solution manual for Introduction to Linear Algebra for Science and Engineering 3rd Edition by Norman

All 9 Chapters Covered
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SOLUTION MANUAL
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,Tableofcontents
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1. Euclidean Vector Spaces ss ss




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




5. Determinants

6. Eigenvectors and Diagonalization ss ss




7. Inner Products and Projections
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8. Symmetric Matrices and Quadratic Forms ss s s ss




9. Complex Vector Spaces ss ss

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CHAPTER 1 Euclidean Vector Spaces ss ss ss ss




1.1 Vectors in R2 and R3 ss ss ss ss




Practice Problems ss




s s s s s s s s s s s s s s s s s s s s s s s s ss

1 2 1+2 3 4 3− 4 −1
−
ss ss ss ss
s s ss
A1 3 (a)
s s s s + = (b) = =
=
4 3 4+3 ss ss 7 2 1 2− 1 ss ss 1
x2 sss s

1 2
sss s
sss s
sss s

1 4 3 sss s
3
3 4
sss s 2 sss s

4 4
2
sss s
1
3 sss s

4



s s s s s s s s s
x1
s s s s s s s s s s s s s s s s s s s s s s s s s s s ss

−1 3(−1) 2 3 4 6 −2
(c) 3
−3
ss
= = (d) 2 −2 ss = − =
4 3(4) 12 1 −1 2 −2 4
sss s



3 2 3
4 2
1
sss s



3
ss
2 sss s

2
1 2 s s


1

4 ss
x1
3

x1
s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s ss

4 −1 4 + (−1) 3 −3 −2 −3 − (−2) −1
(b) −4 − 5
ss ss ss ss

A2 (a) −2s s +3 = +3
−2 1 = ss ss
=−4 − 5 −9= ss ss
s s s s s s s s s s s s

3 ( −6
s s s s s s s s s s s s s s s s s s s s s s ss


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




−2 (−2)(−2) 4 3 3 3 12s s s s 6
4 s s s s




√
√
ss ssss ss sss s sssss s ssss ss sss s ss ss ssss ssss ssss sss s ssss ssss ssssssss sss s ssssssss

3 1/4 2 1/2 3/2 2 1 2 3 5
2ssss
1 − 2 1/3 = 2/3 − 2/3 = 0 (f) 2 √ + 3 √ 6 = √ 6 + 3√ 6 = 4 √6
ssss

(e) 3 ss s ss ss ss ssss s ss s ss ss ss
3


Copyright ⃝c 2013 Pearson Canada Inc. ss ss ss




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2 Chapter 1 ss Euclidean Vector Spaces ss ss




⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤
⎥⎢ 2⎥ ⎥ ⎤ 2– ⎥–3 ⎥
ss s s

s s
⎥5 ss ss




⎥ 5 = 2
(a) ⎥3⎥ – ⎥ 1 ⎥ = ⎥ 3⎥– 1 ⎥ ⎥ ⎥
ss s s


A3 ss
ss
ss
ss
ss
ss
ss ss


⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦ ss

4 –2 ss ss ss

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
2
⎥ ⎥ ⎥–3 ⎥ ⎢⎥ 2 + (–3) ⎥⎥ ⎥ –1 ⎥
ss
ss
ss ss ss ss
ss

(b) ⎥ 1 ⎥ +⎥ 1⎥ = ⎥ 1 + 1 ⎥ = ⎥ 2 ⎥ ss ss
ss




⎣ ⎦ ⎣ ⎦ ⎣–6 + (–4)⎦ ⎣–10⎦
s s s s s s s s



–6 –4 ss ss



⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥ 4⎥ ⎢⎥ (–6)4 ⎥⎥ ⎥ –24
(c) –6 ⎥–5⎥ = ⎥ (–6)(–5) ⎦ = ⎥⎣ 30 ⎥⎥⎥
⎣ ⎦ ⎣(–6)(–6)⎥ 36 ⎦
s s ss ss ss ss
ss ss sss
s

–6
⎡ ⎤ ⎡ ⎤ ⎡ 10 ⎤ ⎡ ⎤ ⎡ ⎤
⎥⎢–5 ⎥ ⎥–1 ⎥ ⎢⎥ ⎥⎥ ⎥⎢–3⎥⎥ ⎥ ⎥
7
(d) –2 ⎥ 1 ⎥ + 3 ⎥ 0⎣ ⎥ ⎦= ⎥–2 ⎣ ⎥ ⎦+ ⎥ ⎣0 ⎥⎦ = ⎥–2
⎣ ⎦ ⎣⎥ ⎦
ss ss ss ss ss ss ss ss ss ss


–1 –2
ss ss ss ss ss ss ss

–3 –5
ss

1 ss


⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤
⎥ 2/3⎥ 1 ⎢⎢ 3 ⎥⎥ ⎥ 4/3 ⎥ ⎥ 1
ss ss


⎥ 7/3 ⎥ ss s s
s ss




(e) 2 ⎥–1/3⎥ + 3 ⎥
s s ss ⎢ –2⎥⎥ = ⎥–2/3⎥ + ⎥⎤⎥–2/3⎥ = ⎥–4/3⎥ ss
ss
ss ss ss ss s ss ss



⎣⎢ s s⎥⎦ ⎣ ⎣⎢ ⎣⎢ ⎥⎦ ⎢⎣ ⎥⎦ ss s s
ss


2 1 s4 s
1/3 13/3 ss
s
ss



⎦ ⎥⎦
⎡⎤ ⎡, ⎤
⎥–1⎥⎥ ⎢⎡,
s



, 2 ⎤⎥ ⎥⎡–π⎤⎥ ⎢ 2– π⎥
s



, ⎥⎡1⎤⎥ ss s ss



(f) 2⎥1 ⎥+ π ⎥ 0 ⎥ = ⎥ 2⎥⎥ + ⎥ 0⎥ = ⎥
ss
ss s s
, ⎥ ss s s s s ss s s
ss
ss

⎢ , 2⎥⎦
⎣ ⎦ ⎣ ⎢⎣, ⎦ ⎣ ⎣ ss ss s

1 1 2 π 2+ π ss
s
ss s ss


⎦ ⎡ ⎤ ⎦
⎡ ⎤ ⎢⎥ ⎥ ⎡ ⎤
⎢⎥2 ⎥ 6 ⎥ –4 ⎥ ss


A4 (a) 2˜v – 3 w̃ = ⎥ 4 ⎥ – ⎥–3⎥ = ⎥ 7 ⎥
s s ss ss ss ss ss ss ss
ss



⎣ ⎦ ⎣ 9 ⎦ ⎣–13⎦
ss ss ss
ss

–4 ss



⎛⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤ ss
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
4
⎥
⎥ ⎢ ⎥ ⎞ ⎥⎥ ⎥ ⎥⎥
⎥ 1 ⎥ ⎢⎢5⎥⎥ ⎢ ⎥⎥ ⎢⎢–15⎥⎥ ⎥⎥ ⎢⎢–10 ⎥
ss

ss
ss
ss ⎥
5
s s
⎥ ss

5 5
(b) –3(˜v + 2w̃ ) + 5˜v = –3 ⎥⎥ 2 ⎥ + ⎥–2⎥⎥ + ⎥ 10 ⎥ = –3 ⎥0⎥ + ⎥ 10 ⎥ = ⎥ 0
s s ss ss + 10 = 10
ss ss ss ss ss ss ss ss ss ss ss ss ss ss s s s s


⎝⎣ ⎦ ⎣ ⎦ ⎣–10⎦ ⎣ ⎦ ⎣–10⎦ ⎣–12⎦⎥ ⎥⎣–10⎦⎥ ⎥⎣–22⎦⎥
ss ss ss ss ss ss ss ss s s



–2
ss

6 4 ss

⎠
(c) We have w̃ – 2˜u = 3˜v, so 2˜u = w̃ – 3˜v or ˜u = 12( w̃ – 3˜v). This gives
ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss s s s ss ss ss




⎛ ⎡⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤
⎜2 3 ⎟ –1/2 ⎥
ss

⎢ –1
1 ⎥ ⎥ ⎥ ⎥ ⎥⎥⎥ 1 ⎥ ⎥ ⎥
ss




ss⎜⎢⎣ –1⎥⎥ – ⎥⎢⎣6 ⎦⎥⎥
˜u = 2 ⎝⎥⎥ ⎥⎟⎠= 2 ⎢⎥⎣ –7⎥⎥ ⎢⎣–7/2⎥⎥⎦
⎦ =⎥ ss ss ss ss ss ss ss ss


–6 9/2 ss ss

3 ss
9 ss
ss
ss
ss



⎦
⎡ ⎤
–3
(d) We have ˜u – 3˜v = 2˜u, so ˜u = –3˜v = ⎥ ⎥ –6⎥ .
⎣ ⎥
ss ss ss ss ss ss ss ss ss ss ss ss

ss

6 ss

⎡ ⎤ ⎡ ⎤ ⎡ ⎦⎤
⎥ 3/2 ⎥ ⎥ 5/2 ⎥ ⎢⎢ 4 ⎥ s s s s

(a) 1˜v + 1w̃ = ⎥1/2⎥ + ⎥–1/2⎥ = ⎥ 0 ⎥
s s s s
A5 ss s s s ss ss ss

⎢⎣ ⎥⎦
s s ss ss ss s s s s


2 2 ⎢⎣ ⎥⎦ ⎢⎣
1/2 – –1/2 ss




⎥⎦
1 ss

⎡ ⎤ ⎛ ⎡⎤ ⎡⎤ ⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥⎢ 8⎥⎥ ⎜⎥⎥6⎥ ⎥15⎥ ⎟⎟ ⎥ 16 ⎥ ⎥⎢–9⎥⎥ ⎥25 ⎥
(b) 2(˜v + w̃ ) – (2˜v – 3w̃) = 2⎥ 0 ⎥ – ⎥⎥2⎝⎣ – –3 = 0 – 5 = –5
⎣ ⎦ ⎥ ⎦⎥ ⎣⎥⎥⎦⎠ ⎥ ⎣ ⎥⎦ ⎥ ⎥ ⎥⎣ ⎥⎦
s s ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss ss
ss ss ss ss ss ss ss ss ss

–1 2 –6 –2 –10
⎣ ss

8 ss



⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎥ ⎥ ⎥⎢
5 6 ⎥–1 ⎥ s s


(c) We have w̃ – ˜u = 2˜v, so ˜u = w̃ – 2˜v. This gives ˜u = ⎥–1⎥ – ⎥2⎥⎥ = ⎥–3⎥.
s s ss ss ss ss ss ss ss ss ss ss s s ss ss ss ss ss ss
ss
ss
ss
ss



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