Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 167
Subsection Four.I.4: Determinants Exist
Four.I.4.10 This is the permutation expansion of the determinant of a 2×2 matrix
∣ a b
∣ ∣ ∣∣∣ c d∣ = ad · 1 0
∣∣∣ 0 1∣ + bc · 0 1
1 0∣
and the permutation expansion of the determinant of its transpose.
∣ a c
∣ ∣ ∣∣∣ b d∣ = ad · 1 0
∣∣∣ 0 1∣ + cb · 0 1
1 0∣
As with the 3×3 expansions described in the subsection, the permutation matrices from corresponding
terms are transposes (although this is disguised by the fact that each is self-transpose).
Four.I.4.11 Each of these is easy to check.
(a) permutation φ 1 φ 2 (b) permutation φ 1 φ 2 φ 3 φ 4 φ 5 φ 6
inverse φ 1 φ 2 inverse φ 1 φ 2 φ 3 φ 5 φ 4 φ 6
Four.I.4.12 (a) sgn(φ 1 ) = +1, sgn(φ 2 ) = −1
(b) sgn(φ 1 ) = +1, sgn(φ 2 ) = −1, sgn(φ 3 ) = −1, sgn(φ 4 ) = +1, sgn(φ 5 ) = +1, sgn(φ 6 ) = −1
Four.I.4.13 The pattern is this.
i 1 2 3 4 5 6 . . .
sgn(φ i ) +1 −1 −1 +1 +1 −1 . . .
So to find the signum of φ n! , we subtract one n! − 1 and look at the remainder on division by four. If
the remainder is 1 or 2 then the signum is −1, otherwise it is +1. For n > 4, the number n! is divisible
by four, so n! − 1 leaves a remainder of −1 on division by four (more properly said, a remainder or 3),
and so the signum is +1. The n = 1 case has a signum of +1, the n = 2 case has a signum of −1 and
the n = 3 case has a signum of −1.
Four.I.4.14 (a) Permutations can be viewed as one-one and onto maps φ: {1, . . . , n} → {1, . . . , n}.
Any one-one and onto map has an inverse.
(b) If it always takes an odd number of swaps to get from P φ to the identity, then it always takes
an odd number of swaps to get from the identity to P φ (any swap is reversible).
(c) This is the first question again.
Four.I.4.15 If φ(i) = j then φ −1 (j) = i. The result now follows on the observation that P φ has a 1 in
entry i, j if and only if φ(i) = j, and P φ −1 has a 1 in entry j, i if and only if φ −1 (j) = i,
Four.I.4.16 This does not say that m is the least number of swaps to produce an identity, nor does it
say that m is the most. It instead says that there is a way to swap to the identity in exactly m steps.
Let ι j be the first row that is inverted with respect to a prior row and let ι k be the first row giving
that inversion. We have this interval of rows.
⎛ ⎞
.
.
ι k
ι r1
.
j < k < r 1 < · · · < r s
ι rs
⎜
⎝ ι j..
⎟
⎠
Swap.
⎛ ⎞
.
.
ι j
ι r1
.
ι rs
⎜
⎝ ι ḳ
⎟
⎠
.