MATLAB: Solution is not matching with actual equation

MATLABoverdetermined equ...overdetermined equationsystem of linear equation

I am solving overdetermined linear equation having 12 equations with 5 unknowns using matlab 2015. I got the solution. to check the solution I multiplied Matrix A with Solution matrix, but it is not matching with matrix B. what to do?

Best Answer

That the solution is not exact is not even remotely surprising, for several reasons. First, an over-determined system will generally not have an exact solution. The best you can do is to minimize the residuals of all of the errors in combination. This is what is done. Don't believe me? Try a simple example.
A = [1 1;1 2;3 1]
A =
   1     1
   1     2
   3     1
b = [-8;7;6];
Surely, how easy can this 2 variable system be to solve? I can hardly imagine something easier that that. And even if the answer is not exact, it will surely be close, no?
x = A\b
x =
          1
          1
[A*x,b]
ans =
          2           -8
          3            7
          4            6
Hmm. So when we multiply A*x, we are pretty far off indeed. But then, I did say this is not even remotely a surprise. In fact, it is easily shown that any other choice for a solution vector x will yield residuals with a larger norm.
norm(A*x - b)
ans =
       10.954
norm(A*(x + randn(2,1)) - b)
ans =
       11.235
norm(A*(x + randn(2,1)) - b)
ans =
       11.021
Now, this silly example is not a proof. Merely a counter-example to the idea that it is trivial to create a system of over-determined equations where the solution still has significantly large residual errors. In the case of a 12x5 system, so 12 equations, 5 unknowns, that is just as trivial.
Ok, now as far as the spreadsheet that you show, again, you seem to be a bit clueless. (Sorry, but true.)
    A
    A =
                        770.55                    776.45                    765.05                       771                     772.3
                        636.05                       641                     630.8                       634                    636.95
                       1404.95                    1409.3                      1382                    1397.3                      1401
                        894.65                     897.4                     886.8                    889.35                    894.65
                       1421.96                      1438                      1418                    1429.9                    1433.4
    >> b
    b =
                           784                     766.5                     773.5
                           639                     631.2                    633.65
                          1401                      1371                   1375.25
                        907.25                     881.1                     896.9
                        1441.4                   1416.95                    1420.7
So, first, you show a 5x5 matrix, NOT a 12x5 matrix. A is not even close to singular, so a solution will exist that will be reasonably accurate.
x = A\b
x =
          1.34306951538017        -0.735590758162936         0.825063315954427
         -2.83318049316987          1.94002770486605         -1.22605398297668
          4.17833546820366           2.5252718327165          5.24318794991482
          3.10947548609037         0.720519361180246          1.63716711244305
         -4.71982438285554         -3.44491091004859         -5.41737628875096
Be careful. Do NOT copy only a 5 digit approximation to that solution. I would guess that is what you did. It is also perhaps the most common novice mistake that people make.
A*x - b
ans =
                         0                         0      9.09494701772928e-13
      4.54747350886464e-13      6.82121026329696e-13      1.02318153949454e-12
                         0                         0                         0
                         0     -1.13686837721616e-13       5.6843418860808e-13
     -4.54747350886464e-13     -2.27373675443232e-13      6.82121026329696e-13
So the residuals here are tiny, as I predicted. Had you copied rounded 5 digit approximations to x into a spreadsheet,
xhat = round(x,5)
xhat =
                   1.34307                  -0.73559                   0.82506
                  -2.83318                   1.94003                  -1.22605
                   4.17834                   2.52527                   5.24319
                   3.10948                   0.72052                   1.63717
                  -4.71982                  -3.44491                  -5.41738
A*xhat - b
ans =
        0.0110884999994596       0.00215949999983422       0.00146600000061881
       0.00913649999938571       0.00178199999913886       0.00120399999980236
        0.0200864999997066       0.00393449999955919       0.00262300000213145
        0.0128304999998363       0.00249499999961245       0.00167349999958333
        0.0205611999986104       0.00399759999959315       0.00272860000109176
Still not immense errors, but consistent with my expectations.
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