Solutions Manual: Conv for Introduction to Linear Algebra for Science and Engineering (3rd Edition) - Graded A+ ...

STUDY NOTES & REFERENCE GUIDE




All 9 Chapters Covered
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SOLUTION MANUAL




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Table of contents
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1. Euclidean Vector Spaces
··




2. Systems of Linear Equations
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3. Matrices, Linear Mappings, and Inverses
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4. Vector Spaces
··




5. Determinants
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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
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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 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


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2 Chapter 1 Euclidean Vector Spaces
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥⎢ 2⎥ 3 – ⎥ 15 ⎥ = ⎥ 2 – 5 ⎥ ⎥–3 ⎥
A3 (a) ⎥ ⎥ ⎥ ⎥ ⎥ 3 – 1 ⎥ = ⎥ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦
4 –2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
2
⎥ ⎥ ⎥–3 ⎥ ⎢⎥ 2 + (–3) ⎥⎥ ⎥ –1 ⎥
(b) ⎥ 1 ⎥ + ⎥ 1 ⎥ = ⎥ 1 + 1 ⎥ = ⎥ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣–6 + (–4)⎦ ⎣–10⎦
–6 –4
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥ 4⎥ ⎢⎥ (–6)4 ⎥⎥ ⎥ –24
(c) –6 ⎥–5⎥ = ⎥ (–6)(–5) ⎦ = ⎥ ⎥
⎣ ⎦ ⎣(–6)(–6)⎥ ⎣ 30 ⎥⎥⎦
–6 36
⎡ ⎤ ⎡ ⎤ ⎡ 10 ⎤ ⎡ ⎤ ⎡7⎤
⎥⎢–5 ⎥ ⎥–1 ⎥ ⎢⎥ ⎥⎥ ⎥⎢–3⎥⎥ ⎥ ⎥
(d) –2 ⎥ 1 ⎥ + 3 ⎥⎣0 ⎥⎦ = ⎥⎣–2⎥⎦ + ⎥ 0 ⎥ = ⎥–2⎥
⎣ ⎦ –1 –2 ⎣–3⎦ ⎣–5⎦
1
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥ 2/3⎥ 1 ⎢⎢ 3 ⎥⎥ ⎥ 4/3 ⎥ ⎥ 1 ⎥⎥ ⎥ 7/3 ⎥
(e) 2 ⎥ ⎣⎢ –1/3⎥ ⎥⎦ + 3 ⎢⎣⎥–2⎦⎥⎥ = ⎣⎥
⎢ –2/3⎥ ⎥⎦ + ⎣⎥⎢ –2/3⎥ ⎥⎦ = ⎥⎢⎣–4/3⎥ ⎥⎦
2 1 4 1/3 , 13/3
⎡⎤ ⎡ ⎤
, ⎡1⎤ ⎥–1⎥⎥ ⎢⎡,, 2 ⎤ ⎡⎤ ⎢ 2 – π⎥
⎥ 1⎥ + π 0 = 2⎥⎥ + ⎥–π ⎥
···




,
(f) 2 ⎥ ⎥ ·




⎥ ⎥ ⎥ ⎥ ⎥0 ⎥=⎥ 2⎥⎥
⎣ ⎦ ⎣ ⎦ ⎢⎣, ⎦ ⎣ ⎦ ⎣ ⎢ , ⎦
1 1 2 π 2+π
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎥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⎦ ⎣ ⎦ ⎣–10⎦ ⎣–12⎦⎥ ⎣⎥–10⎦⎥ ⎣⎥ ⎦⎥
–22
–2 6 4
(c) We have w̃ – 2˜u = 3˜v, so 2˜u = w̃ – 3˜v or ˜u = 12( w̃ – 3˜v). This gives
⎛ ⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤
⎥⎜⎥2⎥ ⎥ 3 ⎥⎥⎥⎟ ⎢⎥–1⎥ ⎥ –1/2 ⎥
1 1
⎜⎢⎣ –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⎥⎥ ⎜⎥⎥6⎥ ⎥15⎥ ⎟⎟ ⎥ 16 –9 25
⎥ ⎥⎢ ⎥⎥ ⎥ ⎥
(b) 2(˜v + w̃ ) – (2˜v – 3 w̃) = 2 ⎥ 0 ⎣⎥ –⎦⎥⎥⎝⎣
2⎥ ⎦– ⎥–3⎣ ⎥⎥ ⎦ – ⎥⎣ 5 ⎥
⎦⎠ = ⎥⎣ 0 ⎥ ⎦ = ⎣⎥ –5 ⎦
⎥ –1 2 –6 –2 8 –10
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
5 6 –1
⎥ ⎥ ⎥⎢ ⎥ ⎥ ⎥
(c) We have w̃ – ˜u = 2˜v, so ˜u = w̃ – 2˜v. This gives ˜u = ⎥–1⎥ – ⎥2⎥ = ⎥–
3⎥. ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–2 2 –4


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