Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 71
4 (a) 195.08/6.02 × 10 23 = 3.239 × 10 −22
(b) 4
(c) 4 · 3.239 × 10 −22 = 1.296 × 10 −21
(d) 1.296 × 10 −21 /21.45 = 6.042 × 10 −23 cubic centimeters
(e) 3.924 ⎛ × 10 −8 centimeters. ⎞ ⎛
⎞ ⎛
⎞
3.924 × 10−8 0
0
(f) 〈 ⎝ 0 ⎠ , ⎝3.924 × 10 −8 ⎠ , ⎝ 0 ⎠〉
0
0 3.924 × 10 −8
Topic: Voting Paradoxes
1 The mock election corresponds to the table on page 150 in the way shown in the first table, and after
cancellation the result is the second table.
positive spin negative spin positive spin negative spin
D > R > T
5 voters
R > T > D
8 voters
T > D > R
8 voters
T > R > D
2 voters
D > T > R
4 voters
R > D > T
2 voters
D > R > T
3 voters
R > T > D
4 voters
T > D > R
6 voters
T > R > D
–
D > T > R
–
R > D > T
All three come from the same side, the left, as the result from this Topic says must happen. Tallying
the election can now proceed, using the cancelled numbers
−1 voter
3 ·
T
D
R
1 voter
to get the same outcome.
1 voter
1 voter
+ 4 ·
T
D
R
1 voter
−1 voter
1 voter
+ 6 ·
T
D
R
−1 voter
–
1 voter
=
7 voter
T
D
R
1 voter
5 voter
2 (a) The two can be rewritten as −c ≤ a − b and −c ≤ b − a. Either a − b or b − a is nonpositive
and so −c ≤ −|a − b|, as required.
(b) This is immediate from the supposition that 0 ≤ a + b − c.
(c) A trivial example starts with the zero-voter election and adds any one voter. A more interesting
example is to take the Political Science mock election and add two T > D > R voters (they
can be added one at a time, to satisfy the “addition of one more voter” criteria in the question).
Observe that the additional voters have positive spin, which is the spin of the votes remaining after
cancellation in the original mock election. This is the resulting table of voters, and next to it is the
result of cancellation.
positive spin negative spin positive spin negative spin
D > R > T
5 voters
R > T > D
8 voters
T > D > R
T > R > D
2 voters
D > T > R
4 voters
R > D > T
10 voters 2 voters
The election, using the cancelled numbers, is this.
−1 voter
3 ·
T
D
R
1 voter
1 voter
−1 voter
+ 4 ·
T
D
R
1 voter
1 voter
D > R > T
3 voters
R > T > D
4 voters
T > D > R
8 voters
1 voter
+ 8 ·
T
D
R
−1 voter
T > R > D
–
D > T > R
–
R > D > T
–
1 voter
=
9 voters
T
D
R
−1 voter
7 voters
The majority cycle has indeed disappeared.
(d) One such condition is that, after cancellation, all three be nonnegative or all three be nonpositive,
and: |c| < |a + b| and |b| < |a + c| and |a| < |b + c|. This follows from this diagram.