Matrix inversion by elementary row operations

knitr::opts_chunk$set( warning = FALSE, message = FALSE ) options(digits=4) par(mar=c(5,4,1,1)+.1)

The following examples illustrate the steps in finding the inverse of a matrix using elementary row operations (EROs):

  • Add a multiple of one row to another (rowadd())
  • Multiply one row by a constant (rowmult())
  • Interchange two rows (rowswap())

These have the properties that they do not change the inverse. The method used here is sometimes called the Gauss-Jordan method, a form of Gaussian elimination. Another term is (row-reduced) echelon form.

Steps:

  1. Adjoin the identity matrix to the right side of A, to give the matrix $[A | I]$
  2. Apply row operations to this matrix until the left ($A$) side is reduced to $I$
  3. The inverse matrix appears in the right ($I$) side

Why this works: The series of row operations transforms $$ [A | I] \Rightarrow [A^{-1} A | A^{-1} I] = [I | A^{-1}]$$

If the matrix is does not have an inverse (is singular) a row of all zeros will appear in the left ($A$) side.

Load the matlib package

library(matlib)

Create a 3 x 3 matrix

A <- matrix( c(1, 2, 3, 2, 3, 0, 0, 1,-2), nrow=3, byrow=TRUE)

Join an identity matrix to A

(AI <- cbind(A, diag(3)))

Apply elementary row operations to reduce A to an identity matrix.

The right three cols will then contain inv(A). We will do this three ways:

  1. first, just using R arithmetic on the rows of AI
  2. using the ERO functions in the matlib package
  3. using the echelon() function

1. Using R arithmetic

(AI[2,] <- AI[2,] - 2*AI[1,]) # row 2 <- row 2 - 2 * row 1 (AI[3,] <- AI[3,] + AI[2,]) # row 3 <- row 3 + row 2 (AI[2,] <- -1 * AI[2,]) # row 2 <- -1 * row 2 (AI[3,] <- -(1/8) * AI[3,]) # row 3 <- -.25 * row 3

Now, all elements below the diagonal are zero

AI #--continue, making above diagonal == 0 AI[2,] <- AI[2,] - 6 * AI[3,] # row 2 <- row 2 - 6 * row 3 AI[1,] <- AI[1,] - 3 * AI[3,] # row 1 <- row 1 - 3 * row 3 AI[1,] <- AI[1,] - 2 * AI[2,] # row 1 <- row 1 - 2 * row 2 AI #-- last three cols are the inverse (AInv <- AI[,-(1:3)]) #-- compare with inv() inv(A)

2. Do the same, using matlib functions rowadd(), rowmult() and rowswap()

AI <- cbind(A, diag(3)) AI <- rowadd(AI, 1, 2, -2) # row 2 <- row 2 - 2 * row 1 AI <- rowadd(AI, 2, 3, 1) # row 3 <- row 3 + row 2 AI <- rowmult(AI, 2, -1) # row 1 <- -1 * row 2 AI <- rowmult(AI, 3, -1/8) # row 3 <- -.25 * row 3 # show result so far AI #--continue, making above-diagonal == 0 AI <- rowadd(AI, 3, 2, -6) # row 2 <- row 2 - 6 * row 3 AI <- rowadd(AI, 2, 1, -2) # row 1 <- row 1 - 2 * row 2 AI <- rowadd(AI, 3, 1, -3) # row 1 <- row 1 - 3 * row 3 AI

3. Using echelon()

echelon() does all these steps row by row, and returns the result

echelon( cbind(A, diag(3)))

It is more interesting to see the steps, using the argument verbose=TRUE. In many cases, it is informative to see the numbers printed as fractions.

echelon( cbind(A, diag(3)), verbose=TRUE, fractions=TRUE)