Chapter D Section PDM Exercise C30

Each of the archetypes below is a system of equations with a square coefficient matrix, or is a square matrix itself. Compute the determinant of each matrix, noting how Theorem SMZD indicates when the matrix is singular or nonsingular.

Archetype A, Archetype B, Archetype F, Archetype K, Archetype L

Archetype A

\[ \begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix} \]

Calculate determinant by expanding on the third row.

\[ \begin{split} det\ A & = 1 \begin{bmatrix} -1 & 2 \\ 1 & 1 \\ \end{bmatrix} -1 \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix} +0 \begin{bmatrix} 1 & -1 \\ 2 & 1 \\ \end{bmatrix} \\ &= 1 (-1-2) - 1 (1-4) + 0 \\ &= -3 + 3 + 0 \\ &= 0 \end{split} \]

A <- matrix(c(1,2,1,-1,1,1,2,1,0), nrow=3)
A
##      [,1] [,2] [,3]
## [1,]    1   -1    2
## [2,]    2    1    1
## [3,]    1    1    0
det(A)
## [1] 0

Since \(det\ A = 0\), per Theorem SMZD \(A\) is singular.

Calculate all other determinants using R.

B <- matrix(c(-7,5,1,-6,5,0,-12,7,4), nrow=3)
F <- matrix(c(33,99,78,-9,-16,-47,-36,2,10,27,17,3,-2,-7,-6,4), nrow=4)
K <- matrix(c(10,12,-30,27,18,18,-2,-21,30,24,24,-6,-23,36,30,24,0,-30,37,30,-12,-18,39,-30,-20), nrow=5)
L <- matrix(c(-2,-6,10,-7,-4,-1,-5,7,-5,-3,-2,-4,7,-6,-4,-4,-4,10,-9,-6,4,6,-13,10,6), nrow=5)

Archetype B

B
##      [,1] [,2] [,3]
## [1,]   -7   -6  -12
## [2,]    5    5    7
## [3,]    1    0    4
det(B)
## [1] -2

Since \(det\ B \ne 0\), per Theorem SMZD \(B\) is nonsingular.

Archetype F

F
##      [,1] [,2] [,3] [,4]
## [1,]   33  -16   10   -2
## [2,]   99  -47   27   -7
## [3,]   78  -36   17   -6
## [4,]   -9    2    3    4
det(F)
## [1] -18

Since \(det\ F \ne 0\), per Theorem SMZD \(F\) is nonsingular.

Archetype K

K
##      [,1] [,2] [,3] [,4] [,5]
## [1,]   10   18   24   24  -12
## [2,]   12   -2   -6    0  -18
## [3,]  -30  -21  -23  -30   39
## [4,]   27   30   36   37  -30
## [5,]   18   24   30   30  -20
det(K)
## [1] 16

Since \(det\ K \ne 0\), per Theorem SMZD \(K\) is nonsingular.

Archetype L

L
##      [,1] [,2] [,3] [,4] [,5]
## [1,]   -2   -1   -2   -4    4
## [2,]   -6   -5   -4   -4    6
## [3,]   10    7    7   10  -13
## [4,]   -7   -5   -6   -9   10
## [5,]   -4   -3   -4   -6    6
det(L)
## [1] -4.437343e-30

R calculates \(det\ L\) as very close to zero, but not exactly zero. I have a suspicion that this is just a rounding error. Check if inverse matrix exists.

library(matlib)
inv(L)
## Error in Inverse(X, tol = sqrt(.Machine$double.eps), ...): X is numerically singular

\(L\) is not inversible. Therefore, it is singular. Therefore, \(det\ L = 0\).