearle]

What do you mean by "extract the equation"? As a string? Not really -- you might as well just do it by hand. As a function you can evaluate? Yes: see above for edit.