ERROR USING VERTCAT HELP PLEASE

I keep getting this error

"Error using vertcat Dimensions of matrices being concatenated are not consistent."

P=(100:100:1000);

for i=1:P

D = [1 0 0 -0.857 -0.832 -0.812;

    0 1 0 0.286 0.555 0.542;
    0 0 1 0.429 0 -0.217;
    0 0 0 0 0 0.0379;
    0 0 0 -0.1286 0 0.1218;
    0 0 0 0.0857 0.1664 0.1625;];

E= [0;

    P;
    0;
    0.15.*P;
    0;
    0.3.*P;];

X=D\E;

end

[José-Luis]

Please type in your code instead of pasting an image. It would make it easier for people to help you.

[David]

Ok, i did that.

Answer 1

[James Tursa]

It is unclear what you are trying to do with the loop. P is a vector [100 200 ... 1000], so the for loop indexing i=1:P doesn't seem to make sense. And then inside your matrix building you have scalars mixed in with this P vector for the concatenation, hence the error. Did you mean to do this instead?

P=(100:100:1000);
for k=P
D = [1 0 0 -0.857 -0.832 -0.812;
    0 1 0 0.286 0.555 0.542;
    0 0 1 0.429 0 -0.217;
    0 0 0 0 0 0.0379;
    0 0 0 -0.1286 0 0.1218;
    0 0 0 0.0857 0.1664 0.1625;];
E= [0;
    k;
    0;
    0.15.*k;
    0;
    0.3.*k;];
X=D\E;
end

[David]

This works thank you so much! i guess i was just following a template on my professors example problem (bad idea).

The only thing is that when i run the program the X only gives me one value rather than the 10 that are in (100:100:1000).

[James Tursa]

Option 1) Preallocate X and save each iteration in X, e.g.

P=(100:100:1000);
X = zeros(6,numel(P));
m = 1;
for k=P
D = [1 0 0 -0.857 -0.832 -0.812;
    0 1 0 0.286 0.555 0.542;
    0 0 1 0.429 0 -0.217;
    0 0 0 0 0 0.0379;
    0 0 0 -0.1286 0 0.1218;
    0 0 0 0.0857 0.1664 0.1625;];
E= [0;
    k;
    0;
    0.15.*k;
    0;
    0.3.*k;];
X(:,m)=D\E;
m = m + 1;
end

Option 2) Do all of the calculation in one fell swoop without a loop. E.g.,

P=(100:100:1000);
D = [1 0 0 -0.857 -0.832 -0.812;
    0 1 0 0.286 0.555 0.542;
    0 0 1 0.429 0 -0.217;
    0 0 0 0 0 0.0379;
    0 0 0 -0.1286 0 0.1218;
    0 0 0 0.0857 0.1664 0.1625;];
E = zeros(6,numel(P));
E(2,:) = P;
E(4,:) = P*0.15;
E(6,:) = P*0.3;
X=D\E;