solution manual for Introduction to Linear Algebra for Science and Engineering 3rd Edition by Norman

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 Diagonalization

7. Inner Products and Projections

8. Symmetric Matrịces and Quadratịc Forms

9. Complex Vector Spaces

, ✐

✐




CHAPTER 1 Euclịdean Vector Spaces

1.1 Vectors ịn R2 and R3
Practịce Problems
1 2 1+2 3 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 3 x1

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 (−2)3 −6 2 4 1 4 /3 7/3
(c) −2 = = (d) 1
+ 13 = + =
−2 (−2)(−2) 4 6 2 3 3 1 4
√
3 1 /4 2 1 /2 3 /2 √ 2 1 2 3 5
(e) 2
3 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
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢32 ⎥– 1 ⎢ 5=⎥ 2–5 ⎥ –3
A3 (a) ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 3 – 1 ⎥ = ⎢ ⎢ 2⎥⎥
⎣⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
4 –2 4 – (–2) 6
⎡ ⎤
2
⎢ ⎥ –3
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ 2 + (–3) ⎥⎥
⎢ ⎢ –1 ⎥
(b) ⎢ 1 ⎥ + ⎢ 1 ⎥ = ⎢ 1 + 1 ⎥ = ⎢ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–6 –4 –6 + (–4) –10
⎡ ⎤ ⎡ (–6)4 ⎤ ⎡ –24 ⎤
4
⎢ ⎥ ⎢⎢ ⎥⎥ ⎢
(c) –6 ⎥–5⎥ = ⎥ (–6)(–5) = ⎣ 30 ⎥⎥⎥
⎣ ⎦ ⎣(–6)(–6)⎦ ⎥ ⎥ 36
⎦
–6
⎡ ⎤ ⎡ ⎤ ⎡ 10 ⎤ ⎡ ⎤ ⎡7⎤
⎢⎢–5 ⎥ ⎢–1 ⎥ ⎢⎢ ⎥⎥ ⎢⎢–3⎥⎥ ⎢ ⎥
(d) –2 ⎥ ⎣1 ⎥ ⎦+ 3 ⎥ ⎣0 ⎥⎦ = ⎥⎣–2⎥⎦ + ⎥⎣ 0 ⎦⎥ = ⎥ ⎣ –2⎥⎦
1 –1 –2 –3 –5
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 2/3 ⎥ ⎢⎢ 3 ⎥⎥ ⎢ 4/3 ⎥ ⎢ 1 ⎥⎥ ⎢ 7/3 ⎥
1
(e) 2 ⎣⎥⎢ –1/3⎥ ⎥⎦ + 3 ⎢⎣⎥–2⎥⎥ ⎦ = ⎣⎥ ⎢ –2/3⎥ + ⎣⎥⎢ –2/3⎥⎥⎦ = ⎥ ⎢⎣–4/3⎥⎥⎦
2 1 4 1/3 13/3
⎥⎦
⎡⎤ ⎡, ⎤
⎡ ⎤ ⎢–1⎥⎥ ⎡ ,⎤ ⎡⎤ ⎢ 2 – π⎥
, 1 ⎢ ,2 –π
(f) 2 ⎢ 1⎢⎥ ⎥+ π ⎢ 0 ⎥ = ⎢ 2⎥⎥⎥ +⎢⎢ ⎥0 ⎥ = ⎢ , ⎥
⎢ , 2⎥
⎣⎦ ⎣ ⎦ ⎢⎣ , ⎦ ⎣ ⎣ ⎦
1 1 2 π 2 + π
⎡ ⎤ ⎡ ⎤ ⎦
⎢⎢ 2 ⎥ ⎢⎢ 6 ⎥ ⎡ ⎤
A4 (a) 2˜v – 3 w̃ = ⎢ 4 ⎥ – ⎢–3⎥ = ⎢ ⎢ –47 ⎥
⎥
⎣ ⎦ ⎣ 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 ⎥
⎝⎣ ⎦ ⎣ ⎣ ⎦ ⎣⎦ ⎣ ⎦ ⎣ ⎦ ⎥ ⎣⎥ ⎦⎥ ⎣⎥ ⎦⎥
–2 6 –10 4 –10 –12 –10 –22
⎦⎠
(c) We have w̃ – 2˜u = 3˜v, so 2˜u = w̃ – 3˜v or ˜u = 1 ( w̃ –2 3˜v). Thịs gịves
⎛ ⎡ ⎤ ⎡ ⎞⎤ ⎡ ⎤ ⎡ ⎤
⎜2
1 ⎜ ⎢ ⎥ ⎢ ⎥⎟⎥
3⎟
1 ⎢⎢–1 ⎥ ⎢ –1/2 ⎥
⎜⎢⎣ –1⎥⎥⎦ – ⎥⎢⎣ 6⎦⎥⎟ ⎢⎣–7/2 ⎥⎦
˜u = 2 ⎝⎥⎥ ⎠ = 2 ⎢⎥⎣ –7 ⎥
⎥⎥ ⎥⎦ = ⎥
9/2⎥
3 –6 9
⎡ ⎤
–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 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
1/2 –1 –1/2
⎡ 8 ⎤ ⎛ ⎡⎤ ⎡⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ 25 ⎤
⎢⎢ ⎥⎥ ⎜⎜⎢6⎥ ⎢15⎥ ⎟⎟ ⎢ ⎥ ⎢⎢–9⎥⎥
16
⎢ ⎥
(b) 2(˜v + w̃ ) – (2˜v – 3w̃) = 2 ⎥ 0 ⎥⎣ – ⎥⎥ ⎥ ⎦– ⎥–3⎣ ⎥⎥
⎦ 2⎝⎣ ⎦⎠= ⎥⎣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


Copyrịght ⃝c 2013 Pearson Canada Ịnc.