Hi Moohwan,
On Dec 11, 2009, at 11:26 AM, Moohwan Kim wrote:
> Dear R family
>
> I have a following question.
> Suppose I have a matrix as follows, for instance:
> tau=
> 0 0 0 0 1
> 1 0 0 0 0
> 0 1 0 0 0
> 0 0 1 0 0
> 0 0 0 1 0
>
> I want to have the inverse of the above matrix and then add some
> exponent to it. That is, I want to calculate tau to the (-m). For
> example, m=893.
If you *really* want the inverse, use the `solve` function without a second
parameter:
R> tau <- c(0, 0, 0, 0, 1,
1, 0, 0, 0, 0,
0, 1, 0, 0, 0,
0, 0, 1, 0, 0,
0, 0, 0, 1, 0)
R> tau <- matrix(tau, ncol=5, byrow=TRUE)
R> tau %*% solve(tau)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 0 1 0 0 0
[3,] 0 0 1 0 0
[4,] 0 0 0 1 0
[5,] 0 0 0 0 1
-steve
--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact
______________________________________________
R-help@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.
