Inverse of matrix in R

RMatrix Inverse

R Problem Overview


I was wondering what is your recommended way to compute the inverse of a matrix?

The ways I found seem not satisfactory. For example,

> c=rbind(c(1, -1/4), c(-1/4, 1))  
> c  
      [,1]  [,2]  
[1,]  1.00 -0.25  
[2,] -0.25  1.00  
> inv(c)  
Error: could not find function "inv"  
> solve(c)    
          [,1]      [,2]  
[1,] 1.0666667 0.2666667  
[2,] 0.2666667 1.0666667  
> solve(c)*c  
            [,1]        [,2]  
[1,]  1.06666667 -0.06666667  
[2,] -0.06666667  1.06666667  
> qr.solve(c)*c  
            [,1]        [,2]  
[1,]  1.06666667 -0.06666667  
[2,] -0.06666667  1.06666667  

Thanks!

R Solutions


Solution 1 - R

solve(c) does give the correct inverse. The issue with your code is that you are using the wrong operator for matrix multiplication. You should use solve(c) %*% c to invoke matrix multiplication in R.

R performs element by element multiplication when you invoke solve(c) * c.

Solution 2 - R

You can use the function ginv() (Moore-Penrose generalized inverse) in the MASS package

Solution 3 - R

Note that if you care about speed and do not need to worry about singularities, solve() should be preferred to ginv() because it is much faster, as you can check:

require(MASS)
mat <- matrix(rnorm(1e6),nrow=1e3,ncol=1e3)

t0 <- proc.time()
inv0 <- ginv(mat)
proc.time() - t0 

t1 <- proc.time()
inv1 <- solve(mat)
proc.time() - t1 

Solution 4 - R

solve(matrix) = Inverse of the matrix, it does the job perfectly.

Solution 5 - R

Use solve(matrix) if the matrix is larger than 1820x1820. Using inv() from matlib or ginv() from MASS takes longer or will not solve at all because of RAM limits.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionTimView Question on Stackoverflow
Solution 1 - Ruser28View Answer on Stackoverflow
Solution 2 - RdougView Answer on Stackoverflow
Solution 3 - RMatthias SchmidtblaicherView Answer on Stackoverflow
Solution 4 - RMostak AhammedView Answer on Stackoverflow
Solution 5 - Rgrad studentView Answer on Stackoverflow