(1) If ||v|| = 7 and ||w|| = 4, what are the smallest and largest possible values
of ||v − w||? What are the smallest and largest values of v · w?
Solution: ||v − w|| ≤ ||v|| + ||w|| = 11 and ||v − w|| ≥ ||v|| − ||w||| = 3, while
|v · w| = ||v||||w|| cos θ ≤ ||v|| · ||w|| = 28, so −28 ≤ v · w ≤ 28.
(2) What values of c

1
 1
1
give dependent columns?
 
 
3 5
1 0 c
c
2 4 , 1 1 0 , 2
1 c
0 1 1
3
c
1
3

c
5 
6
Solution: For the first matrix, one wants to express (5, 4, c) as a linear combination of (1, 1, 1) and (3, 2, 1): taking twice the first plus the second gives
(5, 4, 3), so c = 3. Similarly, c = −1 for the second (making the first
and third columns add up to the second) while c = 0 is the solution
for the last one (dependent columns = dependent rows).
(3) Invent a 3x3 magic matrix M3 with entries 1, . . . , 9. All rows, columns,
and diagonals add up to the same number. What is M3 times (1, 1, 1)? Do
the same for a 4x4 matrix M4 with entries 1, . . . , 16.
Solution: The choice of matrix does not matter: the sum of all the entries is
1 + · · · + 9 = 45, each row and column must sum to 45/3 = 15.
Multiplying by (1, 1, 1) sums each row, thus giving you (15, 15, 15).
Similarly, in the 4x4 case, one gets (34, 34, 34, 34).
(4) For which three numbers a will elimination fail to give three pivots for the
following matrix?


a 2 3
A= a a 4 
a a a
Solution: Running elimination, one obtains






a
2
3
a
2
3
a 2 3
 a a 4 → 0 a−2
1 → 0 a−2
1 
0 a−2 a−3
0
0
a−4
a a a
Thus, elimination fails if a = 0, 2, 4.
(5) Write these problems as matrix problems Ax = b and solve them.
• Alice is twice as old as Bob now, but was three times as old as Bob 10
years ago. What are the ages of Alice and Bob now?
• The line y = mx + c contains points (2, 5) and (3, 7). Find m and c.
Solution: For the first problem, we have the equations x − 2y = 0 and (x − 10) −
3(y − 10) = 0. Rearranging the latter equation, one obtains
1 −2
x
0
·
=
1 −3
y
−20
Solving gives x = 40, y = 20. For the second problem, we have the
equations 5 = 2m + c and 7 = 3m + c, giving the matrix equation
2 1
m
5
·
=
3 1
c
7
Solving gives m = 2, c = 1.
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