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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?

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.

Steven Lord
on 17 Oct 2016

Now the cause of the problem is clear. You wrote:

This yields the following system of linear equations:

0.8*SS1+0.3*SS2=SS1 0.25*SS1+0.7*SS2=SS2 SS1+SS2=1

You have a typo in your second equation. You wrote 0.25 instead of 0.2. So linsolve is solving the problem you told it to solve, not the problem you wanted it to solve. When I write the correct coefficient matrix, both linsolve and backslash (\) return the answer you expected.

A = [0.8-1 0.3; 0.2 0.7-1; 1 1];

b = [0; 0; 1];

x1 = linsolve(A, b)

x2 = A\b

Since you're interested in Markov chains you may find the PageRank and Markov Chains section of the chapter on Linear Equations in Cleve's Numerical Computing in MATLAB interesting and informative.

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