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
@LECTSOLUTIONSSTUVIA
, ✐
✐
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
3
1 4 3 3 4 2 4
4
1
2
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
2 3
2
3 1
4
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) 2
3 1 − 2 1/3 = 2/3 − 2/3 = 0 (f) 2 √3 + 3 √ 6 = √ 6 + 3 √ 6 = 4 √ 6
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, 2 Chapter 1 Euclidean Vector Spaces
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢32 –⎥ 1 ⎢ 5= ⎥ 2–5 ⎥ –3
A3 (a) ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 3 – 1 ⎥ = ⎢ ⎢ 2 ⎥⎥
⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦
4 –2
⎡ ⎤
2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢–3 ⎥ ⎢ –1 ⎥
⎢ ⎥ ⎢ 2 + (–3) ⎥⎥
⎢
(b) ⎢ 1 ⎥ + ⎢ 1 ⎥ = ⎢ 1 + 1 ⎥ = ⎢ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣–6 + (–4)⎦ ⎣–10⎦
–6 –4
⎡ ⎤ ⎡ ⎤ ⎡
(–6)4 –24 ⎤
⎢ 4 ⎥ ⎢⎢ ⎥⎥ ⎢
(c) –6 ⎥–5 ⎥ = ⎥(–6)(–5)⎥ = ⎥ 30 ⎥⎥⎥
⎣ ⎦ ⎣(–6)(–6)⎦ ⎣ 36 ⎦
–6
⎡ ⎤ ⎡ ⎤ ⎡ 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 ⎥
(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 = 21 (w̃ – 3˜v). This gives
⎛ ⎡ ⎤ ⎡ ⎞⎤ ⎡ ⎤ ⎡ ⎤
⎜2 3⎟ ⎢ –1 ⎢ –1/2 ⎥
1 ⎜ ⎢ ⎥ ⎢ ⎥⎟⎥ 1 ⎢ ⎥
⎜⎢⎣ –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 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 ⎥
⎣ – ⎥⎥ ⎥ –⎦ ⎥–3⎣ ⎥⎥
⎦ 2⎝⎣ ⎦⎠= ⎥⎣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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, ✐
✐
Section 1.1 Vectors in R2 and R3
⎡ ⎤ ⎡ 2⎤ ⎡ ⎤
⎢ 10 ⎥ ⎢ ⎥ ⎢⎢ 8 ⎥⎥
1 1 1 1 2
(d) We have ˜u2 + ˜v3 = w̃ , so ˜u2 = w̃ – ˜v,3or ˜u = 2 w̃ – ˜v =
3 ⎥–2⎢⎣⎥⎥–⎦ ⎥⎢2/3
⎣ ⎥ ⎥⎦ = ⎥
⎢⎣–8/3 ⎥⎥. ⎦
–4 2/3 –14/3
A6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥ ⎢⎢ 2⎥ ⎢ 1 ⎥
P˜Q = O˜Q – O˜P = ⎥ ⎣1 ⎥⎦– ⎥⎣3⎥ ⎦ = ⎣–2⎥⎦
–2 1 ⎥–3
⎡ 1⎤ ⎡ 2⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢⎢ ⎥ ⎢–1 ⎥
P˜R = O˜R – O˜P = ⎥⎣4⎥⎦ – ⎥⎣3⎥⎦ = ⎥⎣ 1 ⎥⎦
0 1 –1
⎡⎤ ⎡⎤ ⎡–7 ⎤
⎢ –5⎥ ⎢
2⎥
⎥ ⎢ ⎥
P˜S = O˜S – O˜P = ⎥ ⎣1 ⎥⎦– ⎥⎣3⎥ ⎦ =⎥ ⎣ –2⎥⎦
5 1 4
⎡ 1⎤ ⎡ 3 ⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢⎢ ⎥ ⎢–2 ⎥
Q˜R = O˜R – O˜Q = ⎥4⎣⎥⎦– ⎥⎣ 1 ⎥ ⎦ = ⎥⎣ 3 ⎥⎦
0 –2 2
⎡⎤ ⎡⎤ ⎡ ⎤
⎢1⎥ ⎢–5⎥⎥ ⎢6⎥
S˜R = O˜R – O˜S = ⎥4⎣⎥⎦ – ⎥⎣ 1 ⎥⎦ = ⎥⎣ 3 ⎥⎦
0 5 –5
Thus, ⎡ 1⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢–2⎥⎥ ⎢⎢–1⎥ ⎢⎢–7⎥ ⎢⎢ 6 ⎥
P˜Q + Q˜R = ⎥–2 ⎣ –2⎥
⎣ ⎥⎦+ ⎥⎣3 ⎥⎦= ⎥⎣ 1 ⎥⎦ = ⎥ ⎣ 3 ⎦⎥ = P˜S + S˜R
⎦ +⎥
–3 2 –1 4 –5
3 –5
A7 (a) The equation of the line is ˜x = +t ,teR
4 1
2 –4
(b) The equation of the line is ˜x = +t ,teR
3 –6
⎡⎤ ⎡ ⎤
⎢⎢ 2⎥ ⎢ 4 ⎥
(c) The equation of the line is ˜x = ⎥0⎣⎥⎦+ t ⎥⎣ –2 ⎥ ⎦,t e R
5 –11
⎡⎤ ⎡ ⎤
⎢⎢ 4⎥ ⎢ –2 ⎥
(d) The equation of the line is ˜x = ⎥1⎥ + t ⎥ 1 ⎥, t e R
⎣ ⎦ ⎣ ⎦
5 2
A8 Note that alternative correct answers are possible.
(a) The direction vector d˜ of the line is given by the directed line segment joining the two point
–1 3
= . This, along with one of the points, may be used to obtain an equation for the lin
2 –5
–1 3
˜x = +t , teR
2 –5
Copyright ⃝c 2013 Pearson Canada Inc.
, 4 Chapter 1 Euclidean Vector Spaces
(b) The direction vector d˜ of the line is given by the directed line segment joining the two point
4 –6
= . This, along with one of the points, may be used to obtain an equation for the line
1 –2
4 –6
˜x = +t , teR
1 –2
(c) The direction vector d˜ of the line is given by the directed line segment joining the two points: d˜ = ⎥ 1⎣
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢–3⎥
1
⎣⎥ 3⎦⎥ =⎣ ⎥–2
⎦ ⎥. This, along with one of the points, may be used to obtain an equation for the line
–5 5
⎡ ⎤
⎡ ⎤ ⎢⎢–3⎥
⎢⎢ 1 ⎥
˜x = ⎥⎣ 3 ⎥⎦ + t ⎥⎣ –2⎥⎦ , t e R
–5 5
(d) The direction vector d ˜of the line is given by the directed line segment joining the two points: d = ˜⎥2
⎡ ⎤ ⎡ ⎤
⎢–2⎥
6
⎢⎥
⎣⎥ 1 ⎥⎦ = ⎣⎥ ⎦⎥. This, along with one of the points, may be used to obtain an equation for the line
1
1 1
⎡⎤
⎡ ⎤ ⎢⎢ 6 ⎥
⎢–2
⎢ ⎥
˜x = ⎥⎣ 1 ⎥⎦ + t ⎥⎣1 ⎥⎦ , t e R
1 1
⎡
˜ ⎢
(e) The direction vector d of the line is given by the directed line segment joining the two points: d =˜ ⎥ 1
⎣
⎡ ⎤ ⎡ ⎤
1/2 –3/2⎥
⎥ ⎥ ⎢
1/4 = 3/4 . This, along with one of the points, may be used to obtain an equation for the line
⎣⎥ ⎥⎦ ⎥⎣ ⎦⎥
1 –2/3
⎡ ⎤
⎡ ⎤ ⎢–3/2 ⎥
⎢1/2 ⎥
˜x = ⎥ ⎣ 3/4 ⎥⎦ , t e R
⎣ 1/4⎥⎦ + t ⎥
1 –2/3
A9 (a) We
have
x2 = 3x1 + 2
x2 + 1 = 3x1 + 3
x2 + 1 = 3(x1 + 1)
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, ✐
✐
Section 1.1 Vectors in R2 and R3
Let t = x1 + 1. Then, from the equation above we get x2 + 1 = 3t. Solving the equations f
we find that the par amet ric equations are x1 = –1 + t, x2 = –1 + 3t, t e R and the corresponding v
–1 1
equation is ˜x + t , t e R.
–1 3
=
(b) We have
2x1 + 3x2 = 5
2x1 – 2 = –3x2 + 3
2(x1 – 1) = –3(x2 – 1)
1 1
(x1 – 1) = – (x2 – 1)
3 2
Let t = – x2 –1 . Then, 1 (x1 – 1) = t. Solving the equations for x1 and x2 we find that the pa
2 3
equations are x = 1 + 3t, x = 1 – 2t, t e R and the corresponding vector equation is ˜x =1 3+ t
1 2
1 –2
A10 (a) Let P, Q, and R be three points in Rn, with corresponding vectors ˜p, ˜q, and ˜r. If P, Q, and
collinear, then the directed line segments P˜Q and P˜R should define the same line. That
direction vector of one should be a non-zero scalar multiple of the direction vector of the
e
Therefore, P˜Q = tP˜R, for some t R.
4 1 3 –5 1 –6
(b) We have P˜ Q = – = and P˜ R = – = = –2P˜ Q, so they are collinear.
1 2 –1 4 2 2
⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥ ⎢ 1⎥ ⎢⎢ 2 ⎥ ⎢⎢–3⎥⎥ ⎢⎢ 1⎥⎥ ⎢–4 ⎥
(c) We have S T = ⎥–2⎥ – ⎥0⎥ = ⎥–2⎥ and S U = ⎥ 4 ⎥ – ⎥0⎥ = ⎥ 4 ⎥. Therefore, the points S , T , and U a
˜ ˜
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
3 1 2 –1 1 –2
collinear because S˜U ≠ tS˜T for any real number t.
Homework
Problems
(b) 2
x
B1 (a) 3
3 1 0
x
3 1
2 4
1
3
1
4 2 2
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, 6 Chapter 1 Euclidean Vector
Spaces (c) 1 3 (d) 3
6 3 5
2
5
2 3
1 5 3 18
2 3
0 0 –4 ,
B2 (a)
0 (b) (c) (d) , (e) 2 3
,
0 0 –2 27/5 2 – 3 3/2
⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥⎥ ⎢⎢6⎥ ⎡⎤
⎢⎢ 8 ⎥⎥ ⎢⎢0⎥⎥ ⎢7/3 ⎥⎥ ⎢ 0⎥
B3 (a) –3 (b) 0 (c) –20 (d) 0 (e) –4/3 (f) 0
⎥ ⎣ ⎥⎦ ⎥⎣ ⎥⎦ ⎣⎥ ⎥⎦ ⎥⎣ ⎥⎦ ⎣⎥ ⎦⎥
13/3 ⎣⎥ ⎦⎥
–6 0 –4 0 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
21 ⎢ –1 ⎥ ⎢ –7 ⎥ ⎢ 9 ⎥
B4 (a)⎢⎢ 4 ⎥ (b) ⎢–12⎥ (c) ⎢–13/2⎥ (d) ⎢–1/2⎥
⎣⎥ ⎦⎥ ⎣
1
⎦ ⎣ ⎦ ⎣ ⎦
–5 2 –2
⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤
⎢–2 ⎥ ⎢⎢ 0 ⎥⎥ ⎢0⎥⎥ ⎢ 4 ⎥
B5 (a) ⎢ ⎥0 (b) 1/4 (c) ˜u = 0 (d) ˜u = –7/2
⎥⎣ ⎥⎦ ⎣⎥ ⎦ ⎣⎥ ⎦⎥ ⎣⎥ ⎦⎥
4 –3/4 ⎥ 0 5/2
⎡⎤
⎡ ⎤ ⎢–2⎥ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 3 ⎥⎥ ˜ ˜
⎢⎢ 7 ⎥⎥
PR = 0 , P S = 2 , QR = 1 , S˜R = ⎢ –2⎥ ˜ ⎢ ⎢–5⎥ –9
B6 (a) P˜Q = ⎢–1⎥,
⎣ ⎦
–2
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎥ ⎥⎦
1 –6 3 7
⎡ 1⎤ ⎡⎤ ⎡ 5⎤ ⎡ ⎤ ⎡ ⎤
– 0 1⎥ ⎢ 5⎥
P˜Q = ⎥9 ⎥⎥, ˜R = 3 ,
⎢–
P˜S = ⎥6 ⎥⎥, Q˜R = ⎢–6⎥ , ˜
(b) P S R = ⎢–3
⎣⎢ ⎥⎦
–4 ⎣⎢ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎢ ⎥ ⎦ ⎣⎢ ⎥⎦
4 –2 8 6
⎡⎤
–3 ⎢ 1⎥
(a) ˜x = + t 4 , t e R. (b) ˜x = t 3 , t e R.
⎥⎥
⎡4 –3 ⎣ ⎦
1
B7 2 ⎤⎥ ⎡2⎤ ⎡3⎤ ⎡ 2⎤
⎢ ⎥ ⎢⎢ ⎥⎥
(c) ˜x = ⎥ 3 ⎥ + t ⎥–4⎥, t e R. (d) ˜x = ⎥1⎥ + t ⎥⎢ –3⎥⎥ , t e R.
⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–1 8 2 2
⎡ ⎤
–1
(a) ˜x = 3 + t –2 , t e R (b) ˜x = t ⎢ ⎣2⎢ ⎥ , ⎦⎥ teR
1 1 –1
B8 ⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤
⎤
⎢ 2⎥ ⎢–3⎥ 1 ⎢–1/2⎥
⎢ ⎥
(c) ˜x = ⎥⎣–6⎥⎦ + t ⎥⎣11⎥⎦ , t e R (d) ˜x = ⎥⎣–1 ⎥⎦ + t ⎥⎣ 4/3 ⎥⎦ , t e R
3 –1 1/2 1/2
B9 Other correct answers are possible.
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✐
Section 1.1 Vectors in R2 and R3
3/2 –1/2
(a) x1 = – 12 t + 32 , x 2 = t; ˜x =+t
0 1
0 1
(b) x = t, x = – 1 t + 3 ; ˜x = +t
1 2 2 2 3/2 –1/2
⎡ ⎤
2
B10 (a) Since –2P˜Q = ⎢ –2⎥ = P˜R, the point P, Q, and R must be collinear.
⎥
⎣ ⎦⎥
–4
⎡ ⎤
–5
(b) Since –S˜T = ⎥ –1⎥ = S˜U, the point S , T , and U must be collinear.
⎥
⎣ ⎦⎥
–1
Computer Problems
⎡ ⎤
–2322 ⎡ 0⎤
⎢⎢
C1 (a) –1761 ⎥ (b) 0
⎣⎥ ⎥⎦ ⎣⎥ ⎦⎥
1667 0
Conceptual
Problems
13 1 1 t +t
D1 (a) We need to find t1 and t2 such that =t +t = 1 2.
1 2
–12 1 –1 t1 – t2
That is, we need to solve the two equations in two unknowns t1 + t2 = 13 and
– t1 –t2 =
substitution and/or elimination we find that t1 = 1 and t2 = 25 .
2 2
1
1
x1
1
13
(b) We use the same approach as in part (a). We need to find t1 and t2 such that
x1 1 1 t1 + t2
= t1 + t2 =
x2 1 –1 t1 – t2
Solving t1 + t2 = x1 and t1 – t2 = x2 by substitution and/or elimination gives t1 2= 1 (x1 + x2)
1
2
(x1 – x2).
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, 8 Chapter 1 Euclidean Vector Spaces
, , ,
(c) We have x1 = 2 and x2 = π, so we get t12= 1 ( 2 + π) and t22 = 1 ( 2 – π).
D2 (a) P˜Q + Q˜R + R˜P can be described informally as “start at P and move to Q, then move from
R, then from R to P; the net result is a zero change in position.”
(b) We have P˜Q = ˜q – ˜p, Q˜R = ˜r – ˜q, and R˜P = ˜p – ˜r. Thus,
P˜Q + Q˜R + R˜P = ˜q – ˜p +˜r – ˜q + ˜p – ˜r = ˜0
D3 Assu m e that ˜x = ˜p + td˜, t e R, is a line in R2 passing through the origin. Then, there exists a real nu
0
that = ˜p + t1d˜. Hence, ˜p = –t1d˜ and so ˜p is a scalar multiple of d˜. On the other hand, assume that ˜
0
scalar multiple of d˜. Then, there exists a real number t1 such that ˜p = t1d˜. Hence, if we take t = – t
0
the line with vector equation ˜x = ˜p + td˜ passes through the point ˜p + (–t )d1˜ = t d˜1 – t d˜1 = ˜0 = as requi
0
⎡ ⎤ ⎡ y⎤1
⎢⎢x1⎥ ⎢ ⎥
D4 Let ˜x = ⎣ x2⎦ and ˜y = ⎣ y2⎦ . Then,
⎥x3 ⎥ ⎥y3 ⎥
⎡ ⎤
⎡ x1 + ⎤ ⎡ ⎤ ⎡ tx1 + ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢y1⎥
y1
⎢ ⎥ t(x
⎢ 1 1⎥ + y ) ⎥ ⎥ ⎢ x ⎥
1⎥ ⎢y
⎢ 1⎥ ⎥ ⎢x ⎢ 1⎥
⎢ ty1
t(˜x + ˜y) =⎣⎥t x 2 + y⎥⎦2 =⎥⎣ t(x 2 + y 2⎥) =⎥⎣ tx 2 + ty 2⎥ = t ⎥⎣x 2⎦⎥ + t ⎥⎣y 2 ⎦⎥= t ⎥⎣x 2 ⎦⎥+ t y⎥⎣ 2 ⎦⎥= t ˜x + t˜y
x +y 3 t(x +
3 3 tx + 3 x 3 y 3 x 3 3 y
⎦ ⎦
y3) ty3
1.2 Vectors in Rn
Practice Problems
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤
⎢1⎥ ⎢ 2 ⎥⎥ ⎢⎢ 1 ⎥ ⎢ 4 ⎥ ⎢ 5⎥
A1 (a) ⎢ 3 ⎥ + 2 ⎢ 3 ⎥ = ⎢ 3 ⎥ + ⎢ 6 ⎥ = ⎢ ⎥ 9
1 ⎢ 2⎥ 2
⎣ – ⎥⎦ ⎣–1 ⎦ ⎣ – ⎥⎦ ⎣1 ⎦
⎢⎣ ⎥2⎦ ⎥ ⎥ ⎢ 0⎥
–1 1 2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 1⎥ ⎢–1 ⎥⎥ ⎢⎢ 3 ⎥ ⎢ 1 ⎥ ⎢–3⎥
⎡ ⎤ ⎡ ⎤
⎢6 ⎥⎥ ⎢10 ⎥
–2 1 –1 –2 3
(b) ⎢ ⎥ – 3 ⎢ ⎥ + 2 ⎢ ⎥ = ⎢ 4⎥ – ⎢ ⎥ + ⎢ ⎥ = ⎢ 3⎥–2 –7
⎢⎣ 5⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ 5⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ 8⎥⎦ ⎢⎣10⎥⎦
1 2 0 1 6 0 –5
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤
1 2 ⎢ 2⎥⎥ ⎢⎢ 2⎥ ⎢ 4⎥⎥ 6⎥⎥ 0
⎢ ⎥⎥ ⎥–2 ⎥ ⎢ ⎢ ⎢
2 ⎢ ⎥ 0 4 –4 0
⎢ ⎥ 0
⎢⎥
⎢ ⎥
(c) 2 ⎥⎥1 ⎥⎥+ 2 ⎥ 1 ⎥ – 3 ⎥⎥1 ⎥⎥= ⎥ 2 ⎥ + ⎥⎥ 2 ⎥⎥ – ⎥ 3 ⎥ = ⎥1⎥
⎢⎣ 0 ⎥⎦ ⎢⎣ 2⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 4⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣1⎥⎦
–1 1 –1 –2 2 –3 3
A2 (a) Since the condition of the set contains the square of a variable in it, we suspect that it is no
subspace. To prove it is not a subspace we just need to find one example where the set is n
closed under addition.
Copyright ⃝c 2013 Pearson Canada Inc.
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
@LECTSOLUTIONSSTUVIA
, ✐
✐
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
3
1 4 3 3 4 2 4
4
1
2
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
2 3
2
3 1
4
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) 2
3 1 − 2 1/3 = 2/3 − 2/3 = 0 (f) 2 √3 + 3 √ 6 = √ 6 + 3 √ 6 = 4 √ 6
Copyright ⃝c 2013 Pearson Canada Inc.
, 2 Chapter 1 Euclidean Vector Spaces
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢32 –⎥ 1 ⎢ 5= ⎥ 2–5 ⎥ –3
A3 (a) ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 3 – 1 ⎥ = ⎢ ⎢ 2 ⎥⎥
⎣ ⎦ ⎣ ⎦ ⎣4 – (–2)⎦ ⎣ 6 ⎦
4 –2
⎡ ⎤
2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢–3 ⎥ ⎢ –1 ⎥
⎢ ⎥ ⎢ 2 + (–3) ⎥⎥
⎢
(b) ⎢ 1 ⎥ + ⎢ 1 ⎥ = ⎢ 1 + 1 ⎥ = ⎢ 2 ⎥
⎣ ⎦ ⎣ ⎦ ⎣–6 + (–4)⎦ ⎣–10⎦
–6 –4
⎡ ⎤ ⎡ ⎤ ⎡
(–6)4 –24 ⎤
⎢ 4 ⎥ ⎢⎢ ⎥⎥ ⎢
(c) –6 ⎥–5 ⎥ = ⎥(–6)(–5)⎥ = ⎥ 30 ⎥⎥⎥
⎣ ⎦ ⎣(–6)(–6)⎦ ⎣ 36 ⎦
–6
⎡ ⎤ ⎡ ⎤ ⎡ 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 ⎥
(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 = 21 (w̃ – 3˜v). This gives
⎛ ⎡ ⎤ ⎡ ⎞⎤ ⎡ ⎤ ⎡ ⎤
⎜2 3⎟ ⎢ –1 ⎢ –1/2 ⎥
1 ⎜ ⎢ ⎥ ⎢ ⎥⎟⎥ 1 ⎢ ⎥
⎜⎢⎣ –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 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 ⎥
⎣ – ⎥⎥ ⎥ –⎦ ⎥–3⎣ ⎥⎥
⎦ 2⎝⎣ ⎦⎠= ⎥⎣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
Copyright ⃝c 2013 Pearson Canada Inc.
, ✐
✐
Section 1.1 Vectors in R2 and R3
⎡ ⎤ ⎡ 2⎤ ⎡ ⎤
⎢ 10 ⎥ ⎢ ⎥ ⎢⎢ 8 ⎥⎥
1 1 1 1 2
(d) We have ˜u2 + ˜v3 = w̃ , so ˜u2 = w̃ – ˜v,3or ˜u = 2 w̃ – ˜v =
3 ⎥–2⎢⎣⎥⎥–⎦ ⎥⎢2/3
⎣ ⎥ ⎥⎦ = ⎥
⎢⎣–8/3 ⎥⎥. ⎦
–4 2/3 –14/3
A6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥ ⎢⎢ 2⎥ ⎢ 1 ⎥
P˜Q = O˜Q – O˜P = ⎥ ⎣1 ⎥⎦– ⎥⎣3⎥ ⎦ = ⎣–2⎥⎦
–2 1 ⎥–3
⎡ 1⎤ ⎡ 2⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢⎢ ⎥ ⎢–1 ⎥
P˜R = O˜R – O˜P = ⎥⎣4⎥⎦ – ⎥⎣3⎥⎦ = ⎥⎣ 1 ⎥⎦
0 1 –1
⎡⎤ ⎡⎤ ⎡–7 ⎤
⎢ –5⎥ ⎢
2⎥
⎥ ⎢ ⎥
P˜S = O˜S – O˜P = ⎥ ⎣1 ⎥⎦– ⎥⎣3⎥ ⎦ =⎥ ⎣ –2⎥⎦
5 1 4
⎡ 1⎤ ⎡ 3 ⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢⎢ ⎥ ⎢–2 ⎥
Q˜R = O˜R – O˜Q = ⎥4⎣⎥⎦– ⎥⎣ 1 ⎥ ⎦ = ⎥⎣ 3 ⎥⎦
0 –2 2
⎡⎤ ⎡⎤ ⎡ ⎤
⎢1⎥ ⎢–5⎥⎥ ⎢6⎥
S˜R = O˜R – O˜S = ⎥4⎣⎥⎦ – ⎥⎣ 1 ⎥⎦ = ⎥⎣ 3 ⎥⎦
0 5 –5
Thus, ⎡ 1⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ ⎥ ⎢–2⎥⎥ ⎢⎢–1⎥ ⎢⎢–7⎥ ⎢⎢ 6 ⎥
P˜Q + Q˜R = ⎥–2 ⎣ –2⎥
⎣ ⎥⎦+ ⎥⎣3 ⎥⎦= ⎥⎣ 1 ⎥⎦ = ⎥ ⎣ 3 ⎦⎥ = P˜S + S˜R
⎦ +⎥
–3 2 –1 4 –5
3 –5
A7 (a) The equation of the line is ˜x = +t ,teR
4 1
2 –4
(b) The equation of the line is ˜x = +t ,teR
3 –6
⎡⎤ ⎡ ⎤
⎢⎢ 2⎥ ⎢ 4 ⎥
(c) The equation of the line is ˜x = ⎥0⎣⎥⎦+ t ⎥⎣ –2 ⎥ ⎦,t e R
5 –11
⎡⎤ ⎡ ⎤
⎢⎢ 4⎥ ⎢ –2 ⎥
(d) The equation of the line is ˜x = ⎥1⎥ + t ⎥ 1 ⎥, t e R
⎣ ⎦ ⎣ ⎦
5 2
A8 Note that alternative correct answers are possible.
(a) The direction vector d˜ of the line is given by the directed line segment joining the two point
–1 3
= . This, along with one of the points, may be used to obtain an equation for the lin
2 –5
–1 3
˜x = +t , teR
2 –5
Copyright ⃝c 2013 Pearson Canada Inc.
, 4 Chapter 1 Euclidean Vector Spaces
(b) The direction vector d˜ of the line is given by the directed line segment joining the two point
4 –6
= . This, along with one of the points, may be used to obtain an equation for the line
1 –2
4 –6
˜x = +t , teR
1 –2
(c) The direction vector d˜ of the line is given by the directed line segment joining the two points: d˜ = ⎥ 1⎣
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢–3⎥
1
⎣⎥ 3⎦⎥ =⎣ ⎥–2
⎦ ⎥. This, along with one of the points, may be used to obtain an equation for the line
–5 5
⎡ ⎤
⎡ ⎤ ⎢⎢–3⎥
⎢⎢ 1 ⎥
˜x = ⎥⎣ 3 ⎥⎦ + t ⎥⎣ –2⎥⎦ , t e R
–5 5
(d) The direction vector d ˜of the line is given by the directed line segment joining the two points: d = ˜⎥2
⎡ ⎤ ⎡ ⎤
⎢–2⎥
6
⎢⎥
⎣⎥ 1 ⎥⎦ = ⎣⎥ ⎦⎥. This, along with one of the points, may be used to obtain an equation for the line
1
1 1
⎡⎤
⎡ ⎤ ⎢⎢ 6 ⎥
⎢–2
⎢ ⎥
˜x = ⎥⎣ 1 ⎥⎦ + t ⎥⎣1 ⎥⎦ , t e R
1 1
⎡
˜ ⎢
(e) The direction vector d of the line is given by the directed line segment joining the two points: d =˜ ⎥ 1
⎣
⎡ ⎤ ⎡ ⎤
1/2 –3/2⎥
⎥ ⎥ ⎢
1/4 = 3/4 . This, along with one of the points, may be used to obtain an equation for the line
⎣⎥ ⎥⎦ ⎥⎣ ⎦⎥
1 –2/3
⎡ ⎤
⎡ ⎤ ⎢–3/2 ⎥
⎢1/2 ⎥
˜x = ⎥ ⎣ 3/4 ⎥⎦ , t e R
⎣ 1/4⎥⎦ + t ⎥
1 –2/3
A9 (a) We
have
x2 = 3x1 + 2
x2 + 1 = 3x1 + 3
x2 + 1 = 3(x1 + 1)
Copyright ⃝c 2013 Pearson Canada Inc.
, ✐
✐
Section 1.1 Vectors in R2 and R3
Let t = x1 + 1. Then, from the equation above we get x2 + 1 = 3t. Solving the equations f
we find that the par amet ric equations are x1 = –1 + t, x2 = –1 + 3t, t e R and the corresponding v
–1 1
equation is ˜x + t , t e R.
–1 3
=
(b) We have
2x1 + 3x2 = 5
2x1 – 2 = –3x2 + 3
2(x1 – 1) = –3(x2 – 1)
1 1
(x1 – 1) = – (x2 – 1)
3 2
Let t = – x2 –1 . Then, 1 (x1 – 1) = t. Solving the equations for x1 and x2 we find that the pa
2 3
equations are x = 1 + 3t, x = 1 – 2t, t e R and the corresponding vector equation is ˜x =1 3+ t
1 2
1 –2
A10 (a) Let P, Q, and R be three points in Rn, with corresponding vectors ˜p, ˜q, and ˜r. If P, Q, and
collinear, then the directed line segments P˜Q and P˜R should define the same line. That
direction vector of one should be a non-zero scalar multiple of the direction vector of the
e
Therefore, P˜Q = tP˜R, for some t R.
4 1 3 –5 1 –6
(b) We have P˜ Q = – = and P˜ R = – = = –2P˜ Q, so they are collinear.
1 2 –1 4 2 2
⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥ ⎢ 1⎥ ⎢⎢ 2 ⎥ ⎢⎢–3⎥⎥ ⎢⎢ 1⎥⎥ ⎢–4 ⎥
(c) We have S T = ⎥–2⎥ – ⎥0⎥ = ⎥–2⎥ and S U = ⎥ 4 ⎥ – ⎥0⎥ = ⎥ 4 ⎥. Therefore, the points S , T , and U a
˜ ˜
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
3 1 2 –1 1 –2
collinear because S˜U ≠ tS˜T for any real number t.
Homework
Problems
(b) 2
x
B1 (a) 3
3 1 0
x
3 1
2 4
1
3
1
4 2 2
Copyright ⃝c 2013 Pearson Canada Inc.
, 6 Chapter 1 Euclidean Vector
Spaces (c) 1 3 (d) 3
6 3 5
2
5
2 3
1 5 3 18
2 3
0 0 –4 ,
B2 (a)
0 (b) (c) (d) , (e) 2 3
,
0 0 –2 27/5 2 – 3 3/2
⎡ ⎤ ⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢ 3 ⎥⎥ ⎢⎢6⎥ ⎡⎤
⎢⎢ 8 ⎥⎥ ⎢⎢0⎥⎥ ⎢7/3 ⎥⎥ ⎢ 0⎥
B3 (a) –3 (b) 0 (c) –20 (d) 0 (e) –4/3 (f) 0
⎥ ⎣ ⎥⎦ ⎥⎣ ⎥⎦ ⎣⎥ ⎥⎦ ⎥⎣ ⎥⎦ ⎣⎥ ⎦⎥
13/3 ⎣⎥ ⎦⎥
–6 0 –4 0 0
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
21 ⎢ –1 ⎥ ⎢ –7 ⎥ ⎢ 9 ⎥
B4 (a)⎢⎢ 4 ⎥ (b) ⎢–12⎥ (c) ⎢–13/2⎥ (d) ⎢–1/2⎥
⎣⎥ ⎦⎥ ⎣
1
⎦ ⎣ ⎦ ⎣ ⎦
–5 2 –2
⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤
⎢–2 ⎥ ⎢⎢ 0 ⎥⎥ ⎢0⎥⎥ ⎢ 4 ⎥
B5 (a) ⎢ ⎥0 (b) 1/4 (c) ˜u = 0 (d) ˜u = –7/2
⎥⎣ ⎥⎦ ⎣⎥ ⎦ ⎣⎥ ⎦⎥ ⎣⎥ ⎦⎥
4 –3/4 ⎥ 0 5/2
⎡⎤
⎡ ⎤ ⎢–2⎥ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 3 ⎥⎥ ˜ ˜
⎢⎢ 7 ⎥⎥
PR = 0 , P S = 2 , QR = 1 , S˜R = ⎢ –2⎥ ˜ ⎢ ⎢–5⎥ –9
B6 (a) P˜Q = ⎢–1⎥,
⎣ ⎦
–2
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎥ ⎥⎦
1 –6 3 7
⎡ 1⎤ ⎡⎤ ⎡ 5⎤ ⎡ ⎤ ⎡ ⎤
– 0 1⎥ ⎢ 5⎥
P˜Q = ⎥9 ⎥⎥, ˜R = 3 ,
⎢–
P˜S = ⎥6 ⎥⎥, Q˜R = ⎢–6⎥ , ˜
(b) P S R = ⎢–3
⎣⎢ ⎥⎦
–4 ⎣⎢ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎢ ⎥ ⎦ ⎣⎢ ⎥⎦
4 –2 8 6
⎡⎤
–3 ⎢ 1⎥
(a) ˜x = + t 4 , t e R. (b) ˜x = t 3 , t e R.
⎥⎥
⎡4 –3 ⎣ ⎦
1
B7 2 ⎤⎥ ⎡2⎤ ⎡3⎤ ⎡ 2⎤
⎢ ⎥ ⎢⎢ ⎥⎥
(c) ˜x = ⎥ 3 ⎥ + t ⎥–4⎥, t e R. (d) ˜x = ⎥1⎥ + t ⎥⎢ –3⎥⎥ , t e R.
⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
–1 8 2 2
⎡ ⎤
–1
(a) ˜x = 3 + t –2 , t e R (b) ˜x = t ⎢ ⎣2⎢ ⎥ , ⎦⎥ teR
1 1 –1
B8 ⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤
⎤
⎢ 2⎥ ⎢–3⎥ 1 ⎢–1/2⎥
⎢ ⎥
(c) ˜x = ⎥⎣–6⎥⎦ + t ⎥⎣11⎥⎦ , t e R (d) ˜x = ⎥⎣–1 ⎥⎦ + t ⎥⎣ 4/3 ⎥⎦ , t e R
3 –1 1/2 1/2
B9 Other correct answers are possible.
Copyright ⃝c 2013 Pearson Canada Inc.
, ✐
✐
Section 1.1 Vectors in R2 and R3
3/2 –1/2
(a) x1 = – 12 t + 32 , x 2 = t; ˜x =+t
0 1
0 1
(b) x = t, x = – 1 t + 3 ; ˜x = +t
1 2 2 2 3/2 –1/2
⎡ ⎤
2
B10 (a) Since –2P˜Q = ⎢ –2⎥ = P˜R, the point P, Q, and R must be collinear.
⎥
⎣ ⎦⎥
–4
⎡ ⎤
–5
(b) Since –S˜T = ⎥ –1⎥ = S˜U, the point S , T , and U must be collinear.
⎥
⎣ ⎦⎥
–1
Computer Problems
⎡ ⎤
–2322 ⎡ 0⎤
⎢⎢
C1 (a) –1761 ⎥ (b) 0
⎣⎥ ⎥⎦ ⎣⎥ ⎦⎥
1667 0
Conceptual
Problems
13 1 1 t +t
D1 (a) We need to find t1 and t2 such that =t +t = 1 2.
1 2
–12 1 –1 t1 – t2
That is, we need to solve the two equations in two unknowns t1 + t2 = 13 and
– t1 –t2 =
substitution and/or elimination we find that t1 = 1 and t2 = 25 .
2 2
1
1
x1
1
13
(b) We use the same approach as in part (a). We need to find t1 and t2 such that
x1 1 1 t1 + t2
= t1 + t2 =
x2 1 –1 t1 – t2
Solving t1 + t2 = x1 and t1 – t2 = x2 by substitution and/or elimination gives t1 2= 1 (x1 + x2)
1
2
(x1 – x2).
Copyright ⃝c 2013 Pearson Canada Inc.
, 8 Chapter 1 Euclidean Vector Spaces
, , ,
(c) We have x1 = 2 and x2 = π, so we get t12= 1 ( 2 + π) and t22 = 1 ( 2 – π).
D2 (a) P˜Q + Q˜R + R˜P can be described informally as “start at P and move to Q, then move from
R, then from R to P; the net result is a zero change in position.”
(b) We have P˜Q = ˜q – ˜p, Q˜R = ˜r – ˜q, and R˜P = ˜p – ˜r. Thus,
P˜Q + Q˜R + R˜P = ˜q – ˜p +˜r – ˜q + ˜p – ˜r = ˜0
D3 Assu m e that ˜x = ˜p + td˜, t e R, is a line in R2 passing through the origin. Then, there exists a real nu
0
that = ˜p + t1d˜. Hence, ˜p = –t1d˜ and so ˜p is a scalar multiple of d˜. On the other hand, assume that ˜
0
scalar multiple of d˜. Then, there exists a real number t1 such that ˜p = t1d˜. Hence, if we take t = – t
0
the line with vector equation ˜x = ˜p + td˜ passes through the point ˜p + (–t )d1˜ = t d˜1 – t d˜1 = ˜0 = as requi
0
⎡ ⎤ ⎡ y⎤1
⎢⎢x1⎥ ⎢ ⎥
D4 Let ˜x = ⎣ x2⎦ and ˜y = ⎣ y2⎦ . Then,
⎥x3 ⎥ ⎥y3 ⎥
⎡ ⎤
⎡ x1 + ⎤ ⎡ ⎤ ⎡ tx1 + ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢y1⎥
y1
⎢ ⎥ t(x
⎢ 1 1⎥ + y ) ⎥ ⎥ ⎢ x ⎥
1⎥ ⎢y
⎢ 1⎥ ⎥ ⎢x ⎢ 1⎥
⎢ ty1
t(˜x + ˜y) =⎣⎥t x 2 + y⎥⎦2 =⎥⎣ t(x 2 + y 2⎥) =⎥⎣ tx 2 + ty 2⎥ = t ⎥⎣x 2⎦⎥ + t ⎥⎣y 2 ⎦⎥= t ⎥⎣x 2 ⎦⎥+ t y⎥⎣ 2 ⎦⎥= t ˜x + t˜y
x +y 3 t(x +
3 3 tx + 3 x 3 y 3 x 3 3 y
⎦ ⎦
y3) ty3
1.2 Vectors in Rn
Practice Problems
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤
⎢1⎥ ⎢ 2 ⎥⎥ ⎢⎢ 1 ⎥ ⎢ 4 ⎥ ⎢ 5⎥
A1 (a) ⎢ 3 ⎥ + 2 ⎢ 3 ⎥ = ⎢ 3 ⎥ + ⎢ 6 ⎥ = ⎢ ⎥ 9
1 ⎢ 2⎥ 2
⎣ – ⎥⎦ ⎣–1 ⎦ ⎣ – ⎥⎦ ⎣1 ⎦
⎢⎣ ⎥2⎦ ⎥ ⎥ ⎢ 0⎥
–1 1 2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ 1⎥ ⎢–1 ⎥⎥ ⎢⎢ 3 ⎥ ⎢ 1 ⎥ ⎢–3⎥
⎡ ⎤ ⎡ ⎤
⎢6 ⎥⎥ ⎢10 ⎥
–2 1 –1 –2 3
(b) ⎢ ⎥ – 3 ⎢ ⎥ + 2 ⎢ ⎥ = ⎢ 4⎥ – ⎢ ⎥ + ⎢ ⎥ = ⎢ 3⎥–2 –7
⎢⎣ 5⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ 5⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ 8⎥⎦ ⎢⎣10⎥⎦
1 2 0 1 6 0 –5
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎤
1 2 ⎢ 2⎥⎥ ⎢⎢ 2⎥ ⎢ 4⎥⎥ 6⎥⎥ 0
⎢ ⎥⎥ ⎥–2 ⎥ ⎢ ⎢ ⎢
2 ⎢ ⎥ 0 4 –4 0
⎢ ⎥ 0
⎢⎥
⎢ ⎥
(c) 2 ⎥⎥1 ⎥⎥+ 2 ⎥ 1 ⎥ – 3 ⎥⎥1 ⎥⎥= ⎥ 2 ⎥ + ⎥⎥ 2 ⎥⎥ – ⎥ 3 ⎥ = ⎥1⎥
⎢⎣ 0 ⎥⎦ ⎢⎣ 2⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 4⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣1⎥⎦
–1 1 –1 –2 2 –3 3
A2 (a) Since the condition of the set contains the square of a variable in it, we suspect that it is no
subspace. To prove it is not a subspace we just need to find one example where the set is n
closed under addition.
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