From: Robert A LaBudde <ral_at_lcfltd.com>

Date: Sat, 21 May 2011 23:02:55 -0400

> b<- c(0,2,2,0) #rhs

> b

[1] 0 2 2 0

[3,] 0 0 1 0

[4,] 0 0 0 1

>

> giA%*%b #particular solution

Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral_at_lcfltd.com

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sun 22 May 2011 - 03:05:04 GMT

Date: Sat, 21 May 2011 23:02:55 -0400

solve() only works for nonsingular systems of equations.

Use a generalized inverse for singular systems:

> A<- matrix(c(1,2,1,1, 3,0,0,4, 1,-4,-2,-2, 0,0,0,0), ncol=4, byrow=TRUE) > A

[,1] [,2] [,3] [,4]

[1,] 1 2 1 1 [2,] 3 0 0 4 [3,] 1 -4 -2 -2 [4,] 0 0 0 0

> b<- c(0,2,2,0) #rhs

> b

[1] 0 2 2 0

> > require('MASS') > giA<- ginv(A) #M-P generalized inverse > giA [,1] [,2] [,3] [,4] [1,] 0.6666667 1.431553e-16 0.33333333 0 [2,] 0.3333333 -1.000000e-01 -0.03333333 0[3,] 0.1666667 -5.000000e-02 -0.01666667 0 [4,] -0.5000000 2.500000e-01 -0.25000000 0

> > require('Matrix') > I<- as.matrix(Diagonal(4)) #order 4 identity matrix > I [,1] [,2] [,3] [,4] [1,] 1 0 0 0 [2,] 0 1 0 0

[3,] 0 0 1 0

[4,] 0 0 0 1

>

> giA%*%b #particular solution

[,1]

[1,] 6.666667e-01 [2,] -2.666667e-01 [3,] -1.333333e-01 [4,] -2.220446e-16 > giA%*%A - I #matrix for parametric homogeneous solution [,1] [,2] [,3] [,4] [1,] 0.000000e+00 0.0 0.0 5.551115e-16 [2,] 3.469447e-17 -0.2 0.4 4.024558e-16[3,] 4.510281e-17 0.4 -0.8 2.706169e-16 [4,] -3.330669e-16 0.0 0.0 -7.771561e-16

At 09:34 PM 5/21/2011, dslowik wrote:

>I have a simple system of linear equations to solve for X, aX=b:

*> > a
**> [,1] [,2] [,3] [,4]
**>[1,] 1 2 1 1
**>[2,] 3 0 0 4
**>[3,] 1 -4 -2 -2
**>[4,] 0 0 0 0
**> > b
**> [,1]
**>[1,] 0
**>[2,] 2
**>[3,] 2
**>[4,] 0
**>
**>(This is ex Ch1, 2.2 of Artin, Algebra).
**>So, 3 eqs in 4 unknowns. One can easily use row-reductions to find a
**>homogeneous solution(b=0) of:
**>X_1 = 0, X_2 = -c/2, X_3 = c, X_4 = 0
**>
**>and solutions of the above system are:
**>X_1 = 2/3, X_2 = -1/3-c/2, X_3 = c, X_4 = 0.
**>
**>So the Kernel is 1-D spanned by X_2 = -X_3 /2, (nulliity=1), rank is 3.
**>
**>In R I use solve():
**> > solve(a,b)
**>Error in solve.default(a, b) :
**> Lapack routine dgesv: system is exactly singular
**>
**>and it gives the error that the system is exactly singular, since it seems
**>to be trying to invert a.
**>So my question is:
**>Can R only solve non-singular linear systems? If not, what routine should I
**>be using? If so, why? It seems that it would be simple and useful enough to
**>have a routine which, given a system as above, returns the null-space
**>(kernel) and the particular solution.
**>
**>
**>
**>
**>--
**>View this message in context:
**>http://r.789695.n4.nabble.com/Finding-solution-set-of-system-of-linear-equations-tp3541490p3541490.html
**>Sent from the R help mailing list archive at Nabble.com.
**>
**>______________________________________________
**>R-help_at_r-project.org mailing list
**>https://stat.ethz.ch/mailman/listinfo/r-help
**>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
**>and provide commented, minimal, self-contained, reproducible code.
*

Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ral_at_lcfltd.com

Least Cost Formulations, Ltd. URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239 Fax: 757-467-2947

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