Introduction to Linear Algebra for Science and Engineering (3rd Edition) by Norman – Solution Manual | Complete Worked Solutions for Linear Algebra

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 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
4 − 2 4
2 1
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 2

3 2 2
−2 1 2
−1 1

4 x1
3
−1
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)
21
+ 31
4
=
1
+
4/3
=
7/3
−2 (−2)(−2) 4 62 3 3 1 4
√
√ 3 5
(f) 2 √3 + 3 √6 = √6 + 3√6 = 4√ 6
3 1/4 2 1/2 3/2 2 1 2
1 − 2 1/3 = 2/3 − 2/3 = 0
2
(e) 3




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


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