INSTRUCTOR’S SOLUTIONS MANUAL
An Introduction to Linear Algebra for Science and Engineering Third Edition
Daniel Norman
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Dan Wolczuk
Chapter 1 Solutions Section 1.1 A Practice Problems
" # " # " # " # 1 2 1+2 3 A1 + = = 4 3 4+3 7 x2 1 2 + 1 4 3 4 2 3
x1 " # " # " # −1 3(−1) −3 A3 3 = = 4 3(4) 12 x 2 −1 3 4
A2
" # " # " # " # 3 4 3−4 −1 − = = 2 1 2−1 1 x2 3 4 − 2 1
−
x1 # " # " # " # 4 −1 4 + (−1) 3 A5 + = = −2 3 −2 + 3 1 " # " # " # 3 (−2)3 −6 A7 −2 = = −2 (−2)(−2) 4
"
4 1 x1
4 1
" # " # " # " # " # 2 3 4 6 −2 A4 2 −2 = − = 1 −1 2 −2 4 x2 2 3 −2 2 −1 1 −2
−1 4
3 2
3 −1
2 1
2
2 1
x1 3 −1 " # " # " # " # −3 −2 −3 − (−2) −1 A6 − = = −4 5 −4 − 5 −9 " # " # " # " # " # 2 4 1 4/3 7/3 A8 12 + 13 = + = 6 3 3 1 4
1
2
" # " # " # " # " # 3 1/4 2 1/2 3/2 A9 −2 = − = 1 1/3 2/3 2/3 0 2 5 2 − 5 −3 A11 3 − 1 = 3 − 1 = 2 4 −2 4 − (−2) 6 4 (−6)4 −24 A13 −6 −5 = (−6)(−5) = 30 −6 (−6)(−6) 36 2/3 3 7/3 A15 2 −1/3 + 13 −2 = −4/3 2 1 13/3 2 3
"√ # " # " # " # " # √ 1 2 3 5 2 √ √ √ √ A10 2 √ + 3 = + = 6 6 3 6 4 6 3 2 −3 2 + (−3) −1 A12 1 + 1 = 1 + 1 = 2 −6 −4 −6 + (−4) −10 −5 −1 10 −3 7 A14 −2 1 + 3 0 = −2 + 0 = −2 1 −1 −2 −3 −5 √ √ −1 √2 −π 2√− π √ 1 A16 2 1 + π 0 = 2 + 0 = 2 √ √ 1 1 π 2 2+π
2 6 −4 w = 4 − −3 = 7 A17 (a) 2~v − 3~ −4 9 −13 1 4 5 5 5 −15 5 −10 (b) −3(~v + 2~ w) + 5~v = −3 2 + −2 + 10 = −3 0 + 10 = 0 + 10 = 10 −2 6 −10 4 −10 −12 −10 −22 ~ − 2~u = 3~v , so 2~u = w ~ − 3~v or ~u = 12 (~ (c) We have w w − 3~v ). This gives
2 3 −1 −1/2 1 1 ~u = −1 − 6 = −7 = −7/2 2 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 ~ = 1/2 + −1/2 = 0 A18 (a) 21 ~v + 12 w 1/2 −1 −1/2 8 6 15 16 −9 25 ~ ) − (2~v − 3~ w) = 2 0 − 2 − −3 = 0 − 5 = −5 (b) 2(~v + w −1 2 −6 −2 8 −10 5 6 −1 ~ − ~u = 2~v , so ~u = w ~ − 2~v . This gives ~u = −1 − 2 = −3. (c) We have w −2 2 −4 10 2 8 ~ , so 12 ~u = w ~ − 13 ~v , or ~u = 2~ (d) We have 12 ~u + 13 ~v = w w − 23 ~v = −2 − 2/3 = − 8/3. −4 2/3 −14/3
3
A19 3 2 1 ~ = OQ ~ − OP ~ = 1 − 3 = −2 PQ −2 1 −3 −5 2 −7 ~ = OS ~ − OP ~ = 1 − 3 = −2 PS 5 1 4 1 −5 6 ~ ~ ~ S R = OR − OS = 4 − 1 = 3 0 5 −5
1 2 −1 ~ = OR ~ − OP ~ = 4 − 3 = 1 PR 0 1 −1 1 3 −2 ~ = OR ~ − OQ ~ = 4 − 1 = 3 QR 0 −2 2
Thus, 1 −2 −1 −7 6 ~ + QR ~ = −2 + 3 = 1 = −2 + 3 = PS ~ + S~R PQ −3 2 −1 4 −5 " # " # 3 −5 A20 The equation of the line is ~x = +t ,t∈R 4 1 A21 The equation of the line is ~x =
" # " # 2 −4 +t ,t∈R 3 −6
2 4 A22 The equation of the line is ~x = 0 + t −2, t ∈ R 5 −11 4 −2 A23 The equation of the line is ~x = 1 + t 1, t ∈ R 5 2
For Problems A24 - A28, alternative correct answers are possible.
~ A24 The direction " # " vector # "d of# the line is given by the directed line segment joining the two points: 2 −1 3 d~ = − = . This, along with one of the points, may be used to obtain an equation for −3 2 −5 the line " # " # −1 3 ~x = +t , t∈R 2 −5 ~ A25 The "direction # " #vector " d# of the line is given by the directed line segment joining the two points: −2 4 −6 d~ = − = . This, along with one of the points, may be used to obtain an equation for the −1 1 −2 line " # " # 4 −6 ~x = +t , t∈R 1 −2
4 A26 The direction vector d~ of the line is given by the directed line segment joining the two points: −2 1 −3 d~ = 1 − 3 = −2. This, along with one of the points, may be used to obtain an equation for 0 −5 5 the line 1 −3 ~x = 3 + t −2 , t ∈ R −5 5 A27 The direction vector d~ of the line is given by the directed line segment joining the two points: 4 −2 6 d~ = 2 − 1 = 1. This, along with one of the points, may be used to obtain an equation for the 2 1 1 line −2 6 ~x = 1 + t 1 , t ∈ R 1 1
A28 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 − 1/4 = 3/4. This, along with one of the points, may be used to obtain an equation for the 1/3 1 −2/3 line 1/2 −3/2 ~x = 1/4 + t 3/4 , t ∈ R 1 −2/3
~ the line is given by the directed line segment joining the two points: A29 The "direction # " vector # " d of # 2 −1 3 d~ = − = . −3 2 −5 x1 = −1 + 3t Hence, the parametric equation of the line is t ∈ R. x2 = 2 − 5t, 5 1 A scalar equation is x2 = 2 + −5 3 (x1 − (−1)) = − 3 x1 + 3 . " # " # " # 2 1 1 ~ A30 The direction vector is d = − = . 2 1 1 x1 = 1 + t Hence, the parametric equation of the line is t ∈ R. x2 = 1 + t, A scalar equation is x2 = 1 + (x1 − 1) = x1 . " # " # " # 3 1 2 ~ A31 The direction vector is d = − = . 0 0 0 x1 = 1 + 2t Hence, the parametric equation of the line is t ∈ R. x2 = 0 + 0t, A scalar equation is x2 = 0 + 0(x1 − 1) = 0.