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:

- 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.

`matlib`

package```
library(matlib)
```

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

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

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

(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)

`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

`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)

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