How to solve an augmented matrix using rref(A)?

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Gina  Mangold
Gina Mangold 2016 年 4 月 23 日
回答済み: Steven Lord 2016 年 4 月 23 日
Given this set of equations:
0.25pF + 0.15pT + 0.25pC + 0.18pCM + 0.20pSB = pF
0.15pF + 0.28pT + 0.18pC + 0.17pCM + 0.05pSB = pT
0.22pF + 0.19pT + 0.22pC + 0.22pCM + 0.10pSB = pC
0.20pF + 0.15pT + 0.20pC + 0.28pCM + 0.15pSB = pCM
0.18pF + 0.23pT + 0.15pC + 0.15pCM + 0.50pSB = pSB
we were asked to use Matlab to find pF, pT, pC, pCM, pSB.
the code I have so far is: C=[0.25 0.15 0.25 0.18 0.20; 0.15 0.28 0.18 0.17 0.05; 0.22 0.19 0.22 0.22 0.10; 0.20 0.15 0.20 0.28 0.15; 0.18 0.23 0.15 0.15 0.5]
%If the price of the items are denoted by the matrix p, then the linear relationship would be Cp=p %or equivalently Cp-p= Cp-Ip= (C-I)p=0 (where I is an identity matrix with 1's on the diagonal and o's everywhere else.)
I=eye(5)
d=[0; 0; 0; 0; 0]
%Therefor this can be represented as the augmented matrix [(C-I)|d]
M=rref([(C-I)d])
But this produces an error: of an unexpected MatLab expression, what am I doing wrong? how do I use rref to solve the augmented matrix?

回答 (2 件)

Star Strider
Star Strider 2016 年 4 月 23 日
You must have a typo somewhere.
This runs for me without error:
C=[0.25 0.15 0.25 0.18 0.20; 0.15 0.28 0.18 0.17 0.05; 0.22 0.19 0.22 0.22 0.10; 0.20 0.15 0.20 0.28 0.15; 0.18 0.23 0.15 0.15 0.5];
I=eye(5);
d=[0; 0; 0; 0; 0];
Ca = [C-I d];
M = rref(Ca);
The solved coefficients appear to be in ‘M(:,5)’.

Steven Lord
Steven Lord 2016 年 4 月 23 日
Don't use rref, use null instead.

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