4
Solutions to Exercises
26 Two eq uations co me from the two components: c + 3d = 14 and 2c + d = 8. The
solution is c = 2 and d = 4. Then 2(1, 2) + 4(3, 1) = (14, 8).
27 A four-dimensional cube has 2
4
= 16 corners and 2 · 4 = 8 three-dimensional faces
and 24 two-dimensional faces and 32 edges in Worked Example 2.4 A.
28 There are 6 unknown numbers v
1
, v
2
, v
3
, w
1
, w
2
, w
3
. The six equati ons come from the
components of v + w = (4, 5, 6) and v − w = (2, 5, 8). Add to find 2v = (6, 10, 14 )
so v = (3, 5, 7) and w = (1, 0, −1).
29 Fact: For any three vectors u, v, w in the plane, some combination cu + dv + ew is
the zero vector (beyond the obvious c = d = e = 0). So if there is one combination
Cu+Dv +Ew that produces b, there will be m any more—just add c, d, e or 2c, 2d, 2e
to the particular sol ution C, D, E.
The example has 3u − 2v + w = 3(1, 3) − 2(2, 7) + 1(1, 5) = (0, 0). It also has
−2u + 1v + 0w = b = (0, 1 ). Adding gives u − v + w = (0, 1). In this case c, d , e
equal 3, −2, 1 and C, D , E = −2, 1, 0.
Could an other example have u, v, w that could NOT combine to produce b ? Yes. The
vectors (1, 1), (2, 2), (3, 3) are on a line and no combination produces b. We can easily
solve cu + dv + ew = 0 but not Cu + Dv + Ew = b.
30 The combinations of v and w fill the plane unless v and w li e on the same line through
(0, 0). Four vectors whose combinati ons fill 4-dimensional space: one example is the
“standard basis” (1, 0, 0 , 0), (0, 1, 0, 0), (0, 0, 1, 0), an d (0, 0, 0, 1).
31 The equati ons cu + dv + ew = b are
2c −d = 1
−c +2d −e = 0
−d +2e = 0
So d = 2e
then c = 3e
then 4e = 1
c = 3/4
d = 2/4
e = 1/4
