[R] Matlab inv() and R solve() differences

[Kingsford Jones]

I suppose the solution is unstable because x is ill-conditioned:

> x
       [,1]   [,2]   [,3]  [,4]
[1,]  0.133  0.254 -0.214 0.116
[2,]  0.254  0.623 -0.674 0.139
[3,] -0.214 -0.674  0.910 0.011
[4,]  0.116  0.139  0.011 0.180
> cor(x)
           [,1]       [,2]       [,3]       [,4]
[1,]  1.0000000  0.9963557 -0.9883690  0.8548065
[2,]  0.9963557  1.0000000 -0.9976663  0.8084090
[3,] -0.9883690 -0.9976663  1.0000000 -0.7663847
[4,]  0.8548065  0.8084090 -0.7663847  1.0000000

> kappa(x)
[1] 2813.326

hth,

Kingsford Jones

On Thu, Jan 29, 2009 at 7:00 PM, Joseph P Gray <jpgray at uwm.edu> wrote:
> I submit the following matrix to both MATLAB and R
>
> x=  0.133 0.254 -0.214 0.116
>    0.254 0.623 -0.674 0.139
>   -0.214 -0.674 0.910 0.011
>    0.116 0.139  0.011 0.180
>
> MATLAB's inv(x) provides the following
>  137.21 -50.68 -4.70 -46.42
> -120.71  27.28 -8.94 62.19
> -58.15   6.93  -7.89  36.94
>  8.35   11.17 10.42 -14.82
>
> R's solve(x) provides:
> 261.94 116.22 150.92 -267.78
> 116.22 344.30 286.68 -358.30
> 150.92 286.68 252.96 -334.09
> -267.78 =358.30 -334.09 475.22
>
> inv(x)*x = I(4)
> and solve(x)%*%x = I(4)
>
> Is there a way to obtain the MATLAB result in R?
>
> Thanks for any help.
>
>
> Pat Gray
>
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