Solution Manual – Introduction to Linear Algebra for Science & Engineering (Norman & Wolczuk, 3rd Ed.)

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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
1

"
# " # " #x1 " # " # " # " # " #
−1 3(−1) −3 2 3 4 6 −2
(c) 3 = = (d) 2 −2 = − =
4 3(4) 12 1 −1 2 −2 4

1 2 3
3
4 1
" 2 2
3 2
2 1 1
1
4

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 = 12 2 4 1
(d) 6 + 13 3 = 3 + 1 = 4
4/3 7/3
−2 (−2)(−2) = 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 √ = √ + 3 6 = 4 6
3 6 6


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c 2013 Pearson Canada Inc.




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2 Chapter 1 Euclidean Vector Spaces
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥2⎥ ⎥5 ⎥ ⎢⎢ 2 – 5 ⎥ ⎥–3 ⎥
A3 (a) ⎥3⎥ – ⎥ 1 ⎥ = ⎥ 3 – 1 ⎥ = ⎥ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦
4 –2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥ 2 ⎥ ⎥–3⎥ ⎥ ⎥ 2 + (–3) ⎥ ⎥ –1 ⎥
(b) ⎥ 1 ⎥ + ⎥ 1 ⎥ = ⎥ 1 + 1 ⎥ = ⎥ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–6 –4 –6 + (–4) –10
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 4⎥ ⎥ (–6)4
⎥ ⎥–24 ⎥
(c) –6 ⎥⎣–5⎥⎦ = ⎥ ⎦ = ⎥⎣ 30 ⎥⎥⎦
⎣ (–6)(–5)⎥
–6 (–6)(–6) 36
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢–5 ⎥ ⎢⎢–1⎥ ⎥ 10 ⎥ ⎢⎢–3⎥ ⎥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⎥ + 13 ⎥–2⎥ = ⎥–2/3⎥ + ⎢⎥⎣ –2/3⎥ ⎢⎣–4/3⎥ ⎥⎦
⎥⎦ = ⎥
⎣ ⎥⎦ ⎥⎣ ⎦⎥ ⎣⎥ ⎥⎦
2 1/3 13/3
⎡ ⎤1 4 ⎡ , ⎤
,
⎥–1⎥⎥ ⎡⎢, 2 ⎤ ⎡ –π⎤ ⎢2 – π⎥
, ⎡⎢⎢ ⎤⎥ ,
(f) 2⎥1⎥ + π⎥ 0 ⎥ = ⎥⎢ 2⎥⎥ + ⎥⎥0⎥⎥ = ⎢⎥ 2 ⎥⎥
1
⎣ ⎦ ⎣ ⎦ ⎢⎣, ⎦ ⎣ ⎦ ⎢⎣ , ⎥⎦
1 1 2 π 2+π
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥2 ⎥ ⎥ 6 ⎥ ⎥ –4 ⎥
A4 (a) 2˜v – 3 w̃ = ⎥ 4 ⎥ – ⎥–3⎥ = ⎥ 7 ⎥
⎣ ⎦ ⎣ 9 ⎦ ⎣–13⎦
–4
⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎜⎜⎢⎢ 1 ⎥ ⎥ 4 ⎥⎥ ⎥ 5 ⎥ ⎥5⎥ ⎥ 5 ⎥ ⎢⎢–15 ⎥ ⎥ 5 ⎥ ⎥–10 ⎥
(b) –3(˜v + 2w̃ ) + 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⎥⎥
⎦ =⎥
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 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 – 3w̃) = 2 ⎥ 0 ⎣⎥ –⎦⎥⎥⎝⎣
2⎥ ⎦– ⎥–3 ⎦⎠= ⎥⎣0 ⎥⎦ – ⎥⎣ 5 ⎥
⎣ ⎥⎥ ⎦ =⎥ ⎣ –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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