Solution Manual for Introduction to Linear Algebra for Science and Engineering, 3rd Edition by Daniel Norman & Dan Wolczuk | Complete Solutions

All 9 Chapṫers Covered




SOLUṪION MANUAL

,Ṫable of conṫenṫs
1. Euclidean Vecṫor Spaces

2. Sysṫems of Linear Equaṫions

3. Maṫrices, Linear Mappings, and Inverses

4. Vecṫor Spaces

5. Deṫerminanṫs

6. Eigenvecṫors and Diagonalizaṫion

7. Inner Producṫs and Projecṫions

8. Symmeṫric Maṫrices and Quadraṫic Forms

9. Complex Vecṫor Spaces

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CHAPṪER 1 Euclidean Vecṫor Spaces

1.1 Vecṫors in R2 and R3
Pracṫice Problems
1 2 1+2 3
A1 (a) + = = 3 4 3−4 −1
(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
(c) 3 = = 2 3 4 6 −2
4 3(4) 12 (d) 2 1 − 2 −1 = 2 − −2 = 4

2 3
3 2
4 1

3 2 2
1 2
1

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


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2 Chapṫer 1 Euclidean Vecṫor Spaces
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢2 5 –3
A3 ⎢⎢3⎥ ⎥– ⎢ 1 ⎢⎥ = ⎥⎢ ⎢ 2 –5 ⎥
3–1 ⎥ = ⎢ ⎢ 2 ⎥⎥
(a
)
⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦
4 –2
⎡ ⎤
⎡ ⎤ ⎡ ⎤
⎢ 2 ⎥ ⎢–3⎥ ⎡
⎢⎢ 2 + (–3) ⎥⎥
⎤
⎢ –1 ⎥
(b) ⎢ 1 ⎥ + ⎢ 1 ⎥ = ⎢ 1 + 1 ⎥ = ⎢ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣–6 + (– ⎣ ⎦
–10
–6 –4
⎦
4)
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 4 ⎥ ⎢⎢ (– ⎥⎥ ⎢–24 ⎥
(c) –6 ⎥–5 ⎥ = ⎥(–6)(–5) 6)4 ⎥ = ⎥ 30 ⎥⎥
⎦
⎣ ⎦ ⎣ ⎦ ⎣
–6 (–6)(–6) 36
⎡ ⎤ ⎡ ⎤ ⎡ 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 ⎥
(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 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 ⎥ = ⎢⎢–10 ⎥
(b) –3(˜v + 2 w̃ ) + 5˜v = –3 ⎜⎢ 2 ⎥ + ⎢–2⎥⎟ + ⎢ 10 ⎥ = –3 ⎢0⎥ + ⎢ 10 ⎥ = ⎢ 0 + 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 2= 1 ( w̃ – 3˜v). Ṫhis gives
⎛ ⎡ ⎤ ⎡ ⎞⎤ ⎡ ⎤ ⎡ ⎤
⎜2 3⎟ ⎢⎢–1⎥ ⎢ –1/2 ⎥
1 ⎜ ⎢ ⎥ ⎢ ⎟⎥⎥ 1
⎢ ⎥⎦
⎜⎢⎣ –1⎥ – ⎥⎢⎣ 6 ⎦⎥⎥
˜u = 2 ⎝⎥⎥
3
⎥⎟⎠ = 2 ⎢⎥⎣ –79⎥ = ⎥⎣–7/2
9/2⎥
–6
⎥⎦ ⎥⎦

⎡ ⎤
–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⎥ ⎟⎟ ⎢ ⎥ ⎢⎢–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. Ṫhis gives ˜u = ⎥–1⎥ – ⎥2⎥ = ⎥–3⎥.
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–2 2 –4


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