Solution Manual – Introduction to Linear Algebra for Science & Engineering (Norman & Wolczuk, 3rd Ed.)

All 9 Chapters Covered




SOLUTION MANUAL

,Table of contents
1. Euclidean Vector Spaces

2. Systems of Linear Equations

3. Matrices, Linear Mappings, and Inverses

4. Vector Spaces

5. Determinants

6. Eigenvectors and Dịagonalịzatịon

7. Ịnner Products and Projectịons

8. Symmetrịc Matrịces and Quadratịc Forms

9. Complex Vector Spaces

, ✐

✐




CHAPTER 1 Euclịdean Vector Spaces

1.1 Vectors ịn R2 and R3
Practịce Problems

3 1 2 1+2 3 4 3−4 −1
A1 (a) + = = (b) − = =
4 3 4+3 7 2 1 2−1 1
x2
1 2
1 4 3 3
3 4 2
4 4
2 1
3
4


x1
−1 3(−1) −3 2 3 4 6 −2
(c) 3 = = (d) 2 −2 = − =
4 3(4) 12 1 −1 2 −2 4

2 3
3 2
4 1

3 2 2
1 2
1

4 x1
3

x1
4 −1 4 + (−1) 3 −3 −2 −3 − (−2) −1
A2 (a) −2 + 3 = −2 + 3 =1 (b) −4 − 5 = −4 − 5 =−9
3 −6 2 4 1 4/3 7/3
(c) −2 = = (d) 1
+1 = + =
(−2)3
−2 (−2)(−2) 4 2 36 3 1 3 4
√
3 1/4 2 1/2 3/2 √ 2 1 2 3 5
(e) 23 1 − 2 1/3 = 2/3 −2/3 =0 (f) 2 √ + 3 √6 = √6 + 3 √6 = 4 √6
3


Copyrịght ⃝c 2013 Pearson Canada Ịnc.

, ✐

✐


2 Chapter 1 Euclịdean Vector Spaces
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
5 –3
⎢ 2 –5 ⎥
⎢2
A3 ⎢⎢3⎥ ⎥– ⎢ 1⎢⎥ =⎥ ⎢ 3 – ⎥ = ⎢ ⎢ 2 ⎥⎥
(a 1
)
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
4 –2 4 – (–2) 6
⎡ ⎤
⎢ 2 ⎥ –3
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢⎢ 2 + (–3) ⎥⎥ ⎢ –1 ⎥
(b) ⎢ 1 ⎥ + ⎢ 1 ⎥ = ⎢ 1 + ⎥ =⎢ 2 ⎥
1
⎣ ⎦ ⎣ ⎦ ⎣–6 + (– ⎣
–10
⎦
–6 –4
⎦
4)
⎡ ⎤ ⎡ (– ⎤ ⎡ ⎤
4 ⎥⎥ ⎢–24 ⎥
⎢ ⎥ ⎢⎢
(c) –6 ⎥–5⎥ = ⎥(–6)(–5) 6)4 ⎥ = ⎥ 30 ⎥⎥
⎦
⎣ ⎦ ⎣(–6)(– ⎣
36
–6
⎦
6)
⎡ ⎤ ⎡ ⎤ ⎡ 10 ⎤ ⎡ ⎤ ⎡7⎤
⎢⎢–5 ⎥ ⎢–1 ⎥ ⎢⎢ ⎥⎥ ⎢⎢–3⎥⎥ ⎢ ⎥
(d) –2 ⎥ ⎣1 ⎥ ⎦+ 3 ⎥ ⎣0 ⎥⎦= ⎥⎣–2⎥⎦ + ⎥⎣ 0 ⎥ ⎦ =⎥ ⎣ –2⎥⎦
1 –1 –2 –3 –5
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 2/3 ⎥ 1 ⎢⎢ 3 ⎥⎥ ⎢ 4/3 ⎥ ⎢ 1 ⎥⎥ ⎢ 7/3 ⎥
⎣⎢ –1/3⎥
(e) 2 ⎥ ⎥⎦ + 3 ⎢⎣⎥–2⎥ ⎥⎦ = ⎣⎥ ⎢ –2/3⎥ + ⎣⎥⎢ –2/3⎥ = ⎥ ⎢⎣–4/3⎥ ⎥⎦
2 1 4 1/3 13/3
⎥⎦ ⎥⎦
⎡ ⎤ ⎡, ⎤
⎡⎤ ⎢–1⎥⎥ ⎡
⎢ ⎡ ⎤ ⎢ 2 – π⎥
⎢1 ⎥ , ⎤ ⎢–π⎥
, ,2 ⎥
(f) 2⎢ ⎥1 + π⎢ 0⎥ = ⎢ 2 ⎥ + 0 =⎢ , ⎥
⎥ ⎢ ⎥ 2⎥
⎣ ⎦ ⎣ ⎦ ⎢⎣ ⎣ ⎦ ⎣ ⎢ , ⎦
1 1 π 2 +π
, ⎦
⎡ ⎤ 2 ⎡ ⎤
⎢⎢ 2 ⎥ ⎡ 6 ⎤ ⎢ –4 ⎥
A4 (a) 2˜v – 3 w̃ = ⎢ 4 ⎥ – ⎢–3⎥⎢⎢= ⎢ ⎥ 7 ⎥
⎣ ⎦ ⎣ 9 ⎦ ⎣–13⎦
–4
⎡ ⎤⎞ ⎡ ⎤ ⎡
⎛⎡ ⎤ 4 ⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎤ ⎡ ⎤
⎜ ⎢ 1
⎢ ⎥ ⎥ ⎢ ⎥⎟
⎢ 5 ⎥
⎥ ⎢⎢5⎥⎥ ⎢⎢ 5 ⎥⎥ ⎢⎢–15⎥⎥ ⎢ 5 ⎥⎥ ⎢ –10
(b) –3(˜v + 2 w̃) + 5˜v = –3 ⎜⎢ 2 ⎥ + ⎢–2⎥⎟ + ⎢ 10 ⎥ = –3 ⎢0⎥ + ⎢ 10 ⎥ = ⎢ 0 + 10 = ⎢10 ⎥
⎣ ⎦ ⎣ ⎦ ⎣–10⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎝⎣ ⎦ ⎣
–2 6 –10 4 –12⎥ ⎥–10⎥ ⎥–22⎥
⎦⎠
(c) We have w̃ – 2˜u = 3˜v, so 2˜u = w̃ – 3˜v or ˜u =2 1 ( w̃ – 3˜v). Thịs gịves
⎛ ⎡ ⎤ ⎡ ⎞⎤ ⎡ ⎤ ⎡ ⎤
⎜ ⎜⎢2 ⎥ ⎢ 3 ⎟⎥⎥⎟ ⎢ –1 ⎢ –1/2 ⎥
1 1 ⎢ ⎥
˜u = 2 ⎝⎥⎥ ⎥⎟⎠ = 2 ⎢⎥⎣ –7 ⎥ = ⎥⎣–7/2⎥⎥⎦
⎜⎢⎣ –1⎥⎥⎦ – ⎥⎢⎣ 6 ⎦⎥⎥ ⎢
3 –6 9 9/2
⎥⎦
⎡ ⎤
–3
(d) We have ˜u – 3˜v = 2˜u, so ˜u = –3˜v = ⎢–6⎥⎥. ⎥
⎣ ⎦
6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
3/2
⎢ ⎥ ⎢5/2 ⎥⎥ ⎢⎢ 4 ⎥
A5 (a) 1˜v + 1 w̃ = ⎢1/2⎥ + ⎢–1/2 ⎥=⎢ 0 ⎥
2 2 ⎢⎣ ⎥⎦
⎢⎣
⎢⎣ ⎥⎦ –1 –1/2
1/2 ⎥⎦
⎡ 8 ⎤ ⎛ ⎡⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ ⎥⎥ ⎜⎜⎢6⎥ ⎢15⎥ ⎟⎟ ⎢ ⎥ ⎢⎢–9⎥⎥
16
⎢
25
⎥
(b) 2(˜v + w̃ ) – (2˜v – 3w̃) = 2 ⎥ 0 ⎥
⎣ – ⎥⎥
⎦ 2⎝⎣⎥ –⎦ ⎥–3⎣ ⎥⎥ ⎦⎠= ⎥⎣0 ⎥⎦ – ⎥⎣ 5 ⎥ = ⎥ ⎣ –5 ⎥⎦
–1 2 –6 –2 8 –10
⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢–1 ⎥
5 6
⎢ ⎥ ⎢⎢ ⎥
(c) We have w̃ – ˜u = 2˜v, so ˜u = w̃ – 2˜v. This gives ˜u = ⎥–1⎥ – ⎥2⎥ = ⎥–3⎥.
⎣ ⎦ ⎣ ⎦ ⎣ ⎦

–2 2 –4