Solution manual for an introduction to linear algebra for science and engineering 2nd edition by norman

✐ ✐ ✐ ✐ CHAPTER Euclidean Vector Spaces 1.1 Vectors in R2 and R3 Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Practice Problems A1 1+2 + = = 4+3 (a) (b) 3−4 −1 − = = 2−1 x2 x2 + 4 − 2 − x1 −1 3(−1) −3 (c) = = 3(4) 12 (d) x1 −2 −2 = − = −1 −2 x2 −2 −1 −2 −1 −1 −1 + (−1) + = = −2 −2 + 3 (−2)3 −6 (c) −2 = = −2 (−2)(−2) A2 (a) 2 −1 x1 (e) 4 x2 −1 3 1/4 1/2 3/2 −2 = − = 1/3 2/3 2/3 x1 −3 −2 −3 − (−2) −1 − = = −4 −4 − −9 4/3 7/3 (d) 12 + 13 = + = 3 √ √ (f) √ + √ = √ + √ = √ 6 6 (b) Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Chapter Euclidean Vector Spaces ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ − ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ ⎢ ⎥ A3 (a) ⎢⎢3⎥⎥ − ⎢⎢ ⎥⎥ = ⎢⎢ − ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −2 − (−2) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ + (−3) ⎥⎥⎥ ⎢⎢⎢ −1 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ (b) ⎢⎢ ⎥⎥ + ⎢⎢ ⎥⎥ = ⎢⎢ + ⎥⎥ = ⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −6 + (−4) −6 −10 −4 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ (−6)4 ⎥⎥⎥ ⎢⎢⎢−24⎥⎥⎥ ⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ (c) −6 ⎢⎢−5⎥⎥ = ⎢⎢(−6)(−5)⎥⎥ = ⎢⎢ 30 ⎥⎥⎥⎥ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 36 −6 (−6)(−6) ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ 10 ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (d) −2 ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 −2 −3 −5 ⎡ ⎤ ⎤ ⎡ ⎤ ⎡ ⎤ to ⎡ Linear ⎤ ⎡ Solution Manual for An Introduction Algebra for Science and Engineering ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ 4/3 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ 7/3 ⎥⎥⎥ ⎢⎢⎢ 2/3 ⎥⎥⎥ ⎢⎢ ⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥ ⎢⎢⎢ ⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎥ ⎥ (e) ⎢⎢−1/3⎥⎥ + ⎢⎢−2⎥⎥ = ⎢⎢−2/3⎥⎥ + ⎢⎢−2/3⎥⎥ = ⎢⎢−4/3⎥⎥ ⎣ ⎦ ⎣ ⎣ ⎦ ⎦ ⎣ ⎦ ⎣ ⎦ 1/3 13/3 ⎤ ⎡ ⎤ ⎡ ⎤ ⎡√ ⎤ ⎡ ⎤ ⎡√ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ √2⎥⎥⎥ ⎢⎢⎢−π⎥⎥⎥ ⎢⎢⎢ 2√− π⎥⎥⎥ √ ⎢⎢⎢⎢1⎥⎥⎥⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (f) ⎢⎢⎢⎢1⎥⎥⎥⎥ + π ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ 2⎥⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣√ ⎦ ⎣ ⎦ ⎣√ ⎦ π 2+π ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ −4 ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ A4 (a) 2v − 3w = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢−3⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −4 −13 ⎤ ⎤ ⎡ ⎤ ⎡ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢5⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−15⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−10⎥⎥⎥ ⎜⎜⎜⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎟⎟⎟ ⎢⎢⎢ ⎥⎥⎥ ⎥ ⎥ ⎢ ⎥ ⎢ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (b) −3(v + 2w) + 5v = −3 ⎜⎜⎜⎜⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢−2⎥⎥⎥⎥⎟⎟⎟⎟ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = −3 ⎢⎢⎢⎢0⎥⎥⎥⎥ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ ⎦ ⎦ ⎣ ⎦ ⎣ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ −22 −10 −10 −2 −12 −10 2nd Edition by (c) We have w − 2u = 3v, so 2u = w − 3v or u = 12 (w − 3v) This gives ⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢−1⎥ ⎢−1/2⎥⎥⎥ ⎜⎜⎜⎜⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥⎟⎟⎟⎟ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥ u = ⎜⎜⎜⎢⎢⎢−1⎥⎥⎥ − ⎢⎢⎢ ⎥⎥⎥⎟⎟⎟ = ⎢⎢⎢−7⎥⎥⎥ = ⎢⎢⎢−7/2⎥⎥⎥⎥ ⎦ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎣ 9/2 −6 (d) A5 (a) (b) (c) ⎡ ⎤ ⎢⎢⎢−3⎥⎥⎥ ⎢ ⎥ We have u − 3v = 2u, so u = −3v = ⎢⎢⎢⎢−6⎥⎥⎥⎥ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢3/2⎥⎥⎥ ⎢⎢⎢ 5/2 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ ⎢ ⎥ 1 v + w = ⎢⎢⎣1/2⎥⎥⎦ + ⎢⎢⎣−1/2⎥⎥⎥⎥⎦ = ⎢⎢⎢⎢⎣ ⎥⎥⎥⎥⎦ 2 1/2 −1 −1/2 ⎡ ⎤ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎜⎜⎜⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢ 15 ⎥⎥⎥⎟⎟⎟ ⎢⎢⎢ 16 ⎥⎥⎥ ⎢⎢⎢−9⎥⎥⎥ ⎢⎢⎢ 25 ⎥⎥⎥ ⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 2(v + w) − (2v − 3w) = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎜⎜⎜⎜⎢⎢⎢⎢2⎥⎥⎥⎥ − ⎢⎢⎢⎢−3⎥⎥⎥⎥⎟⎟⎟⎟ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ −5 ⎥⎥⎥⎥ ⎣ ⎦ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −6 −2 −1 −10 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ We have w − u = 2v, so u = w − 2v This gives u = ⎢⎢⎢⎢−1⎥⎥⎥⎥ − ⎢⎢⎢⎢2⎥⎥⎥⎥ = ⎢⎢⎢⎢−3⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −2 −4 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.1 Vectors in R2 and R3 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ 10 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (d) We have 12 u + 13 v = w, so 12 u = w − 13 v, or u = 2w − 23 v = ⎢⎢⎢⎢−2⎥⎥⎥⎥ − ⎢⎢⎢⎢2/3⎥⎥⎥⎥ = ⎢⎢⎢⎢ −8/3 ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −4 2/3 −14/3 A6 Solution Manual for An Thus, A7 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ PQ = OQ − OP = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −3 −2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ PR = OR − OP = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ PS = OS − OP = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎦ ⎣ Algebra ⎦ ⎣ ⎦ ⎣ for Introduction to Linear ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ QR = OR − OQ = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ S R = OR − OS = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −5 Science and Engineering 2nd Edition by ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ PQ + QR = ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = PS + S R ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −3 −1 −5 (a) The equation of the line is x = −5 +t ,t∈R −4 +t ,t∈R −6 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (c) The equation of the line is x = ⎢⎢⎢⎢0⎥⎥⎥⎥ + t ⎢⎢⎢⎢ −2 ⎥⎥⎥⎥, t ∈ R ⎣ ⎦ ⎣ ⎦ −11 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢4⎥⎥⎥ ⎢ ⎥ ⎢⎢⎢ ⎥⎥⎥ (d) The equation of the line is x = ⎢⎢1⎥⎥ + t ⎢⎢⎢⎢ ⎥⎥⎥⎥, t ∈ R ⎣ ⎦ ⎣ ⎦ (b) The equation of the line is x = 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 points: d = − −3 −1 = This, along with one of the points, may be used to obtain an equation for the line −5 x= −1 +t , −5 t∈R Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Chapter Euclidean Vector Spaces (b) The direction vector d of the line is given by the directed line segment joining the two points: d = −2 − −1 −6 = This, along with one of the points, may be used to obtain an equation for the line −2 x= −6 +t , −2 t∈R ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢ ⎥ (c) The direction vector d of the line is given by the directed line segment joining the two points: d = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢ ⎥⎦⎥ = ⎢⎣⎢−2⎥⎦⎥ This, along with one of the points, may be used to obtain an equation for the line −5 for An Introduction to Linear Algebra for Science and Engineering Solution Manual ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ x = ⎢⎢ ⎥⎥ + t ⎢⎢⎢⎢−2⎥⎥⎥⎥ , ⎣ ⎦ ⎣ ⎦ −5 2nd Edition by t∈R ⎡ ⎤ ⎢⎢⎢4⎥⎥⎥ ⎢ ⎥ (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 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢ ⎥ ⎢⎢⎢ ⎥⎥⎥ x = ⎢⎢ ⎥⎥ + t ⎢⎢⎢⎢1⎥⎥⎥⎥ , ⎣ ⎦ ⎣ ⎦ 1 t∈R ⎡ ⎤ ⎢⎢⎢ −1 ⎥⎥⎥ ⎢ ⎥ (e) The direction vector d of the line is given by the directed line segment joining the two points: d = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎣ ⎦ 1/3 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢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 −2/3 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−3/2⎥⎥⎥ ⎢⎢⎢1/2⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ x = ⎢⎢⎢⎢1/4⎥⎥⎥⎥ + t ⎢⎢⎢⎢ 3/4 ⎥⎥⎥⎥ , ⎣ ⎦ ⎣ ⎦ −2/3 A9 t∈R (a) We have x2 = 3x1 + x2 + = 3x1 + x2 + = 3(x1 + 1) Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.1 Vectors in R2 and R3 Let t = x1 + Then, from the equation above we get x2 + = 3t Solving the equations for x1 and x2 we find that the parametric equations are x1 = −1 + t, x2 = −1 + 3t, t ∈ R and the corresponding vector −1 equation is x = + t , t ∈ R −1 (b) We have 2x1 + 3x2 = 2x1 − = −3x2 + 2(x1 − 1) = −3(x2 − 1) 1 (x1 − 1) = − (x2 − 1) x2 −1 Let t = − Then, (x1 − 1) = t Solving the equations for x1 and x2 we find that the parametric Solution Manual for An Introduction to Linear Algebra for Science and 1Engineering 2nd Edition by equations are x1 = + 3t, x2 = − 2t, t ∈ R and the corresponding vector equation is x = A10 +t −2 ,t∈R (a) Let P, Q, and R be three points in Rn , with corresponding vectors p, q, and r If P, Q, and R are collinear, then the directed line segments PQ and PR should define the same line That is, the direction vector of one should be a non-zero scalar multiple of the direction vector of the other Therefore, PQ = tPR, for some t ∈ R −5 −6 (b) We have PQ = − = and PR = − = = −2PQ, so they are collinear −1 2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (c) We have S T = ⎢⎢−2⎥⎥ − ⎢⎢0⎥⎥ = ⎢⎢−2⎥⎥ and S U = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢0⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ Therefore, the points S , T , and U are not ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 −2 collinear because S U tS T for any real number t Homework Problems B1 (a) −1 x2 x2 (b) −1 + = −2 x1 x1 −2 − −3 − = −6 −3 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ (c) Chapter 1 −3 = −2 Euclidean Vector Spaces x2 x2 (d) −3 x1 −2 −3 3 x1 − 3 −18 − = −9 √ 3√ −4 0 (e) √ (d) (c) (b) B2 (a) 27/5 −2 0 − 3/2 Solution⎡ Manual for to⎡Linear Algebra ⎤ ⎡ ⎤ An Introduction ⎡ ⎤ ⎤ ⎡ ⎤ for Science ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ 7/3 ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ (b) ⎢⎢0⎥⎥ (c) ⎢⎢−20⎥⎥ (d) ⎢⎢0⎥⎥ (e) ⎢⎢−4/3⎥⎥ (f) ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎥⎥⎦ B3 (a) ⎢⎢⎣−3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −6 −4 13/3 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ 21 ⎥⎥⎥ ⎢⎢⎢ −1 ⎥⎥⎥ ⎢⎢⎢ −7 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) ⎢⎢⎢⎢−12⎥⎥⎥⎥ (c) ⎢⎢⎢⎢−13/2⎥⎥⎥⎥ (d) ⎢⎢⎢⎢−1/2⎥⎥⎥⎥ B4 (a) ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −5 −2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (c) u = ⎢⎢⎢⎢0⎥⎥⎥⎥ (d) u = ⎢⎢⎢⎢−7/2⎥⎥⎥⎥ B5 (a) ⎢⎢⎢⎢ ⎥⎥⎥⎥ (b) ⎢⎢⎢⎢ 1/4 ⎥⎥⎥⎥ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ 5/2 −3/4 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−9⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢−5⎥⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ B6 (a) PQ = ⎢⎢⎢⎢−1⎥⎥⎥⎥, PR = ⎢⎢⎢⎢ ⎥⎥⎥⎥, PS = ⎢⎢⎢⎢ ⎥⎥⎥⎥, QR = ⎢⎢⎢⎢ ⎥⎥⎥⎥, S R = ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −2 −6 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) PQ = ⎢⎢⎢⎢ ⎥⎥⎥⎥, PR = ⎢⎢⎢⎢3⎥⎥⎥⎥, PS = ⎢⎢⎢⎢ ⎥⎥⎥⎥, QR = ⎢⎢⎢⎢−6⎥⎥⎥⎥, S R = ⎢⎢⎢⎢−3⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −4 −2 ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ −3 ⎢ ⎥ , t ∈ R (b) x = t ⎢⎢⎢⎢3⎥⎥⎥⎥, t ∈ R +t (a) x = ⎣ ⎦ −3 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ B7 ⎢⎢⎢3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (d) x = ⎢⎢⎢⎢1⎥⎥⎥⎥ + t ⎢⎢⎢⎢−3⎥⎥⎥⎥, t ∈ R (c) x = ⎢⎢⎢⎢ ⎥⎥⎥⎥ + t ⎢⎢⎢⎢−4⎥⎥⎥⎥, t ∈ R ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2 −1 ⎡ ⎤ ⎢⎢⎢−1⎥⎥⎥ −2 ⎢ ⎥ (a) x = +t , t∈R (b) x = t ⎢⎢⎢⎢ ⎥⎥⎥⎥ , t ∈ R ⎣ ⎦ 1 −1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ B8 ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−1/2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (d) x = ⎢⎢⎢⎢ −1 ⎥⎥⎥⎥ + t ⎢⎢⎢⎢ 4/3 ⎥⎥⎥⎥ , t ∈ R (c) x = ⎢⎢⎢⎢−6⎥⎥⎥⎥ + t ⎢⎢⎢⎢ 11 ⎥⎥⎥⎥ , t ∈ R ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 1/2 1/2 and Engineering 2nd Edition by B9 Other correct answers are possible Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.1 Vectors in R2 and R3 B10 (a) x1 = − 12 t + 32 , x2 = t; x = −1/2 3/2 +t (b) x1 = t, x2 = − 12 t + 32 ; x = +t 3/2 −1/2 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ (a) Since −2PQ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ = PR, 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 Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by ⎡ ⎤ ⎢⎢⎢−2322⎥⎥⎥ ⎢⎢⎢ C1 (a) ⎢⎢⎣−1761⎥⎥⎥⎥⎥⎦ 1667 ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ (b) ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ Conceptual Problems D1 13 = t1 −12 That is, we need to solve the two equations substitution and/or elimination we find that t1 (a) We need to find t1 and t2 such that x2 t +t 1 + t2 = −1 t1 − t2 in two unknowns t1 + t2 = 13 and t1 − t2 = −12 Using = 12 and t2 = 25 1 x1 −1 13 −12 (b) We use the same approach as in part (a) We need to find t1 and t2 such that 1 t +t x1 = t1 + t2 = x2 t1 − t2 −1 Solving t1 + t2 = x1 and t1 − t2 = x2 by substitution and/or elimination gives t1 = 12 (x1 + x2 ) and t2 = (x1 − x2 ) Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Chapter (c) We have x1 = D2 √ Euclidean Vector Spaces √ √ and x2 = π, so we get t1 = 12 ( + π) and t2 = 12 ( − π) (a) PQ + QR + RP can be described informally as “start at P and move to Q, then move from Q to R, then from R to P; the net result is a zero change in position.” (b) We have PQ = q − p, QR = r − q, and RP = p − r Thus, PQ + QR + RP = q − p + r − q + p − r = D3 Assume that x = p + td, t ∈ R, is a line in R2 passing through the origin Then, there exists a real number t1 such that = p + t1 d Hence, p = −t1 d and so p is a scalar multiple of d On the other hand, assume that p is a scalar multiple of d Then, there exists a real number t1 such that p = t1 d Hence, if we take t = −t1 , we get that the line with vector equation x = p + td passes through the point p + (−t1 )d = t1 d − t1 d = = as required 0Engineering Solution Manual for An Introduction to Linear Algebra for Science and ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ x D4 Let x = and y = ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣y2 ⎥⎥⎦ Then, x3 y3 2nd Edition by ⎤ ⎡ ⎤ ⎤ ⎡ ⎡ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 + y1 ⎥⎥⎥ ⎢⎢⎢t(x1 + y1 )⎥⎥⎥ ⎢⎢⎢tx1 + ty1 ⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ t(x + y) = t ⎢⎢⎢⎢ x2 + y2 ⎥⎥⎥⎥ = ⎢⎢⎢⎢t(x2 + y2 )⎥⎥⎥⎥ = ⎢⎢⎢⎢tx2 + ty2 ⎥⎥⎥⎥ = t ⎢⎢⎢⎢ x2 ⎥⎥⎥⎥ + t ⎢⎢⎢⎢y2 ⎥⎥⎥⎥ = t ⎢⎢⎢⎢ x2 ⎥⎥⎥⎥ + t ⎢⎢⎢⎢y2 ⎥⎥⎥⎥ = tx + ty ⎦ ⎣ ⎦ ⎦ ⎣ ⎣ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ t(x3 + y3 ) x3 + y3 tx3 + ty3 x3 y3 x3 y3 1.2 Vectors in Rn Practice Problems A1 A2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢5⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢9⎥⎥⎥ (a) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ + ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ + ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣−1⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣−2⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ −1 −1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ 10 ⎥⎥⎥ ⎢⎢ ⎥⎥ ⎢⎢⎢⎢⎢−2⎥⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥ (b) ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ − ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ + ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ 10 ⎥⎥⎦ −5 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (c) ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢1⎥⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢ ⎥⎦⎥ ⎢⎣⎢1⎥⎦⎥ −2 −1 −3 −1 (a) Since the condition of the set contains the square of a variable in it, we suspect that it is not a subspace To prove it is not a subspace we just need to find one example where the set is not closed under addition Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.2 Vectors in Rn ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ Let x = ⎢⎢1⎥⎥ and y = ⎢⎢⎢⎢1⎥⎥⎥⎥ Observe that x and y are in the set since x12 − x22 = 12 − 12 = = x3 and ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎢⎢⎢3⎥⎥⎥ ⎢ ⎥ 2 2 y1 − y2 = − = = y3 , but x + y = ⎢⎢⎢⎢2⎥⎥⎥⎥ is not in the set since 32 − 22 = ⎣ ⎦ (b) Since the condition of the set only contains linear variables, we suspect that this is a subspace To prove it is a subspace we need to show that it satisfies the definition of a subspace Call the set S First, observe that S is a subset of R3 and is non-empty since the zero vector satisfies the ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ conditions of the set Pick any vectors x = ⎢⎢⎢⎢ x2 ⎥⎥⎥⎥ and y = ⎢⎢⎢⎢y2 ⎥⎥⎥⎥ in S Then they must satisfy the condition of ⎣ ⎦ ⎣ ⎦ x3 y3 ⎡ ⎤ ⎢⎢⎢ x1 + y1 ⎥⎥⎥ ⎢ ⎥ Solution Manual An Engineering + y2 ⎥⎥⎥⎥Science x3 and y1 Introduction = y3 We now needtotoLinear show thatAlgebra x + y = ⎢⎢⎢⎢ x2 for S , so x1 = for satisfies theand conditions of the set ⎣ ⎦ x3 + y3 x + y equals its third entry Since x1 = x3 and y1 = y3 st entry of we get x1 + y1 = x3 + y3 as required Thus, S is closed under addition Similarly, to show S is closed under ⎡ ⎤ ⎢⎢⎢tx1 ⎥⎥⎥ ⎢ ⎥ scalar multiplication, we let t be any real number and show that tx = ⎢⎢⎢⎢tx2 ⎥⎥⎥⎥ satisfies the conditions of the ⎣ ⎦ tx3 set Using x1 = x3 we get tx1 = tx3 as required Thus, S is a subspace of R3 2nd Edition by (c) Since the condition of the set only contains linear variables, we suspect that this is a subspace Call the set S First, observe that S is a subset of R2 and is non-empty since the zero vector satisfies the x y conditions of the set Pick any vectors x = and y = in S Then they must satisfy the condition x2 y2 x + y1 of S , so x1 + x2 = and y1 + y2 = Then x + y = satisfies the conditions of the set since x2 + y2 (x1 + y1 ) + (x2 + y2 ) = x1 + x2 + y1 + y2 = + = Thus, S is closed under addition Similarly, for any tx real number t we have that tx = and tx1 + tx2 = t(x1 + x2 ) = t(0) = 0, so S is also closed under scalar tx2 multiplication Thus, S is a subspace of R2 (d) The condition of the set involves multiplication of entries, so we suspect that it is not a subspace Observe ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ that if x = ⎢⎢⎢⎢ x2 ⎥⎥⎥⎥ = ⎢⎢⎢⎢1⎥⎥⎥⎥, then x is in the set since x1 x2 = 1(1) = = x3 , but 2x = ⎢⎢⎢⎢2⎥⎥⎥⎥ is not in the set since ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ x3 2(2) = Therefore, the set is not a subspace ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢ ⎥ (e) At first glance this might not seem like a subspace since we are adding the vector ⎢⎢⎢⎢3⎥⎥⎥⎥ However, the key ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ observation to make is that ⎢⎢⎢⎢3⎥⎥⎥⎥ is equal to ⎢⎢⎢⎢1⎥⎥⎥⎥ + ⎢⎢⎢⎢2⎥⎥⎥⎥ In particular, this is the equation of a plane in R3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ which passes through the origin So, this should be a subspace of R3 We could use the definition of a subspace to prove this, but the point of proving theorems is to make problems easier Therefore, we instead Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 10 Chapter Euclidean Vector Spaces ⎧⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎢1⎥ ⎢2⎥⎪ ⎪ ⎪ ⎪ ⎨⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥⎪ ⎬ ⎢⎢⎢1⎥⎥⎥ , ⎢⎢⎢3⎥⎥⎥⎪ observe that this is a vector equation of the set Span ⎪ and hence is a subspace by Theorem 1.2.2 ⎪ ⎪ ⎪ ⎣ ⎦ ⎣ ⎦ ⎩1 4⎪ ⎭ (f) The set is a subspace of R4 by Theorem 1.2.2 A3 (a) Since the condition of the set only contains linear variables, we suspect that this is a subspace Call the set S By definition S is a subset of R4 and is non-empty since the zero vector satisfies the conditions of the ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢ x ⎥⎥⎥ ⎢⎢⎢y ⎥⎥⎥ set Pick any vectors x = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ and y = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ in S , then x1 + x2 + x3 + x4 = and y1 + y2 + y3 + y4 = We ⎢⎢⎣ x3 ⎥⎥⎦ ⎢⎢⎣y3 ⎥⎥⎦ x4 y4 ⎡ ⎤ + y x ⎢⎢⎢ 1⎥ ⎢⎢⎢ x + y ⎥⎥⎥⎥⎥ 2⎥ ⎢ ⎥⎥ satisfies the conditions of the set since (x1 + y1 ) + (x2 + y2 ) + (x3 + y3 ) + (x4 + y4 ) = have x + y = ⎢⎢⎢⎢ ⎢⎢⎣ x3 + y3 ⎥⎥⎥⎥⎦ Solution Manual forx An Introduction to Linear Algebra for Science and Engineering + y4 x1 + x2 + x3 + x4 + y1 + y2 + y3 + y4 = + = Thus, S is closed under addition Similarly, for any real ⎡ ⎤ ⎢⎢⎢tx1 ⎥⎥⎥ ⎢⎢⎢tx ⎥⎥⎥ number t we have that tx = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ and tx1 + tx2 + tx3 + tx4 = t(x1 + x2 + x3 + x4 ) = t(0) = 0, so S is also ⎢⎢⎣tx3 ⎥⎥⎦ tx4 closed under scalar multiplication Thus, S is a subspace of R4 2nd Edition by (b) The set clearly does not contain the zero vector and hence cannot be a subspace (c) The conditions of the set only contain linear variables, but we notice that the first equation x1 + 2x3 = excludes x1 = x3 = Hence the zero vector is not in the set so it is not a subspace (d) The conditions of the set involve a multiplication of variables, so we suspect that it is not a subspace Using ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ the knowledge gained from problem A2(d) we take x = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ Then, x is in the set since x1 = = 1(1) = x3 x4 ⎢⎢⎣1⎥⎥⎦ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ and x2 − x4 = − = But, 2x = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ is not in the set since 2(2) ⎢⎢⎣2⎥⎥⎦ (e) Since the conditions of the set only contains linear variables, we suspect that this is a subspace Call the set S By definition S is a subset of R4 and is non-empty since the zero vector satisfies the conditions ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢⎢ x2 ⎥⎥⎥⎥ ⎢⎢⎢y ⎥⎥⎥ of the set Pick any vectors x = ⎢⎢⎢⎢ ⎥⎥⎥⎥ and y = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ in S , then 2x1 = 3x4 , x2 − 5x3 = 0, 2y1 = 3y4 , and ⎢⎢⎣ x3 ⎥⎥⎦ ⎢⎢⎣y3 ⎥⎥⎦ x4 y4 ⎡ ⎤ ⎢⎢⎢ x1 + y1 ⎥⎥⎥ ⎢⎢⎢ x + y2 ⎥⎥⎥⎥ ⎥⎥ satisfies the conditions of the set since 2(x1 + y1 ) = 2x1 + 2y1 = y2 − 5y3 = We have x + y = ⎢⎢⎢⎢⎢ ⎢⎢⎣ x3 + y3 ⎥⎥⎥⎥⎦ x4 + y4 3x4 + 3y4 = 3(x4 + y4 ) and (x2 + y2 ) − 5(x3 − y3 ) = x2 − 5x3 + y2 − 5y3 = + = Thus, S is closed under Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.2 Vectors in Rn 11 ⎡ ⎤ ⎢⎢⎢tx1 ⎥⎥⎥ ⎢⎢⎢tx ⎥⎥⎥ addition Similarly, for any real number t we have that tx = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥, 2(tx1 ) = t(2x1 ) = t(3x4 ) = 3(tx4 ), and ⎢⎢⎣tx3 ⎥⎥⎦ tx4 (tx2 ) − t(5x3 ) = t(x2 − 5x3 ) = t(0) = Therefore, S is also closed under scalar multiplication Thus, S is a subspace of R4 (f) Since x3 = the zero vector cannot be in the set, so it is not a subspace A4 Alternative correct answers are possible ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (a) ⎢⎢0⎥⎥ + ⎢⎢0⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ ⎢ ⎥⎥⎥ An ⎢ ⎥ ⎢⎢⎢ ⎥⎥⎥ Solution 2⎥⎥ + ⎢⎢⎢⎢4⎥⎥⎥⎥ Introduction = (b) Manual ⎢⎢⎢⎢−1⎥⎥⎥⎥ − ⎢⎢⎢⎢for ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢⎢⎣ ⎥⎥⎦ ⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤ t.eu/ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (c) ⎢⎢⎢⎢1⎥⎥⎥⎥ + ⎢⎢⎢⎢1⎥⎥⎥⎥ − ⎢⎢⎢⎢2⎥⎥⎥ = ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 (d) to Linear Algebra for Science and Engineering 2nd Edition by 1 −2 +1 = A5 Alternative correct answers are possible (a) We observe that neither vector is a scalar multiple of the⎧ other Hence, this is a linearly independent set of ⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎪ 1 ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢⎢2⎥⎥⎥⎥⎪ ⎬ 4 two vectors in R Hence, it is a plane in R with basis ⎪ , ⎢ ⎥ ⎢ ⎥ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ 1 ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩1 3⎭ a linearly independent set of three vectors in R4 (b) This is a subset of the standard basis for R⎧4 and hence is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ Hence, it is a hyperplane in R with basis ⎪ ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎩ ⎭ 0 (c) Observe that the second and third vectors are just scalar multiples of the first vector Hence, by Theorem 1.2.3, we can write ⎧⎡ ⎤⎫ ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎪ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎢⎢ ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎨⎢⎢⎢⎢ ⎥⎥⎥⎥⎪ ⎨⎢⎢ ⎥⎥ ⎢⎢ ⎥⎥ ⎢⎢ ⎥⎥⎬ ⎬ , = Span Span ⎪ , ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎪ ⎪ ⎪ −2 −1 −1 ⎢ ⎥ ⎢ ⎢⎢⎣ ⎥⎥⎦⎪ ⎢⎢⎣ ⎥⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ ⎩ ⎩ 0 0 ⎭ ⎧⎡ ⎤⎫ ⎪ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎨⎢⎢⎢⎢ ⎥⎥⎥⎥⎪ ⎬ Therefore, it is a line in R with basis ⎪ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎩⎣ ⎦⎪ ⎭ Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 12 Chapter Euclidean Vector Spaces (d) Observe that the third vector is the sum of the first two vectors hence, by Theorem 1.2.3 we can write ⎧⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎧⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢⎢⎢⎢1⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢1⎥⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥⎪ = Span ⎪⎢⎢⎢⎢ ⎥⎥⎥⎥ , ⎢⎢⎢⎢ ⎥⎥⎥⎥⎬ Span ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎪ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦⎪ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −1 ⎭ ⎭ ⎩ −1 ⎪ ⎧⎡ ⎤ ⎡ ⎤⎫ ⎧⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨⎢⎢⎢⎢1⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥⎪ ⎨⎢⎢ ⎥⎥ ⎢⎢ ⎥⎥⎬ ⎬ with basis Since ⎪ , , is linearly independent, we get that it spans a plane in R ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎥ ⎪ ⎪ ⎪ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎪ ⎪ ⎪ 0 0 ⎢ ⎥ ⎢ ⎢ ⎥ ⎥ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −1 ⎭ ⎩ −1 ⎭ A6 If x = p + td is a subspace of Rn , then it contains the zero vector Hence, there exists t1 such that = p + t1 d Thus, p = −t1 d and so p is a scalar multiple of d On the other hand, if p is a scalar multiple of d, say p = t1 d, Solution Manual to Linear Algebra for Science and Engineering then we have x =for p + tAn d = tIntroduction d + td = (t1 + t)d Hence, the set is Span{d} and thus is a subspace A7 , Assume there is a non-empty subset B1 = {v1 2nd Edition by v } of B that is linearly dependent Then there exists ci not all = c1 v1 + · · · + c v = c1 v1 + · · · + c v + 0v +1 + · · · + 0vn which contradicts the fact that B is linearly independent Hence, B1 must be linearly independent Homework Problems ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ B1 (a) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣−2⎥⎥⎦ B2 ⎡ ⎤ ⎢⎢⎢ 10 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (b) ⎢⎢⎢⎢ 13 ⎥⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢ 10 ⎥⎦⎥ −1 ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (c) ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢1⎥⎦⎥ (a) It is not a subspace of R2 (b) It is not a subspace of R3 (c) It is a subspace of R3 (d) It is not a subspace of R3 (e) It is a subspace of R3 (f) It is a subspace of R3 B3 (a) It is not a subspace of R4 (b) It is not a subspace of R4 (c) It is a subspace of R4 (d) It is a subspace of R4 (e) It is not a subspace of R4 (f) It is a subspace of R4 B4 Alternative correct answers are possible Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.2 Vectors in Rn 13 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (a) (−2) ⎢⎢⎢⎢1⎥⎥⎥⎥ + ⎢⎢⎢⎢2⎥⎥⎥⎥ = ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−6⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −1 −8 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢5⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (c) ⎢⎢⎢⎢1⎥⎥⎥⎥ − ⎢⎢⎢⎢4⎥⎥⎥⎥ + ⎢⎢⎢⎢1⎥⎥⎥⎥ = ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (d) −1 −7 +5 = 2 B5 Alternative correct answers are possible ⎧ ⎡ ⎤⎫ Solution Manual for An Introduction to Linear ⎪ ⎪ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢−1⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ ⎢⎢⎢ ⎥⎥⎥ (a) The line in R4 with basis ⎪ ⎢⎢⎢ ⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎩ ⎣ ⎦⎪ ⎭ ⎧ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥⎬ (b) The hyperplane in R4 with basis ⎪ ⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥⎪ ⎪ ⎪ ⎢ ⎪ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩1 1⎭ ⎧⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥⎬ (c) The plane in R4 with basis ⎪ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩0 1⎭ ⎧⎡ ⎤ ⎡ ⎤⎫ ⎪ ⎢⎢⎢3⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎢⎢⎢ ⎥⎥⎥ , ⎢⎢⎢ ⎥⎥⎥⎬ (d) The plane in R4 with basis ⎪ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎪ ⎪ ⎪ ⎪ ⎩1 1⎭ Computer Problems ⎡ ⎤ ⎢⎢⎢−53.72⎥⎥⎥ ⎢⎢⎢ 17.45 ⎥⎥⎥ ⎥⎥⎥ C1 (a) ⎢⎢⎢⎢⎢ ⎥ ⎢⎢⎣ 13.16 ⎥⎥⎥⎦ 2.88 Algebra for Science and Engineering 2nd Edition by ⎡ ⎤ ⎢⎢⎢−16.66902697⎥⎥⎥ ⎢⎢⎢ 11.82240799 ⎥⎥⎥ ⎥⎥⎥ (b) ⎢⎢⎢⎢⎢ ⎥ ⎢⎢⎣ 3.147235505 ⎥⎥⎥⎦ −1.434240231 Conceptual Problems ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢ ⎥⎥ D1 Let x = ⎢⎢⎢⎢ ⎥⎥⎥⎥ and let s, t ∈ R Then ⎣⎢ ⎦⎥ xn ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢(s + t)x1 ⎥⎥⎥ ⎢⎢⎢ sx1 + tx1 ⎥⎥⎥ ⎢⎢⎢ sx1 ⎥⎥⎥ ⎢⎢⎢tx1 ⎥⎥⎥ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢ ⎥⎥ ⎢⎢ ⎥⎥⎥ = ⎢⎢⎢ ⎥⎥⎥ = ⎢⎢⎢ ⎥⎥⎥ + ⎢⎢⎢ ⎥⎥⎥ = (s + t) ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎣⎢ ⎦⎥ ⎢⎣ xn (s + t)xn sxn + txn sxn txn ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢ ⎥⎥ s ⎢⎢⎢ ⎥⎥⎥ + t ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎣ ⎥⎦ ⎣⎢ ⎦⎥ xn xn Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 14 Chapter Euclidean Vector Spaces ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢ ⎥⎥ D2 Let x = ⎢⎢⎢ ⎥⎥⎥, y = ⎢⎢⎢⎢ ⎥⎥⎥⎥, and let t ∈ R ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ xn yn ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢ x1 + y1 ⎥⎥⎥ ⎢⎢⎢t(x1 + y1 )⎥⎥⎥ ⎢⎢⎢tx1 + ty1 ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ ⎢ ⎥⎥⎥ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ t(x + y) = t ⎢⎢⎢ ⎥⎥⎥ = ⎢⎢⎢ = ⎢⎢ ⎥ ⎥ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ xn + yn t(xn + yn ) txn + tyn ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢tx1 ⎥⎥⎥ ⎢⎢⎢ty1 ⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢y1 ⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎢⎢ ⎥⎥ = ⎢⎢ ⎥⎥ + ⎢⎢ ⎥⎥ = t ⎢⎢ ⎥⎥ + t ⎢⎢⎢⎢ ⎥⎥⎥⎥ = tx + ty ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ txn tyn xn yn (a) By definition U ∩ V is a subset of Rn , and ∈ U and ∈ V since they are both subspaces Thus, ∈ U ∩ V Let x, y ∈ U ∩ V Then x, y ∈ U and x, y ∈ V Since U is a subspace it is closed under addition and scalar Solution Manual for An Introduction to Linear Algebra for Science and Engineering multiplication, so x + y ∈ U and tx ∈ U for all t ∈ R Similarly, V is a subspace, so x + y ∈ V and tx ∈ V for all t ∈ R Hence, x + y ∈ U ∩ V and tx ∈ U ∩ V, so U ∩ V is closed under addition and scalar multiplication D3 2nd Edition by (b) Consider the subspaces U = y= x1 | x1 ∈ R and V = 0 x | x2 ∈ R of R2 Then x = ∈ U and x2 0 ∈ V, but x + y is not in U and not in V, so it is not in U ∪ V Thus, U ∪ V is not a subspace y (c) Since U and V are subspaces of Rn , u, v ∈ Rn for any u ∈ U and v ∈ V, so u + v ∈ Rn since Rn is closed under addition Hence, U + V is a subset of Rn Also, since U and V are subspace of Rn , we have ∈ U and ∈ V, thus = + ∈ U + V Pick any vectors x, y ∈ U + V Then, there exists vectors u1 , u2 ∈ U and v1 , v2 ∈ V such that x = u1 + v1 and y = u2 + v2 We have x + y = (u1 + v1 ) + (u2 + v2 ) = (u1 + u2 ) + (v1 + v2 ) with u1 + u2 ∈ U and v1 + v2 ∈ V since U and V are both closed under addition Hence, x + y ∈ U + V Similarly, for any t ∈ R we have tx = t(u1 + v1 ) = tu1 + tv1 where tu1 ∈ U and tv1 ∈ V, so tx ∈ U + V Therefore, U + V is a subspace of Rn D4 There are many possible solutions ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢⎢1⎥⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ (a) Pick b = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v1 = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v2 = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥, and v3 ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣1⎥⎥⎦ 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ (b) Pick b = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v1 = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v2 = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥, and v3 ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢⎢3⎥⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ (c) Pick b = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v1 = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v2 = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥, and v3 ⎢⎢⎣1⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ 0 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢⎢0⎥⎥⎥⎥ (d) Pick b = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v1 = ⎢⎢⎢⎢ ⎥⎥⎥⎥, v2 = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥, and v3 ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣0⎥⎥⎦ 0 ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣0⎥⎥⎦ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣0⎥⎥⎦ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣0⎥⎥⎦ ⎡ ⎤ ⎢⎢⎢3⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣0⎥⎥⎦ Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.2 Vectors in Rn 15 D5 By definition of spanning, every x ∈ Span B can be written as a linear combination of the vectors in B Now, assume that we have x = s1 v1 + · · · + sk vk and x = t1 v1 + · · · + tk vk Then, we have s1 v1 + · · · + sk vk = t1 v1 + · · · + tk vk (s1 v1 + · · · + sk vk ) − (t1 v1 + · · · + tk vk ) = (s1 − t1 )v1 + · · · + (sk − tk )vk = Since {v1 , , vk } is linearly independent, this implies that si − ti = for ≤ i ≤ k That is, si = ti Therefore, there is a unique linear combination of the vectors in B which equals x D6 Assume that {v1 , , vk , x} is linearly dependent Then, there exist coefficients t1 , , tk , tk+1 not all zero such that t1 v1 + · · · + tk vk + tk+1 x = Solution Ant1Introduction Linear Science andthat Engineering = 0, then for we have with some tAlgebra 0, whichfor contradicts the fact {v1 , , vk } is 2nd Edition by v1 + · · · + tk vk = to If tk+1Manual i Hence, we can solve the equation for x to get linearly independent Thus, tk+1 x=− This contradicts our assumption that x tk t1 v1 − · · · − vk tk+1 tk+1 Span B Therefore, B must be linearly independent D7 If x ∈ Span{v1 , sv1 + tv2 }, then x = c1 v1 + c2 (sv1 + tv2 ) = (c1 + sc2 )v1 + c2 tv2 ∈ Span{v1 , v2 } Hence, Span{v1 , sv1 + tv2 } ⊆ Span{v1 , v2 } Since t we get that v2 = −st v1 + 1t (sv1 + tv2 ) Hence, if v ∈ Span{v1 , v2 }, then v = b1 v1 + b2 v2 = b1 v1 + b2 = b1 − −s v1 + (sv1 + tv2 ) t t b2 s b2 v1 + (sv1 + tv2 ) ∈ Span{v1 , sv1 + tv2 } t t Thus, Span{v1 , v2 } ⊆ Span{v1 , sv1 + tv2 } Hence Span{v1 , v2 } = Span{v1 , sv1 + tv2 } D8 (a) TRUE We can rearrange the equation to get −tv1 + v2 = with at least one coefficient (the coefficient of v1 ) non-zero Hence {v1 , v2 } is linearly dependent by definition (b) FALSE If v2 = and v1 is any non-zero vector, then v1 is not a scalar multiple of v2 and {v1 , v2 } is linearly dependent by Theorem 1.2.4 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ (c) FALSE If v1 = ⎢⎢1⎥⎥, v2 = ⎢⎢0⎥⎥, and v3 = ⎢⎢⎢⎢0⎥⎥⎥⎥ Then, {v1 , v2 , v3 } is linearly dependent, but v1 cannot be written ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 as a linear combination of v1 and v2 (d) TRUE If v1 = sv2 + tv3 , then we have v1 − sv2 − tv3 = with at least one coefficient (the coefficient of v1 ) non-zero Hence, by definition, the set is linearly dependent (e) FALSE The set {0} = Span{0} is a subspace by Theorem 1.2.2 (f) TRUE By Theorem 1.2.2 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 16 Chapter Euclidean Vector Spaces 1.3 Length and Dot Products Practice Problems A1 (a) −5 = 22 + (−5)2 = √ 29 √ √ √ √ 2/ √29 = (2/ 29)2 + (−5/ 29)2 = 4/29 + 25/29 = (b) −5/ 29 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ √ ⎢ ⎥ (c) ⎢⎢⎢⎢ ⎥⎥⎥⎥ = 12 + 02 + (−1)2 = ⎣ ⎦ −1 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ Solution Manual for An Introduction to Linear Algebra √ ⎢ ⎥ (d) ⎢⎢⎢⎢ ⎥⎥⎥⎥ = 22 + 32 + (−2)2 = 17 ⎣ ⎦ −2 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ √ ⎢ ⎥ (e) ⎢⎢⎢⎢1/5⎥⎥⎥⎥ = 12 + (1/5)2 + (−3)2 = 251/5 ⎣ ⎦ −3 ⎡ √ ⎤ ⎢⎢⎢ 1/ ⎥⎥⎥ √ √ √ ⎢ √ ⎥ (f) ⎢⎢⎢⎢⎢ 1/ ⎥⎥⎥⎥⎥ = (1/ 3)2 + (1/ 3)2 + (−1/ 3)2 = √ ⎦ ⎣ −1/ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ √ (g) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 12 + (−1)2 + 02 + 22 = ⎢⎢⎣ ⎥⎥⎦ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ √ (h) 12 ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 12 ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 12 12 + 12 + 12 + 12 = ⎢⎢⎣1⎥⎥⎦ ⎢⎢⎣1⎥⎥⎦ 1 A2 is x= −4 x for Science and Engineering 2nd Edition by 3 3/5 = = −4/5 −4 32 + (−4)2 −4 √ 1 1 1/ √2 = √ x= √ = (b) A unit vector in the direction of x = is x 1/ 2 12 + 12 √ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎡ ⎤ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢−1/ 5⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ 1 ⎢ ⎥ ⎢⎢ ⎥⎥ ⎢ ⎥ ⎢⎢⎢ ⎥⎥⎥ (c) A unit vector in the direction of x = ⎢⎢⎢⎢ ⎥⎥⎥⎥ is x= ⎢⎢⎣ ⎥⎥⎦ = √ ⎢⎢⎢⎢⎣ ⎥⎥⎥⎥⎦ = ⎢⎢⎢⎢ 0√ ⎥⎥⎥⎥ ⎣ ⎦ x ⎦ ⎣ (−1)2 + 02 + 22 2 2/ (a) A unit vector in the direction of x = ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ x= (d) A unit vector in the direction of x = ⎢⎢⎢⎢−3⎥⎥⎥⎥ is ⎣ ⎦ x ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢0⎥ ⎢0⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ −3 = ⎢⎢−3⎥⎥ = ⎢⎢−1⎥⎥ ⎢⎢ ⎥⎥ 02 + (−3)2 + 02 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.3 Length and Dot Products ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ (e) A unit vector in the direction of x = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ is ⎢⎢⎣ ⎥⎥⎦ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (f) A unit vector in the direction of x = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ is ⎢⎢⎣ ⎥⎥⎦ −1 x= x x= x 17 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−2/3⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−2/3⎥⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ = ⎢⎢⎢ ⎥⎥⎥ = ⎢⎢⎢ ⎥⎥⎥ (−2)2 + (−2)2 + 12 + 02 ⎢⎢⎢⎣ ⎥⎥⎥⎦ ⎢⎢⎢⎣ ⎥⎥⎥⎦ ⎢⎢⎢⎣ 1/3 ⎥⎥⎥⎦ 0 ⎡ ⎤ ⎡ ⎤ ⎡ √ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ 1/ ⎥⎥⎥ ⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢⎢ 1 ⎢⎢⎢ ⎥⎥⎥ = √ ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ ⎥⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎢⎢⎣ ⎥⎥⎥⎦ ⎢⎢⎢⎣ 0√ ⎥⎥⎥⎦ 12 + 02 + 02 + (−1)2 ⎢⎢⎢⎣ ⎥⎥⎥⎦ −1 −1 −1/ √ −4 −6 − = = (−6)2 + (−2)2 = 10 −2 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ (b) The distance between P and Q is PQ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ = (−4)2 + 02 + 32 = ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ for Science and Engineering Solution Manual for An Introduction to Linear −2Algebra ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥ √ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ we esetnbPanankdDQiirsect.eu/ PQ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢−6⎥⎥⎥⎥ = ⎢⎢⎢⎢ 11 ⎥⎥⎥⎥ = (−7)2 + 112 + 02 = 170 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ √ (d) The distance between P and Q is PQ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ − ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 22 + 52 + (−3)2 + (−4)2 = ⎢⎢⎣−2⎥⎥⎦ ⎢⎢⎣1⎥⎥⎦ ⎢⎢⎣−3⎥⎥⎦ −4 ⎡ ⎤ ⎢⎢⎢6⎥⎥⎥ √ √ √ √ √ ⎢ ⎥ A4 (a) We have x = 42 + 32 + 12 = 26, y = 22 + 12 + 52 = 30, x + y = ⎢⎢⎢⎢4⎥⎥⎥⎥ = 62 + 42 + 62 = ⎣ ⎦ √ 22, and |x · y| = 4(2) + 3(1) + 1(5) = 16 The triangle inequality is satisfied since √ √ √ 22 ≈ 9.38 ≤ 26 + 30 ≈ 10.58 A3 (a) The distance between P and Q is PQ = The Cauchy-Schwarz inequality is also satisfied since 16 ≤ √ 26(30) ≈ 27.93 ⎡ ⎤ ⎢⎢⎢−2⎥⎥⎥ ⎢ ⎥ (b) We have x = 12 + (−1)2 + 22 = 6, y = (−3)2 + 22 + 42 = 29, x+y = ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎣ ⎦ √ 41, and |x · y| = 1(−3) + (−1)(2) + 2(4) = The triangle inequality is satisfied since √ √ √ 41 ≈ 6.40 ≤ + 29 ≈ 7.83 √ A5 2nd Edition by √ (−2)2 + 12 + 62 = √ The Cauchy-Schwarz inequality is satisfied since ≤ 6(29) ≈ 13.19 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (a) ⎢⎢⎢⎢3⎥⎥⎥⎥ · ⎢⎢⎢⎢−2⎥⎥⎥⎥ = 1(2) + 3(−2) + 2(2) = Hence these vectors are orthogonal ⎣ ⎦ ⎣ ⎦ 2 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (b) ⎢⎢⎢⎢ ⎥⎥⎥⎥ · ⎢⎢⎢⎢−1⎥⎥⎥⎥ = (−3)(2) + 1(−1) + 7(1) = Hence these vectors are orthogonal ⎣ ⎦ ⎣ ⎦ Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 18 Chapter Euclidean Vector Spaces ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (c) ⎢⎢⎢⎢1⎥⎥⎥⎥ · ⎢⎢⎢⎢ ⎥⎥⎥⎥ = 2(−1) + 1(4) + 1(2) = Therefore, these vectors are not orthogonal ⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (d) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ · ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 4(−1) + 1(4) + 0(3) + (−2)(0) = Hence these vectors are orthogonal ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ −2 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ x ⎥⎥⎥ (e) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ · ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 0(x1 ) + 0(x2 ) + 0(x3 ) + 0(x4 ) = Hence these vectors are orthogonal ⎢⎢⎣0⎥⎥⎦ ⎢⎢⎣ x3 ⎥⎥⎦ x4 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ 1/3 ⎥⎥⎥ ⎢⎢⎢ 3/2 ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢⎢ 2/3 ⎥⎥⎥ ⎢⎢⎢ for Solution Manual Introduction to Linear Algebra for Science and Engineering ⎥⎥⎥ · ⎢⎢⎢ ⎥⎥⎥⎥⎥ An (f) ⎢⎢⎢⎢⎢ = + 23 (0) + − 13 − 32 + 3(1) = Therefore, these vectors are not orthogonal ⎢⎢⎣−1/3⎥⎥⎥⎦ ⎢⎢⎢⎣−3/2⎥⎥⎥⎦ / A6 2nd Edition by · = 3(2) + (−1)k = − k −1 k Thus, the vectors are orthogonal only when k = (a) The vectors are orthogonal when = k · = 3(k) + (−1)(k2 ) = 3k − k2 = k(3 − k) −1 k Thus, the vectors are orthogonal only when k = or k = ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ (c) The vectors are orthogonal when = ⎢⎢⎢⎢2⎥⎥⎥⎥ · ⎢⎢⎢⎢−k⎥⎥⎥⎥ = 1(3) + 2(−k) + 3(k) = + k ⎣ ⎦ ⎣ ⎦ k Thus, the vectors are orthogonal only when k = −3 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ k ⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ k ⎥⎥⎥ (d) The vectors are orthogonal when = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ · ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ = 1(k) + 2(k) + 3(−k) + 4(0) = ⎢⎢⎣3⎥⎥⎦ ⎢⎢⎣−k⎥⎥⎦ Therefore, the vectors are always orthogonal (b) The vectors are orthogonal when = A7 (a) The scalar equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x1 + 1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ = n · (x − p) = ⎢⎢⎢⎢ ⎥⎥⎥⎥ · ⎢⎢⎢⎢ x2 − 2⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ −1 x3 + = 2(x1 + 1) + 4(x2 − 2) + (−1)(x3 + 3) = 2x1 + + 4x2 − − x3 − = 2x1 + 4x2 − x3 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.3 Length and Dot Products 19 (b) The scalar equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢3⎥⎥⎥ ⎢⎢⎢ x1 − 2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ = n · (x − p) = ⎢⎢0⎥⎥ · ⎢⎢ x2 − 5⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ x3 − = 3(x1 − 2) + 0(x2 − 5) + 5(x3 − 4) = 3x1 − + 5x3 − 20 26 = 3x1 + 5x3 (c) The scalar equation of the plane is Solution Manual for An ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x1 − 1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ = n · (x − p) = ⎢⎢−4⎥⎥ · ⎢⎢ x2 + 1⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ x3 − Introduction to Linear1 Algebra for Science and Engineering 2nd Edition by = 3(x1 − 1) + (−4)(x2 + 1) + 1(x3 − 1) − − 4x2 − + x3 − = 3x1 − 4x2 + x3 (d) The scalar equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−4⎥⎥⎥ ⎢⎢⎢ x1 − 2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ = n · (x − p) = ⎢⎢−2⎥⎥ · ⎢⎢ x2 − 1⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ −2 x3 − = (−4)(x1 − 2) + (−2)(x2 − 1) + (−2)(x3 − 1) = −4x1 + − 2x2 + − 2x3 + −12 = −4x1 − 2x2 − 2x3 A8 (a) The scalar equation of the hyperplane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢3⎥⎥⎥ ⎢⎢⎢ x1 − 1⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ = n · (x − p) = ⎢⎢⎢⎢1⎥⎥⎥⎥ · ⎢⎢⎢⎢ x2 − 1⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ x3 + = 3(x1 − 1) + 1(x2 − 1) + 4(x3 + 1) = 3x1 − + x2 − + 4x3 + = 3x1 + x2 + 4x3 (b) The scalar equation of the hyperplane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ x1 − 2⎥⎥⎥ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ x + 2⎥⎥⎥ ⎥⎥⎥ = n · (x − p) = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ · ⎢⎢⎢⎢⎢ ⎥ ⎢⎢⎣3⎥⎥⎦ ⎢⎢⎣ x3 ⎥⎥⎥⎦ x4 − = 0(x1 − 2) + 1(x2 + 2) + 3x3 + 3(x4 − 1) = x2 + + 3x3 + 3x4 − = x2 + 3x3 + 3x4 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 20 Chapter Euclidean Vector Spaces (c) The scalar equation of the hyperplane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢⎢⎢ x ⎥⎥⎥ = n · (x − p) = ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ · ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ ⎢⎢⎣ x3 ⎥⎥⎦ −2 x4 = 1x1 − 4x2 + 5x3 − 2x4 (d) The scalar equation of the hyperplane is Solution Manual for An ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x1 − 1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ = n · (x − p) = ⎢⎢⎢⎢ ⎥⎥⎥⎥ · ⎢⎢⎢⎢ x3 − 1⎥⎥⎥⎥ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ Introduction to⎢⎢⎢⎢⎢⎣−1Linear ⎥⎥⎦ ⎥⎥⎦ ⎢⎢⎣ x4 − 2Algebra x5 − 1 for Science and Engineering 2nd Edition by x2 + 2(x3 − 1) + (−1)(x4 − 2) + 1(x5 − 1) = x2 + 2x3 − − x4 + + x5 − 1 = x2 + 2x3 − x4 + x5 A9 The components of a normal vector for a hyperplane are the coefficients of the variables in the scalar equation of the hyperplane Thus: ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢ ⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢ ⎥⎥⎥ ⎢⎢ ⎥⎥⎥ ⎢ ⎢ (a) n = (b) n = ⎢⎢−2⎥⎥ (c) n = ⎢⎢ ⎥⎥ (d) n = ⎢⎢⎢⎢ ⎥⎥⎥⎥ (e) n = ⎢⎢⎢⎢−1⎥⎥⎥⎥ ⎣ ⎦ ⎣ ⎦ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢ ⎥⎦⎥ −5 ⎢⎢⎣ ⎥⎥⎦ −3 −1 A10 (a) A normal vector of any plane parallel to the given plane must be a scalar multiple of the normal vector for ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ the given plane Hence, a normal vector for the required plane is n = ⎢⎢⎢⎢−3⎥⎥⎥⎥ Therefore, an equation for the ⎣ ⎦ plane passing through P(1, −3, −1) with normal vector n is ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ x1 − 1⎥⎥⎥ ⎥ ⎢ ⎥ ⎢ = n · (x − p) = ⎢⎢⎢⎢−3⎥⎥⎥⎥ · ⎢⎢⎢⎢ x2 + 3⎥⎥⎥⎥ ⎦ ⎣ ⎦ ⎣ x3 + = 2(x1 − 1) + (−3)(x2 + 3) + 5(x3 + 1) = 2x1 − − 3x2 − + 5x3 + = 2x1 − 3x2 + 5x3 ⎡ ⎤ ⎢⎢⎢0⎥⎥⎥ ⎢ ⎥ (b) A normal vector for the required plane is n = ⎢⎢⎢⎢1⎥⎥⎥⎥ Hence, an equation for the plane passing through ⎣ ⎦ Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.3 Length and Dot Products 21 P(0, −2, 4) with normal vector n is ⎤ ⎡ ⎤ ⎡ ⎢⎢⎢0⎥⎥⎥ ⎢⎢⎢ x1 ⎥⎥⎥ ⎥ ⎢ ⎥ ⎢ = n · (x − p) = ⎢⎢⎢⎢1⎥⎥⎥⎥ · ⎢⎢⎢⎢ x2 + 2⎥⎥⎥⎥ ⎦ ⎣ ⎦ ⎣ x3 − = 0x1 + 1(x2 + 2) + 0(x3 − 4) −2 = x2 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ (c) A normal vector for the required plane is n = ⎢⎢⎢⎢−1⎥⎥⎥⎥ Hence, an equation for the plane passing through ⎣ ⎦ P(1, 2, 1) with normal vector n is ⎤ ⎡ ⎤Algebra ⎡ Solution Manual for An Introduction to Linear ⎢⎢ ⎥⎥ ⎢⎢ x1 − 1⎥⎥ for Science and Engineering 2nd Edition by ⎥⎥ ⎢⎢ ⎥⎥ ⎢⎢ = n · (x − p) = ⎢⎢⎢⎢−1⎥⎥⎥⎥ · ⎢⎢⎢⎢ x2 − 2⎥⎥⎥⎥ ⎦ ⎣ ⎦ ⎣ x3 − https://TestbankDirect.eu/ = 1(x1 − 1) + (−1)(x2 − 2) + 3(x3 − 1) = x1 − − x2 + + 3x3 − = x1 − x2 + 3x3 A11 (a) FALSE One possible counterexample is (b) Our counterexample in part (a) has u 1 · =2= · −97 so the result does not change Homework Problems B1 √ (b) 10 (e) (a) 0√ (d) 30/3 (a) √ 1/ √2 (b) 1/ ⎡ ⎤ ⎢⎢⎢2/3⎥⎥⎥ ⎢⎢⎢2/3⎥⎥⎥ (e) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ 1/3 −3/5 4/5 B2 √ ⎤ ⎡ ⎢⎢⎢−1/ 10⎥⎥⎥ ⎢ √ ⎥ (d) ⎢⎢⎢⎢⎢ 3/ 10 ⎥⎥⎥⎥⎥ ⎣ ⎦ B3 (a) B4 (a) We have x = √ 37 (c) (f) (b) √ 14 ⎡ √ ⎤ ⎢⎢⎢1/ 3⎥⎥⎥ ⎢ √ ⎥ (c) ⎢⎢⎢⎢⎢1/ 3⎥⎥⎥⎥⎥ ⎣ √ ⎦ 1/ √ ⎤ ⎡ ⎢⎢⎢−1/ 12⎥⎥⎥ √ ⎥⎥ ⎢⎢⎢ ⎢−1/ 12⎥⎥⎥⎥ √ ⎥ (f) ⎢⎢⎢⎢⎢ ⎢⎢⎢−1/√ 12⎥⎥⎥⎥⎥ ⎦ ⎣ 3/ 12 √ √ (c) 89 (d) 68 22 + (−6)2 + (−3)2 = 7, y = ⎡ ⎤ ⎢⎢⎢−1⎥⎥⎥ ⎢ ⎥ x + y = ⎢⎢⎢⎢−2⎥⎥⎥⎥ = ⎣ ⎦ (−3)2 + 42 + 52 = √ 50, and (−1)2 + (−2)2 + 22 = Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 22 Chapter Euclidean Vector Spaces Thus, x+y =3≤7+ √ 50 = x + y √ Observe that |x · y| = |2(−3) + (−6)(4) + (−3)(5)| = 45 and x y = 50 > 49 Hence, |x · y| ≤ x y (b) We have x = 42 + 12 + (−2)2 = √ 21, y = √ √ 32 + 52 + 12 = 35, and ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ x + y = ⎢⎢⎢⎢ ⎥⎥⎥⎥ = ⎣ ⎦ −1 Thus, √ 72 + 62 + (−1)2 = √ √ 86 √ x+y = 86 ≤ 21Algebra + 35 = xfor + Science y Solution Manual for An Introduction to Linear and Engineering 2nd Edition by Observe that |x · y| = 4(3) + 1(5) + (−2)(1) = 15 and x y = B5 √ √ √ 21 35 = 735 > 27 Hence, |x · y| ≤ x y (a) (b) (c) (d) (e) (f) −2 · = 1(−2) + 2(2) = 2, so they are not orthogonal 2 −3 · = 4(−3) + 6(2) = 0, so they are orthogonal ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢−4⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎣4⎥⎥⎦ · ⎢⎢⎣ ⎥⎥⎦ = 1(−4) + 4(1) + 1(−4) = −4, so they are not orthogonal −4 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎣3⎥⎥⎦ · ⎢⎢⎣−1⎥⎥⎦ = 1(3) + 3(−1) + 1(0) = 0, so they are orthogonal ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ · ⎢⎢⎢ ⎥⎥⎥ = 1(−3) + 2(0) + 1(5) + 2(1) = 4, so they are not orthogonal ⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢⎢−1⎥⎥⎥⎥ · ⎢⎢⎢⎢−1⎥⎥⎥⎥ = 2(−2) + 1(1) + (−1)(−1) + (−2)(0) + 1(2) = 0, so they are orthogonal ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎣−2⎥⎥⎦ ⎢⎢⎣ ⎥⎥⎦ B6 (a) k = (c) k = −4/5 B7 (a) −x1 + 4x2 + 7x3 = −2 (c) x1 + 2x2 + x3 = B8 (a) 3x1 − 2x2 − 5x3 + x4 = (c) x1 = (b) k = or k = (d) k = or k = (b) x1 − 2x2 + x3 = 33 (d) x1 + x2 + x3 = (b) x1 − 4x2 + x3 − 3x4 = −19 (d) x2 − 2x3 + x4 + x5 = Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.3 Length and Dot Products B9 B10 (a) ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (c) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ −1 ⎡ ⎤ ⎢⎢⎢−1⎥⎥⎥ ⎢ ⎥ (b) ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎣ ⎦ ⎡ ⎢⎢⎢ ⎢⎢⎢ ⎢⎢⎢ (e) ⎢⎢⎢⎢ ⎢⎢⎢ ⎢⎢⎣ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (d) ⎢⎢⎢⎢⎢ ⎥⎥⎥⎥⎥ ⎢⎢⎣ ⎥⎥⎦ −2 ⎤ ⎥⎥ ⎥ ⎥⎥⎥⎥⎥ ⎥ ⎥⎥⎥⎥ ⎥ ⎥⎥⎥⎥ ⎦ −1 23 ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ (f) ⎢⎢⎢⎢−2⎥⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢ ⎥⎦⎥ −1 (a) 5x1 − x2 − 2x3 = (b) 2x2 + 3x3 = −11 (c) 3x1 − 2x2 + 3x3 = (d) x1 − 5x2 + 3x3 = Computer Problems Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by C1 (a) 7.597 (c) (d) 47.0948 Conceptual Problems D1 (a) Intuitively, if there is no point of intersection, the line is parallel to the plane, so the direction vector of the line is orthogonal to the normal to the plane: d · n = Moreover, the point P is not on the plane, so p · n k (b) Substitute x = p + td into the equation of the plane to see whether for some t, x satisfies the equation of the plane n · (p + td) = k Isolate the term in t: t(n · d) = k − n · p There is one solution for t (and thus, one point of intersection of the line and the plane) exactly when n · d If n · d = 0, there is no solution for t unless we also have n · p = k In this case the equation is satisfied for all t and the line lies in the plane Thus, to have no point of intersection, it is necessary and sufficient that n · d = and n · p k D2 Since x = x − y + y, x = x − y + y = (x − y) + y ≤ x − y + y So, x − y ≤ x − y This is almost what we require, but the left-hand side might be negative So, by a similar argument with y, and using the fact that y − x = x − y , we obtain y − x ≤ x − y From this equation and the previous one, we can conclude that x − y ≤ x−y D3 We have v1 + v2 = (v1 + v2 ) · (v1 + v2 ) = v1 · v1 + v1 · v2 + v2 · v1 + v2 · v2 = v1 + + + v2 = v1 + v2 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ 24 Chapter Euclidean Vector Spaces D4 Let x be any point that is equidistant from P and Q Then x satisfies x − p = x − q , or equivalently, x − p = x − q Hence, (x − p) · (x − p) = (x − q) · (x − q) x·x−x· p− p·x− p· p= x·x−x·q−q·x+q·q −2p · x + 2q · x = q · q − p · p 2(q − p) · x = q − p This is the equation of a plane with normal vector 2(q − p) ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥ D5 (a) A point x on the plane must satisfy x − ⎢⎢2⎥⎥ = x − ⎢⎢⎢⎢ ⎥⎥⎥⎥ Square both sides and simplify ⎣ ⎦ ⎣ ⎦ Solution Manual for An Introduction and Engineering 2nd Edition by ⎛ ⎡ ⎤⎞ to ⎛ Linear ⎡ ⎤⎞ ⎛ for⎡ Science ⎡ ⎤⎞ ⎛Algebra ⎤⎞ ⎢⎢⎢2⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢⎢⎢−3⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢⎢⎢2⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢⎢⎢−3⎥⎥⎥⎟⎟⎟ ⎢⎢⎢ ⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢⎢⎢ ⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢ ⎥⎥⎥⎟⎟⎟ ⎜⎜⎜ ⎢ ⎥⎟ ⎢ ⎢ 2 ⎢⎢⎣ ⎥⎥⎦⎟⎟⎠ ⎜⎜⎝ x − ⎢⎢⎣ ⎥⎥⎦⎟⎟⎠ = ⎜⎜⎝ x − ⎢⎢⎣ ⎥⎥⎦⎟⎟⎠ · ⎜⎜⎝ x − ⎢⎢⎢⎢⎣ ⎥⎥⎥⎥⎦⎟⎟⎟⎟⎠ 5 1 ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢ ⎥ ⎢ ⎥ x · x − ⎢⎢⎢⎢2⎥⎥⎥⎥ · x + 33 = x · x − ⎢⎢⎢⎢ ⎥⎥⎥⎥ · x + 26 ⎣ ⎦ ⎣ ⎦ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎜⎜⎜⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥⎟⎟⎟ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎜⎜⎜⎜⎢⎢⎢⎢ ⎥⎥⎥⎥ − ⎢⎢⎢⎢2⎥⎥⎥⎥⎟⎟⎟⎟ · x = 26 − 33 ⎝⎣ ⎦ ⎣ ⎦⎠ ⎜⎜⎜ ⎜⎜⎜ ⎜⎜⎝ 5x1 − 2x2 + 4x3 = 7/2 ⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎜⎜⎜⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥⎟⎟⎟ ⎢⎢⎢−1/2⎥⎥⎥ ⎜ ⎢ ⎥ ⎢ ⎥ ⎟ ⎢ ⎥ (b) A point equidistant from the points is 12 ⎜⎜⎜⎜⎢⎢⎢⎢2⎥⎥⎥⎥ + ⎢⎢⎢⎢ ⎥⎥⎥⎥⎟⎟⎟⎟ = ⎢⎢⎢⎢ ⎥⎥⎥⎥ The vector joining the two points, n = ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎣⎢2⎥⎦⎥ − ⎢⎣⎢ ⎥⎦⎥ = ⎢⎣⎢−2⎥⎦⎥ must be orthogonal to the plane Thus, the equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢−1/2⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ 5x1 − 2x2 + 4x3 = ⎢⎢−2⎥⎥ · ⎢⎢ ⎥⎥⎥⎥ = 7/2 ⎣ ⎦ ⎣ ⎦ D6 Let S denote the set of all vectors orthogonal to u By definition, a vector orthogonal to u must be in Rn , so S is a subset of Rn Moreover, since u · = 0, we have that S is non-empty Let x, y ∈ S Then we have (x + y) · u = x · u + y · u = + = Therefore, S is closed under addition Similarly, for any t ∈ R we have (tx) · u = t(x · u) = t(0) = so S is also closed under scalar multiplication Therefore, S is a subspace of Rn Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐ ✐ ✐ ✐ ✐ Section 1.4 Projections and Minimum Distance 25 D7 By definition, a vector orthogonal to any vector in S must be in Rn , so S ⊥ is a subset of Rn Moreover, since v · = for any v ∈ S , we have that S ⊥ is non-empty Let x, y ∈ S ⊥ This implies that x · v = and y · v = for all v ∈ S Hence, for any v ∈ S we have (x + y) · v = x · v + y · v = + = Therefore,x + y ∈ S ⊥ Similarly, for any t ∈ R and any v ∈ S we have (tx) · v = t(x · v) = t(0) = so (tx) ∈ S ⊥ Therefore, S ⊥ is a subspace of Rn D8 Consider c1 v1 + · · · + ck vk = Then for ≤ i ≤ k we have = (ct1 o + crk vA viebra for Science and Engineering 2nd Edition by 0= v1 + k ) l· g Solution Manual for An Intro du0 c· vtiion L·i·n· ea = c1 (v1 · vi ) + · · · + ci (vi · vi ) + · · · + ck (vk · vi ) + ci vi + + ··· + The fact that vi implies that ci = Since this is valid for ≤ i ≤ k we get {v1 , , vk } is linearly independent ⎡ ⎤ ⎢⎢⎢n1 ⎥⎥⎥ ⎢ ⎥ D9 (a) Let n = ⎢⎢⎢⎢n2 ⎥⎥⎥⎥ We have n · e1 = n e1 cos α But, n = and e1 = 1, so n · e1 = cos α But, n · e1 = n1 , ⎣ ⎦ n3 ⎡ ⎤ ⎢⎢⎢cos α⎥⎥⎥ ⎢ ⎥ so n1 = cos α Similarly, n2 = cos β and n3 = cos γ, so n = ⎢⎢⎢⎢ cos β ⎥⎥⎥⎥ ⎣ ⎦ cos γ (b) cos2 α + cos2 β + cos2 γ = n = 1, because n is a unit vector cos α , where α is the angle between n and e1 and β is the angle between n cos β and e2 But in the plane α + β = π2 , so cos β = cos(π/2 − α) = sin α Now let θ = α, and we have (c) In R2 , the unit vector is n = 1= n = cos2 α + cos2 β = cos2 θ + sin2 θ 1.4 Projections and Minimum Distance Practice Problems A1 (a) We have projv u = u·v −5 0 v= = −5 v perpv u = u − projv u = 3 − = −5 −5 Copyright c 2013 Pearson Canada Inc ✐ ✐ ✐ ✐