Linear Algebra, Theory And Applications, 2012a

Answers To Selected Exercises
G.1 Exercises
1.6
1 (5+i9) −1 = 5
106 − 9
106 i
3 − (1 − i) √ 2, (1 + i) √ 2.
7 x =2− 2t, y = −t, z = t.
9 x = t, y = s +2,z = −s, w = s
G.3 Exercises
1.14
4 This makes no sense at all. You can’t add different
size vectors.
G.4 Exercises
1.17
4
5 If z ≠0, let ω = z
|z|
7 sin(5x) =5cos 4 x sin x − 10 cos 2 x sin 3 x +sin 5 x
cos (5x) =cos 5 x − 10 cos 3 x sin 2 x +5cosx sin 4 x
9 (x +2) ( x − ( i √ 3+1 )) ( x − ( 1 − i √ 3 ))
11 ( x − ( (1 − i) √ 2 )) ( x − ( − (1 + i) √ 2 )) ·
(
x −
(
− (1 − i)
√
2
)) (
x −
(
(1 + i)
√
2
))
15 There is no single √ −1.
G.2 Exercises
1.11
1 x =2− 4t, y = −8t, z = t.
3 These are invalid row operations.
5 x =2,y =0,z =1.
3 | ∑ n
k=1 β ka k b k |≤( ∑n
k=1 β k |a k | 2) 1/2
·
( ∑n
k=1 β k |b k | 2) 1/2
4 The inequality still holds. See the proof of the inequality.
G.5 Exercises
2.2
2 A = A+AT
2
+ A−AT
2
3 You know that A ij = −A ji . Let j = i to conclude
that A ii = −A ii and so A ii =0.
50 ′ =0+0 ′ =0.
60A =(0+0)A =0A +0A. Now add the additive
inverse of 0A to both sides.
70=0A =(1+(−1)) A = A+(−1) A. Hence, (−1) A
is the unique additive inverse of A. Thus −A =
(−1) A. The additive inverse is unique because if A 1
is an additive inverse, then A 1 = A 1 +(A +(−A)) =
(A 1 + A)+(−A) =−A.
487