TEST BANK FOR Introduction to Linear algebra for science and engineering 2nd Edition By Daniel Norman, Dan Wolczuk

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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     x2
  1 2  
+     3
1 4 3
4
3
−
4 2 4
  2 1
2 1
3  
4 x1
−
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
 x2     x2
−1 2 3
3 2 −2
4 1 −1
     
3 2 2
−2 1 2
  −1 1
−1
4   x
3 1

−1
x1
 
             
4 −1 4 + (−1) 3 −3 −2 −3 − (−2) −1
A2 (a) + = = (b) − = =
−2 3 −2 + 3 1 −4 5 −4 − 5 −9
               
3 (−2)3 −6 2 4 1 4/3 7/3
(c) −2 = = (d) 12 + 13 = + =
−2 (−2)(−2) 4 6 3 3 1 4
          √         
3 1/4 2 1/2 3/2 √ 2 1 2 3 5
(e) 23 −2 = − = (f) 2 √ +3 √ = √ + √ = √
1 1/3 2/3 2/3 0 3 6 6 3 6 4 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 4 − (−2) 6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 2 ⎥⎥⎥ ⎢⎢⎢−3⎥⎥⎥ ⎢⎢⎢ 2 + (−3) ⎥⎥⎥ ⎢⎢⎢ −1 ⎥⎥⎥
⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥
(b) ⎢⎢⎣ 1 ⎥⎥⎦ + ⎢⎢⎣ 1 ⎥⎥⎦ = ⎢⎢⎣ 1 + 1 ⎥⎥⎦ = ⎢⎢⎣ 2 ⎥⎥⎥⎥⎦
−6 −4 −6 + (−4) −10
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 4 ⎥⎥⎥ ⎢⎢⎢ (−6)4 ⎥⎥⎥ ⎢⎢⎢−24⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(c) −6 ⎢⎢⎢⎢−5⎥⎥⎥⎥ = ⎢⎢⎢⎢(−6)(−5)⎥⎥⎥⎥ = ⎢⎢⎢⎢ 30 ⎥⎥⎥⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−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 4 1/3 13/3
⎡ ⎤ ⎡ ⎤ ⎡√ ⎤ ⎡ ⎤ ⎡√ ⎤
√ ⎢⎢⎢⎢1⎥⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢ √2⎥⎥⎥ ⎢⎢⎢−π⎥⎥⎥ ⎢⎢⎢ 2√− π⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(f) 2 ⎢⎢⎢⎢1⎥⎥⎥⎥ + π ⎢⎢⎢⎢ 0 ⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ 2⎥⎥⎥⎥⎥ + ⎢⎢⎢⎢ 0 ⎥⎥⎥⎥ = ⎢⎢⎢⎢⎢ 2 ⎥⎥⎥⎥⎥
⎣ ⎦ ⎣ ⎦ ⎣√ ⎦ ⎣ ⎦ ⎣√ ⎦
1 1 2 π 2+π
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 2 ⎥⎥⎥ ⎢⎢⎢ 6 ⎥⎥⎥ ⎢⎢⎢ −4 ⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
A4 (a) 2v − 3 w = ⎢⎢⎢⎢ 4 ⎥⎥⎥⎥ − ⎢⎢⎢⎢−3⎥⎥⎥⎥ = ⎢⎢⎢⎢ 7 ⎥⎥⎥⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−4 9 −13
⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎜⎜⎜⎢⎢⎢ 1 ⎥⎥⎥ ⎢⎢⎢ 4 ⎥⎥⎥⎟⎟⎟ ⎢⎢⎢ 5 ⎥⎥⎥ ⎢⎢⎢5⎥⎥⎥ ⎢⎢⎢ 5 ⎥⎥⎥ ⎢⎢⎢−15⎥⎥⎥ ⎢⎢⎢ 5 ⎥⎥⎥ ⎢⎢⎢−10⎥⎥⎥
⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(b) −3(v + 2 w) + 5v = −3 ⎜⎜⎜⎜⎢⎢⎢⎢ 2 ⎥⎥⎥⎥ + ⎢⎢⎢⎢−2⎥⎥⎥⎥⎟⎟⎟⎟ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = −3 ⎢⎢⎢⎢0⎥⎥⎥⎥ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 0 ⎥⎥⎥⎥ + ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 10 ⎥⎥⎥⎥
⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−2 6 −10 4 −10 −12 −10 −22
 − 2u = 3v, so 2u = w
(c) We have w  − 3v or u = 12 (
w − 3v). 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 − 3v = 2u, so u = −3v = ⎢⎢⎢⎢−6⎥⎥⎥⎥.
⎣ ⎦
6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢3/2⎥⎥⎥ ⎢⎢⎢ 5/2 ⎥⎥⎥ ⎢⎢⎢ 4 ⎥⎥⎥
⎢⎢⎢ ⎥⎥⎥ ⎢⎢⎢ ⎥ ⎢ ⎥
 +  = ⎢⎢⎣1/2⎥⎥⎦ + ⎢⎢⎣−1/2⎥⎥⎥⎥⎦ = ⎢⎢⎢⎢⎣ 0 ⎥⎥⎥⎥⎦
1 1
A5 (a) 2 v 2 w
1/2 −1 −1/2
⎡ ⎤ ⎛⎡ ⎤ ⎡ ⎤⎞ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 8 ⎥⎥⎥ ⎜⎜⎜⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢ 15 ⎥⎥⎥⎟⎟⎟ ⎢⎢⎢ 16 ⎥⎥⎥ ⎢⎢⎢−9⎥⎥⎥ ⎢⎢⎢ 25 ⎥⎥⎥
⎢ ⎥ ⎜⎢ ⎥ ⎢ ⎥⎟ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(b)  ) − (2v − 3
2(v + w w) = 2 ⎢⎢⎢⎢ 0 ⎥⎥⎥⎥ − ⎜⎜⎜⎜⎢⎢⎢⎢2⎥⎥⎥⎥ − ⎢⎢⎢⎢−3⎥⎥⎥⎥⎟⎟⎟⎟ = ⎢⎢⎢⎢ 0 ⎥⎥⎥⎥ − ⎢⎢⎢⎢ 5 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ −5 ⎥⎥⎥⎥
⎣ ⎦ ⎝⎣ ⎦ ⎣ ⎦⎠ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−1 2 −6 −2 8 −10
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 5 ⎥⎥⎥ ⎢⎢⎢6⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(c) We have w  − u = 2v, so u = w  − 2v. This gives u = ⎢⎢⎢⎢−1⎥⎥⎥⎥ − ⎢⎢⎢⎢2⎥⎥⎥⎥ = ⎢⎢⎢⎢−3⎥⎥⎥⎥.
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−2 2 −4


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Section 1.1 Vectors in R2 and R3
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 10 ⎥⎥⎥ ⎢⎢⎢ 2 ⎥⎥⎥ ⎢⎢⎢ 8 ⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
(d) We have 12 u + 13 v = w
 , so 12 u = w w − 23 v = ⎢⎢⎢⎢−2⎥⎥⎥⎥ − ⎢⎢⎢⎢2/3⎥⎥⎥⎥ = ⎢⎢⎢⎢ −8/3 ⎥⎥⎥⎥.
 − 13 v, or u = 2
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−4 2/3 −14/3
A6
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 3 ⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ 1 ⎥⎥⎥
 = OQ
PQ  = ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥
 − OP
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
−2 1 −3
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥
 = OR
PR  = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥
 − OP
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
0 1 −1
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥
 = OS
PS  = ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥ − ⎢⎢⎢⎢3⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥
 − OP
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
5 1 4
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢ 3 ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥
 = OR
QR  = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 3 ⎥⎥⎥⎥
 − OQ
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
0 −2 2
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢1⎥⎥⎥ ⎢⎢⎢−5⎥⎥⎥ ⎢⎢⎢ 6 ⎥⎥⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
S R = OR − OS = ⎢⎢⎢⎢4⎥⎥⎥⎥ − ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 3 ⎥⎥⎥⎥
  
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
0 5 −5

Thus, ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢⎢⎢ 1 ⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥ ⎢⎢⎢−1⎥⎥⎥ ⎢⎢⎢−7⎥⎥⎥ ⎢⎢⎢ 6 ⎥⎥⎥
PQ + QR  = ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ 3 ⎥⎥⎥⎥ = ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥ = ⎢⎢⎢⎢−2⎥⎥⎥⎥ + ⎢⎢⎢⎢ 3 ⎥⎥⎥⎥ = PS
 + SR
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
−3 2 −1 4 −5
   
3 −5
A7 (a) The equation of the line is x = +t ,t∈R
4 1
   
2 −4
(b) The equation of the line is x = +t ,t∈R
3 −6
⎡ ⎤ ⎡ ⎤
⎢⎢⎢2⎥⎥⎥ ⎢⎢⎢ 4 ⎥⎥⎥
⎢ ⎥ ⎢ ⎥
(c) The equation of the line is x = ⎢⎢⎢⎢0⎥⎥⎥⎥ + t ⎢⎢⎢⎢ −2 ⎥⎥⎥⎥, t ∈ R
⎣ ⎦ ⎣ ⎦
5 −11
⎡ ⎤ ⎡ ⎤
⎢⎢⎢4⎥⎥⎥ ⎢⎢⎢−2⎥⎥⎥
⎢⎢⎢ ⎥⎥⎥ ⎢ ⎥
(d) The equation of the line is x = ⎢⎢1⎥⎥ + t ⎢⎢⎢⎢ 1 ⎥⎥⎥⎥, t ∈ 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 points: d =
   
−1 3
= . This, along with one of the points, may be used to obtain an equation for the line
2 −5
   
−1 3
x = +t , t∈R
2 −5


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c 2013 Pearson Canada Inc.