Author: Michael Friendly
Date: 2025-10-05

# 2. Evaluation of determinants

#### Michael Friendly

#### 2025-10-05

This example shows two classical ways to find the determinant, \(\det(A)\) of a square matrix. They each work by reducing the problem to a series of smaller ones which can be more easily calculated.

        library(matlib)

## 1. Calculate `det()` by cofactor expansion

Set up a \(3 \times 3\) matrix, and find its determinant (so we know what the answer should be).

      A <- matrix(c(4, 2, 1,
                    5, 6, 7,
                    1, 0, 3), nrow=3, byrow=TRUE)
      det(A)

    ## [1] 50

### Find cofactors of row 1 elements

The cofactor \(A_{i,j}\) of element \(a_{i,j}\) is the signed determinant of what is left when row i, column j of the matrix \(A\) are deleted. NB: In R, negative subscripts delete rows or columns.

      cat(cofactor(A, 1, 1),  "  ==  ",  1 * det( (A[-1, -1]), "\n" ))

    ## 18   ==   18

      cat(cofactor(A, 1, 2),  "  ==  ", -1 * det( (A[-1, -2]), "\n" ))

    ## -8   ==   -8

      cat(cofactor(A, 1, 3),  "  ==  ",  1 * det( (A[-1, -3]), "\n" ))

    ## -6   ==   -6

### det() = product of row with cofactors

In symbols: \(\det(A) = a_{1,1} * A_{1,1} + a_{1,2} * A_{1,2} + a_{1,3} * A_{1,3}\)

`rowCofactors()` is a convenience function, that calculates these all together

      rowCofactors(A, 1)

    ## [1] 18 -8 -6

Voila: Multiply row 1 times the cofactors of its elements. NB: In R, this multiplication gives a \(1 \times 1\) matrix.

      A[1,] %*% rowCofactors(A, 1)

    ##      [,1]
    ## [1,]   50

      all.equal( det(A), c(A[1,] %*% rowCofactors(A, 1)) )

    ## [1] TRUE

## 2. Finding `det()` by Gaussian elimination (pivoting)

This example follows Green and Carroll, Table 2.2. Start with a 4 x 4 matrix, \(M\), and save `det(M)`.

      M <- matrix(c(2, 3, 1, 2,
                    4, 2, 3, 4,
                    1, 4, 2, 2,
                    3, 1, 0, 1), nrow=4, ncol=4, byrow=TRUE)
      (dsave <- det(M))

    ## [1] 15

    # ### 'pivot' on the leading diagonal element, M[1,1]:

`det()` will be the product of the ‘pivots’, the leading diagonal elements. This step reduces row 1 and column 1 to 0, so it may be discarded. NB: In R, dropping a row/column can change a matrix to a vector, so we use `drop = FALSE` inside the subscript.

      (d <- M[1,1])

    ## [1] 2

        #-- Reduce row 1, col 1 to 0
      (M[1,] <- M[1,, drop=FALSE] / M[1, 1])

    ##      [,1] [,2] [,3] [,4]
    ## [1,]    1  1.5  0.5    1

      (M <- M - M[,1] %*%  M[1,, drop=FALSE])

    ##      [,1] [,2] [,3] [,4]
    ## [1,]    0  0.0  0.0    0
    ## [2,]    0 -4.0  1.0    0
    ## [3,]    0  2.5  1.5    1
    ## [4,]    0 -3.5 -1.5   -2

        #-- Drop first row and column
      M <- M[-1, -1]

        #-- Accumulate the product of pivots
      d <- d * M[1, 1]

### Repeat, reducing new row, col 1 to 0

      (M[1,] <- M[1,, drop=FALSE] / M[1,1])

    ##      [,1]  [,2] [,3]
    ## [1,]    1 -0.25    0

      (M <- M - M[,1] %*%  M[1,, drop=FALSE])

    ##      [,1]   [,2] [,3]
    ## [1,]    0  0.000    0
    ## [2,]    0  2.125    1
    ## [3,]    0 -2.375   -2

      M <- M[-1, -1]
      d = d * M[1, 1]

### Repeat once more. d = det(M)

      (M[1,] <- M[1,, drop=FALSE] / M[1,1])

    ##      [,1]   [,2]
    ## [1,]    1 0.4706

      (M <- M - M[,1] %*%  M[1,, drop=FALSE])

    ##      [,1]    [,2]
    ## [1,]    0  0.0000
    ## [2,]    0 -0.8824

      M <- M[-1, -1, drop=FALSE]
      d <- d * M[1, 1]

      # did we get it right?
      all.equal(d, dsave)

    ## [1] TRUE