Full Solution Manual for An Introduction to Linear Algebra for Science and Engineering (3rd Edition) by Daniel Norman and Dan Wolczuk Complete Coverage (Chapters 1-9) Verified Mathematical Solutions Vector Spaces / Linear Mappings / Eigenvectors / Orthogo

All 9 Chapters Covered




SOLUTION MANUAL

,Table of contents
1. Euclidean Vector Spaces

2. Systems of Linear Equations

3. Matrices, Linear Mappinḡs, and Inverses

4. Vector Spaces

5. Determinants

6. Eiḡenvectors and Diaḡonaliẓation

7. Inner Products and Projections

8. Symmetric Matrices and Quadratic Forms

9. Complex Vector Spaces

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CHAPTER 1 Euclidean Vector Spaces

1.1 Vectors in R2 and R3
Practice 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


3 2 3
4 2
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 (−2)3 −6
(c) −2 = = (d)
2 1
+ 13
4
=
1
+
4/3
=
7/3
−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


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2 Chapter 1 Euclidean Vector Spaces
⎡ ⎡ ⎡ ⎡ ⎡ ⎡
⎥⎡ 2⎥ ⎥ 5
⎥
⎡
2–5 ⎥–3 ⎥
⎡ = 2
3 – ⎥ 1 = ⎥
A3 (a) ⎥ ⎥ ⎥ ⎥ ⎥ 3 – 1 ⎥ ⎥ ⎥
⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡
4 – 4 – (–2) 6
⎡
2
⎡ ⎡ ⎡ ⎡ ⎡ ⎡
2
⎥ ⎥ ⎥– ⎡⎡⎥ 2 + (–3) ⎥ –1 ⎥
⎡ ⎥⎡ = 2
3⎥
(b) ⎥ 1 ⎥ + ⎥ 1 ⎥ = ⎥ 1 + 1 ⎥ ⎥ ⎥
⎡ ⎡ ⎡ ⎡ ⎡ ⎡
–6 – –6 + (– –10
⎡ ⎡
4 4)
⎡ ⎡ ⎡ ⎡ ⎡ ⎡
⎥ 4⎥ ⎡⎥ (–6)4 ⎡ ⎥ ⎥
(c) –6 ⎥–5 ⎥ = ⎥(–6)(–5)⎥ = ⎥–24 30 ⎥⎥
⎡ ⎦
⎡ ⎡ ⎡ ⎡
–6 (–6)(– 36
⎡
6)
⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡
⎥⎡–5 ⎥ ⎥ ⎥–1 ⎥⎡– ⎥ ⎥
(d) –2 ⎥ 1 ⎥ + 3 ⎥ 0 ⎥ = ⎥–2⎥ + ⎥ 0 ⎥ = ⎥–2⎥
⎡10 ⎡ ⎡
⎡⎥⎥⎡ 3 ⎥⎡ 7
⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡
1 –1 – – –5
⎡ ⎡
2 3
⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡
⎥ 2/3⎥ 1 ⎡3⎡ ⎥ 4/3 ⎥1 ⎥ 7/3 ⎥
⎡
(e) 2 ⎥–1/3⎥ + 3 ⎢⎥–2⎥ ⎥ =⎥ ⎡ –2/3⎥ + ⎥⎡⎡–2/3⎥ = ⎥–4/3⎥
⎡ ⎥
⎡⎡
⎡⎡ ⎡⎡ ⎡ ⎡⎡ ⎡⎡ ⎡⎡ ⎡⎡
2 1 4 13/3
⎡ ⎡⎡ ⎡⎡
1/3
⎡ ⎡, ⎡
⎡⎡ ⎥ – ⎡ ⎡ ⎡ 2 – π⎡
⎥ ⎥1 ⎡ ⎡
, ⎡ ⎥⎡ –
, 1 ⎡⎡ ,2 ⎥ π ⎥
(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⎥⎡ ⎥⎡ ⎡⎡– ⎥ ⎥⎡ ⎡⎡–10 ⎥
5 ⎡ ⎡ 5
⎥⎥ ⎥ ⎡ 15⎡ ⎡
(b) –3(˜v + 2 w̃ ) + 5˜v = –3 ⎥⎥ 2 ⎥ + ⎥–2⎥⎥ + ⎥ 10 ⎥ = –3 ⎥0⎥ + ⎥ 10 ⎥ = ⎥ 0 + 10 = 10
⎡⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎡ ⎥ ⎡ ⎥– ⎥ ⎡
⎥ –22⎥⎡
–2 6 – 4 – –
⎡⎡ ⎡ ⎡ ⎡ ⎡ ⎡
10 10 12 10
(c) We have w̃ – 2˜u = 3˜v, so 2˜u = w̃ – 3˜v or ˜u = 12( w̃ – 3˜v). This ḡives
⎡ ⎡⎡ ⎡ ⎡⎡ ⎡ ⎡ ⎡
⎥⎡ ⎥2⎥ ⎥ 3 ⎡⎥⎥⎡ ⎡⎥– ⎥ –1/2 ⎥
1 1
˜u = ⎥⎥–1⎥ – ⎥ 6 ⎥⎥ = ⎡ –7 = –7/2
⎥
1⎥ ⎥ ⎥ ⎥
2 ⎡⎡⎡⎡ (d) We have ˜u – 3˜v = 2˜u, so ˜u3= –3˜v =
⎡
⎣ ⎦ 3 ⎡
6
⎥⎡⎡ ⎥ ⎥–6⎥. –
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