Evaluation of determinants

Michael Friendly

2024-10-03

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