good afternoon everyone can anyone answer me this question ? i have an error in the matlab code

here are some of the programs that I'm working on

sou_y = [5 10]; 
rec_y = [5 25 40]; 
d = [0.1000 0.0977 0.0908
     0.1000 0.0977 0.0908];
pan_ray_total= [ 100.0000  101.9084  105.9481
                            100.1249  101.1187  104.4031];
t_obs = d;	
vsem=zeros(length(sou_y),length(rec_y));
for i= 1 :length(sou_y)
    for j= 1:length(rec_y)
        vsem(i,j) = pan_ray_total(i,j)/t_obs(i,j); 
    end
end
v0_linier = sum (sum (vsem )) ./ n_ray ;
t_cal_linier = pan_ray_total ./ v0_linier;
dt_linier = t_obs - t_cal_linier;
for l = 1 :length(sou_y)
        dt((length(rec_y)*(1-1 ))+1 :length(rec_y)* 1,1 ) =dt_linier(1,:);
end
G = [ 100.0000  0  0 ; 76.4853  25.4951  0; 45.4063  30.2709  0; 100.1249  0  0; 67.4125  33.7062                    0; 34.8010  34.8010  34.8010]; 
G_linier = G ;
misfit = dt.^2;
We = (misfit).^-1;
m_linier = (inv(G_linier'*diag(We)*G_linier))*G_linier'*diag(We)*dt;
dv_linier = zeros(length(m_linier),l);
v_linier = zeros(length(m_linier),1); 
for i = 1 :length(m_linier)
    dv_linier (i) = (-1*m_linier(i)*v0_linier^2)/(1 + m_linier(i) * v0_linier);
    v_linier(i) = v0_linier + dv_linier(i); 
end

and i have problem like this

Error using *

Inner matrix dimensions must agree.

Error in te (line 21)

m_linier = (inv(G_linier'*diag(We)*G_linier))*G_linier'*diag(We)*dt;

how can i fix this?

[Walter Roberson]

What is the difference between this and https://www.mathworks.com/matlabcentral/answers/492720-good-afternoon-everyone-can-anyone-answer-me-this-question-i-have-an-error-in-the-matlab-code ?

[No Freed]

i was input G value

Answer 1

[Walter Roberson]

I think you need

m_linier = (G_linier*diag(We)*G_linier')\G_linier*diag(We)*dt

[No Freed]

thank you...its working sir

[Image Analyst]

Can you "Accept this answer" to give Walter "credit" for the correct answer?

[No Freed]

the formula that was given by walter is about resolved my program however i still found the error at matrix iteration like this

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND =

5.516471e-18.

Thank you

[Walter Roberson]

When you construct matrices based on 6 x 3 * 3 x 3 * 3 * 6, then the resulting matrix has rank no more than 3. It cannot possibly have an inverse.

You can proceed using pseudo-inverses, but it is not obvious that the result will have any meaning.

Question for you:

for l = 1 :length(sou_y)

So l is a loop control variable. After the loop, it will be left at the last value it was set to, so l = length(sou_y) is true after the for loop.

dv_linier = zeros(length(m_linier),l);

And there you set dv_linier to have l columns -- that is, to have length(sou_y) columns.

    dv_linier (i) = (-1*m_linier(i)*v0_linier^2)/(1 + m_linier(i) * v0_linier);    

but your syntax only assigns to dv_linier as if it is a vector. Unless you are doing linear indexing/

There are some valid cases to use linear indexing, but it is not obvious that this is any of those cases. My experiments suggest that it is not a place to use linear indexing to address multiple columns. I would suggest to you that perhaps when you define dv_linier that the number of columns should be 1 (the digit one) rather than l (lower-case L that got more or less accidentally left as length(sou_y))