inverse matrix

Sundar Dorai-Raj <sundar.dorai-raj at pdf.com> writes:

Sam R. Smith wrote:

if solve(a,b) means to calculate an inverse matrix of
a with b, and i wonder why solve(a)%%b will get
different result?

It does? Or perhaps your "%%" is not just a typo. It should be "%*%".

 > a <- matrix(rnorm(16), 4, 4)
 > b <- matrix(rnorm(4), 4, 1)
 > solve(a, b)

            [,1]
[1,] -0.8005768
[2,]  0.5913755
[3,] -1.8256012
[4,]  0.8973716

 > solve(a) %*% b

            [,1]
[1,] -0.8005768
[2,]  0.5913755
[3,] -1.8256012
[4,]  0.8973716

I think the issue is this:

 solve(a, b)

[1] -0.7251033 -0.3903765  0.3212044 -1.2969697

 solve(a)%*% b

[,1]
[1,] -0.7251033
[2,] -0.3903765
[3,]  0.3212044
[4,] -1.2969697

So b gets promoted to a column matrix in one case but not the other. 

This is slightly odd, but it's been that way "forever" and in S(-PLUS)
too, so I think it is unchangeable (it's the sort of thing that
there's a 99% chance that some people have actually been relying on). 

If you want a vector result from a matrix multiply, there's always

drop(solve(a)%*% b)

[1] -0.7251033 -0.3903765  0.3212044 -1.2969697

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