Dear Tom,
I'm not sure that I follow the question correctly, but I think that you want
to solve the over determined system tdata %*% x == sumtd.
Some time ago, I believe that I posted to the list a function for Gaussian
elimination that I use in teaching basic linear algebra. Here's an updated
version of that function, and its application to your problem:
GaussianElimination <- function(A, b, tol=sqrt(.Machine$double.eps),
verbose=FALSE, fractions=FALSE){
# A: coefficient matrix
# b: right-hand size vector
# tol: tolerance for checking for 0 pivot
# verbose: if TRUE, print intermediate steps
# fractions: try to express nonintegers as rational numbers
# If b is absent returns the reduced row-echelon form of A.
# If b is present, reduces A to RREF carrying b along.
if (fractions) {
mass <- require(MASS)
if (!mass) stop("fractions=TRUE needs MASS package")
}
if ((!is.matrix(A)) || (!is.numeric(A))) stop("argument must be a
numeric matrix")
if ((!missing(b)) && ((!(length(b) == nrow(A))) ||
(!is.numeric(b))))
stop("argument must be a numeric vector of length equal to the
number of columns in A")
n <- nrow(A)
m <- ncol(A)
if (!missing(b)) A <- cbind(A, b)
for (i in 1:min(c(m, n))){
currentColumn <- A[,i]
currentColumn[1:n < i] <- 0
which <- which.max(abs(currentColumn)) # find maximum pivot in
current column at or below current row
pivot <- A[which, i]
if (abs(pivot) <= tol) next # check for 0 pivot
if (which > i) A[c(i, which),] <- A[c(which, i),] # exchange rows
A[i,] <- A[i,]/pivot # pivot
row <- A[i,]
A <- A - outer(A[,i], row) # sweep
A[i,] <- row # restore current row
if (verbose) if (fractions) print(fractions(A)) else print(round(A,
round(abs(log(tol,10)))))
}
for (i in 1:n) if (max(abs(A[i,1:m])) <= tol) A[c(i,n),] <- A[c(n,i),]
# 0 rows to bottom
if (fractions) fractions (A) else round(A, round(abs(log(tol,10))))
}
> GaussianElimination(tdata, sumtd)
b
[1,] 1 0 0 0 -3 30
[2,] 0 1 0 0 -1 12
[3,] 0 0 1 0 -2 20
[4,] 0 0 0 1 0 12
A caveat, however: Unlike some other members of this list, I'm not an expert
in numerical linear algebra and therefore can't vouch for the quality of the
algorithm -- it's meant basically for demonstration.
I hope this helps,
John
--------------------------------
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
--------------------------------
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Mulholland, Tom
> Sent: Friday, January 07, 2005 2:00 AM
> To: R-Help (E-mail)
> Subject: [R] Basic Linear Algebra
>
> I don't normally have to go anywhere near this stuff , but it
> seems to me that this should be a straight-forward process in R.
>
> For the purposes of this enquiry I thought I would use
> something I can work out on my own.
>
> So I have my matrix and the right hand results from that matrix
>
> tdata <-
> matrix(c(0,1,0,-1,-1,2,0,0,-5,-6,0,0,3,-5,-6,1,-1,-1,0,0),byro
> w = T,ncol = 5) sumtd <- c(0,0,0,-2)
>
> > tdata
> [,1] [,2] [,3] [,4] [,5]
> [1,] 0 1 0 -1 -1
> [2,] 2 0 0 -5 -6
> [3,] 0 0 3 -5 -6
> [4,] 1 -1 -1 0 0
> > sumtd
> [1] 0 0 0 -2
> >
>
> which I can calculate to give me 3x+30, x+12, 2x+20, 12x, x
>
> Would someone be kind enough to point me in the right
> direction as to which tools I should be using and any sage
> words of advice for the barely informed.
>
> Tom Mulholland
>
> _
> platform i386-pc-mingw32
> arch i386
> os mingw32
> system i386, mingw32
> status
> major 2
> minor 0.0
> year 2004
> month 10
> day 04
> language R
>
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