Matrix inversion by elementary row operations

knitr::opts_chunk$set(
  warning = FALSE,
  message = FALSE
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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:

Adjoin the identity matrix to the right side of A, to give the matrix $[A | I]$
Apply row operations to this matrix until the left ($A$) side is reduced to $I$
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:

first, just using R arithmetic on the rows of AI
using the ERO functions in the matlib package
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)

Load the matlib package