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Correct and incorrect answer from linsolve

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Oddur Bjarnason
Oddur Bjarnason on 14 Oct 2016
Edited: Walter Roberson on 17 Oct 2016
The following gives a correct answer:
A=[-2/5,1/5;2/5,-1/5;1,1]
B=[0;0;1]
linsolve(A,B)
ans =0.3333 0.6667
The following, however, gives an incorrect answer:
C=[-0.2,0.3;0.25,-0.3;1,1]
D=[0;0;1]
C\D
ans = 0.5699 0.4297
The correct answer, found by substitution, is 0.6000 0.4000
How can I get a correct answer by linsolve?

Answers (1)

Walter Roberson
Walter Roberson on 14 Oct 2016
x = sym('x',[2 1]);
C*x
ans =
(3*x2)/10 - x1/5
x1/4 - (3*x2)/10
x1 + x2
You have a system of 3 equations in 2 unknowns. It is overdetermined, and might not have any exact solution. The \ operation will do a least-squared fit to find an answer that is least bad in some sense.
Solving (3*x2)/10 - x1/5 = 0 for x1 gives x1 = (3*x2)/2. Substituting that back into C*x gives
0
(3*x2)/40
(5*x2)/2
solving (3*x2)/40 = 0 for x2 gives x2 = 0. Substituting that back gives (5*0)/2 = 1 which is 0 = 1 which has no solution.
>> C*[0.6;0.4]
ans =
0
0.03
1
So 0.6 0.4 is not a solution after-all.
>> sum((D-C*(C\D)).^2)
ans =
0.000407074042245239
>> sum((D-C*[0.6;0.4]).^2)
ans =
0.0009
so the solution found by C\D gives less of an error than [0.6 0.4] does.
  6 Comments
Oddur Bjarnason
Oddur Bjarnason on 17 Oct 2016
Thank you Steven, When I was filling in the coefficient Matrix I read 0.2S1 to be 0.25. So you are right - it was a typo.

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