Need Help with fixing "Error using vertcat Dimensions of matrices being concatenated are not consistent" for my matrix

Hello All,

I am in need of some help, I am trying to store results where P1, P2, and P3 are dependent on J in matrix B however I keep getting the the vertcat error message. I believe it may have something to do with storing several matrices in one matrix but I don't really know how to get around this since I need to create a matrix for every value of J.

j = 0:10;
B = [0;(P1+P2+P3);(P1+P2+P3);P1;0;0;0;0;0;P3;0;P2];

Any help would be appreciated!

Thank you for your time.

CAAJ

Answer 1

[John BG]

another way would be (let me call j now j0)

syms j0
P2(j0)=3*j0^2-2*j0+.5
P2(j0) =
3*j0^2 - 2*j0 + 1/2
>> P3(j0)=(8-j0^2).^5
P3(j0) =
-(j0^2 - 8)^5
 B = [0;(P1+P2+P3);(P1+P2+P3);P1;0;0;0;0;0;P3;0;P2]
B(j0) =
                                0
 3*j0^2 - (j0^2 - 8)^5 - j0 + 3/2
 3*j0^2 - (j0^2 - 8)^5 - j0 + 3/2
                           j0 + 1
                                0
                                0
                                0
                                0
                                0
                    -(j0^2 - 8)^5
                                0
              3*j0^2 - 2*j0 + 1/2

now you change j0 from symbolic to double

j0=[0:1:10]
 =
     0     1     2     3     4     5     6     7     8     9    10

and translate symbolic B to a cell

C=sym2cell(B)

you can still read the symbolic expression with

C{1}
ans(j0) =
                                0
 3*j0^2 - (j0^2 - 8)^5 - j0 + 3/2
 3*j0^2 - (j0^2 - 8)^5 - j0 + 3/2
                           j0 + 1
                                0
                                0
                                0
                                0
                                0
                    -(j0^2 - 8)^5
                                0
              3*j0^2 - 2*j0 + 1/2

but you have all numeric values, individually

single(C{1}(4))
ans =
   1.0e+04 *
           0
  -3.2722500
  -3.2722500
   0.0005000
           0
           0
           0
           0
           0
  -3.2768000
           0
   0.0040500

or read all values to for instance write them in another variable

for k=1:1:numel(j0) 
     double(C{1}(k)) 
 end

   =
     1.0e+04 *
                     0
     1.681050000000000
     1.681050000000000
     0.000200000000000
                     0
                     0
                     0
                     0
                     0
     1.680700000000000
                     0
     0.000150000000000
   =
     1.0e+03 *
                     0
     1.035500000000000
     1.035500000000000
     0.003000000000000
                     0
                     0
                     0
                     0
                     0
     1.024000000000000
                     0
     0.008500000000000

..

   =
     1.0e+10 *
                     0
    -1.842435143950000
    -1.842435143950000
     0.000000001200000
                     0
                     0
                     0
                     0
                     0
    -1.842435179300000
                     0
     0.000000034150000

if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?

To any other reader, please if you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

Answer 2

[James Tursa]

Not sure what P1, P2, and P3 dimensions are, but maybe something like this is what you need:

row0 = zeros(1,size(P1,2));
B = [row0;(P1+P2+P3);(P1+P2+P3);P1;row0;row0;row0;row0;row0;P3;row0;P2];

[CAAJ]

P1, P2, and P3 are scalar values

[James Tursa]

If P1, P2, and P3 were truly scalar, you would not be getting the vercat error message for the B line. Double check those variable sizes.

[CAAJ]

you were right they are actually 1x11 matrices. Would this be the issue, and if so how should I adjust?

Thank you again for your help!

[James Tursa]

You can use my posted solution if you want to keep the P stuff in rows. Or if you want one big column you could use transposes, e.g.,

B = [0;(P1+P2+P3)';(P1+P2+P3)';P1';0;0;0;0;0;P3';0;P2'];

But not knowing what this is being used for I can't advise you what you should do.

[CAAJ]

How would I be able to display the resulting matrix in regards to my J value?

[James Tursa]

Display in what way? For one thing, your j variable has 11 elements, but your B variable has 12 rows. So not sure how they correlate to each other.

[CAAJ]

I would like to see matrix B(the 12x1 matrix) after every iteration of J(0 to 10). So essentially when J=1, it would result in a different matrix than J=2. Because P1=0, P2=(.1*J)*1000 and P3=(1-(.1*J))*1000.

Hopefully this clarifies this situation.

Thanks again

Answer 3

[CAAJ]

Hello All,

Thank you all for your assistance but I have actually found a way to solve and display the results I needed. I ended up creating a function to solve the B matrix for the 10 different iterations. See the code below.

Created Function:

function [B]= Result_B(P0,P1)

for j = 1:10; 
    P3=(1-(.1*j))*P0;
    P2=(.1*j)*P0;
    B(:,:,j) = [0;(P1+P2+P3);(P1+P2+P3);P1;0;0;0;0;0;P3;0;P2];
end

In the command window:

B = Result_B(P0,P1)
B(:,:,j) %to see the resulting matrix from the jth iteration