Solutions Elementary Linear Algebra Andrilli Hecker

270 Pages·2009·3.95 MB·English

Instructor’s Manual with Sample Tests for

Elementary Linear Algebra

Fourth Edition

Stephen Andrilli and David Hecker
Dedication

To all the instructors who have used the various
editions of our book over the years
Table of Contents

Preface ........................................................................iii

Answers to Exercises………………………………….1
Chapter 1 Answer Key ……………………………….1
Chapter 2 Answer Key ……………………………...21
Chapter 3 Answer Key ……………………………...37
Chapter 4 Answer Key ……………………………...58
Chapter 5 Answer Key ……………………………...94
Chapter 6 Answer Key …………………………….127
Chapter 7 Answer Key …………………………….139
Chapter 8 Answer Key …………………………….156
Chapter 9 Answer Key …………………………….186
Appendix B Answer Key ………………………….203
Appendix C Answer Key ………………………….204

Chapter Tests ……..……………………………….206
Chapter 1 – Chapter Tests …………………………207
Chapter 2 – Chapter Tests …………………………211
Chapter 3 – Chapter Tests …………………………217
Chapter 4 – Chapter Tests …………………………221
Chapter 5 – Chapter Tests …………………………224
Chapter 6 – Chapter Tests …………………………230
Chapter 7 – Chapter Tests …………………………233

Answers to Chapter Tests ………………………….239
Chapter 1 – Answers for Chapter Tests …………...239
Chapter 2 – Answers for Chapter Tests …………...242
Chapter 3 – Answers for Chapter Tests …………...244
Chapter 4 – Answers for Chapter Tests …………...247
Chapter 5 – Answers for Chapter Tests …………...251
Chapter 6 – Answers for Chapter Tests …………...257
Chapter 7 – Answers for Chapter Tests …………...261
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Preface

This Instructor’s Manual with Sample Tests is designed
to accompany Elementary Linear Algebra, 4th edition, by
Stephen Andrilli and David Hecker.

This manual contains answers for all the computational
exercises in the textbook and detailed solutions for
virtually all of the problems that ask students for proofs.
The exceptions are typically those exercises that ask
students to verify a particular computation, or that ask for
a proof for which detailed hints have already been
supplied in the textbook. A few proofs that are extremely
trivial in nature have also been omitted.

This manual also contains sample Chapter Tests for the
material in Chapters 1 through 7, as well as answer keys
for these tests.

Additional information regarding the textbook, this
manual, the Student Solutions Manual, and linear algebra
in general can be found at the web site for the textbook,
where you found this manual.

Thank you for using our textbook.

Stephen Andrilli

David Hecker

August 2009

iii
Andrilli/Hecker - Answers to Exercises Section 1.1
Answers to Exercises
Chapter 1
Section 1.1
(1) (a) [9; 4], distance = p97 (c) [ 1; 1;2; 3; 4], distance =
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
(b) [ 5;1; 2], distance = p30 p31
(cid:0) (cid:0)
(2) (a) (3;4;2) (see Figure 1) (c) (1; 2;0) (see Figure 3)
(cid:0)
(b) (0;5;3) (see Figure 2) (d) (3;0;0) (see Figure 4)
(3) (a) (7; 13) (b) ( 6;3;2) (c) ( 1;3; 1;4;6)
(cid:0) (cid:0) (cid:0) (cid:0)
(4) (a) 16; 13;8 (b) 26;0; 6;6
3 (cid:0) 3 (cid:0) 3 (cid:0)
(cid:0) (cid:1) (cid:2) (cid:3)
(5) (a) 3 ; 5 ; 6 ; shorter, since length of original vector is > 1
p70 (cid:0)p70 p70
h i
(b) 4 ; 1 ;0; 2 ; shorter, since length of original vector is > 1
p21 p21 (cid:0)p21
(c) [h0:6; 0:8]; neither,isince given vector is a unit vector
(cid:0)
(d) 1 ; 2 ; 1 ; 1 ; 2 ; longer, since length of original vector
p11 (cid:0)p11 (cid:0)p11 p11 p11
h=p11 <1 i
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(6) (a) Parallel (c) Not parallel
(b) Parallel (d) Not parallel
(7) (a) [ 6;12;15] (c) [ 3;4;8] (e) [6; 20; 13]
(cid:0) (cid:0) (cid:0) (cid:0)
(b) [2;0; 6] (d) [ 5;1;1] (f) [ 23;12;11]
(cid:0) (cid:0) (cid:0)
(8) (a) x+y=[1;1], x y=[ 3;9], y x=[3; 9] (see Figure 5)
(cid:0) (cid:0) (cid:0) (cid:0)
(b) x+y=[3; 5], x y=[17;1], y x=[ 17; 1] (see Figure 6)
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
(c) x+y=[1;8; 5],x y=[3;2; 1],y x=[ 3; 2;1](seeFigure7)
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
(d) x+y=[ 2; 4;4],x y=[4;0;6],y x=[ 4;0; 6](seeFigure8)
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
(9) With A = (7; 3;6), B = (11; 5;3), and C = (10; 7;8), the length of
(cid:0) (cid:0) (cid:0)
side AB = length of side AC = p29. The triangle is isosceles, but not
equilateral, since the length of side BC is p30.
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Andrilli/Hecker - Answers to Exercises Section 1.1
2
Andrilli/Hecker - Answers to Exercises Section 1.1
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Andrilli/Hecker - Answers to Exercises Section 1.1
(10) (a) [10; 10] (b) [ 5p3; 15] (c) [0;0]=0
(cid:0) (cid:0) (cid:0)
(11) See Figures 1.7 and 1.8 in Section 1.1 of the textbook.
(12) See Figure 9. Both represent the same diagonal vector by the associative
law of addition for vectors.
Figure 9: x+(y+z)
(13) [0:5 0:6p2; 0:4p2] [ 0:3485; 0:5657]
(cid:0) (cid:0) (cid:25) (cid:0) (cid:0)
(14) (a) If x = [a ;a ] is a unit vector, then cos(cid:18) = a1 = a1 = a . Simi-
1 2 1 x 1 1
larly, cos(cid:18) =a . k k
2 2
(b) Analogous to part (a).
(15) Net velocity = [ 2p2; 3+2p2], resultant speed 2.83 km/hr
(cid:0) (cid:0) (cid:25)
(16) Net velocity = 3p2;8 3p2 ; speed 2.83 km/hr
(cid:0) 2 (cid:0)2 (cid:25)
h i
(17) [ 8 p2; p2]
(cid:0) (cid:0) (cid:0)
(18) Acceleration = 1 [12; 344;392] [0:0462; 0:2646;0:3015]
20 13 (cid:0) 65 65 (cid:25) (cid:0)
(19) Acceleration = 13;0;4
2 3
(20) 180[ 2;3;1] [(cid:2) 96:22;(cid:3)144:32;48:11]
p14 (cid:0) (cid:25) (cid:0)
(21) a=[ mg ; mg ]; b=[ mg ;mgp3]
(cid:0)
1+p3 1+p3 1+p3 1+p3
(22) Leta=[a ;:::;a ]. Then, a 2 =a2+ +a2 isasumofsquares,which
1 n k k 1 (cid:1)(cid:1)(cid:1) n
must be nonnegative. The square root of a nonnegative real number is a
nonnegative real number. The sum can only be zero if every a =0.
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Andrilli/Hecker - Answers to Exercises Section 1.2
(23) Let x=[x ;x ;:::;x ]. Then cx = (cx )2+ +(cx )2 =
1 2 n 1 n
k k (cid:1)(cid:1)(cid:1)
c2(x2+ +x2) = c x2+ +x2 = c x :
1 (cid:1)(cid:1)(cid:1) n j j 1 (cid:1)(cid:1)(cid:1) pn j jk k
(24) Ipn each part, suppose thapt x=[x ;:::;x ], y=[y ;:::;y ], and
1 n 1 n
z=[z ;:::;z ].
1 n
(a) x+(y+z)=[x ;:::;x ]+[(y +z );:::;(y +z )]=
1 n 1 1 n n
[(x +(y +z ));:::;(x +(y +z ))] =
1 1 1 n n n
[((x +y )+z );:::;((x +y )+z )]=
1 1 1 n n n
[(x +y );:::;(x +y )]+[z ;:::;z ] =(x+y)+z
1 1 n n 1 n
(b) x+( x)=[x ;:::;x ]+[ x ;:::; x ]=
1 n 1 n
(cid:0) (cid:0) (cid:0)
[(x + ( x ));:::;(x + ( x ))] = [0;:::;0]. Also, ( x) + x =
1 1 n n
(cid:0) (cid:0) (cid:0)
x+( x) (by part (1) of Theorem 1.3) =0, by the above.
(cid:0)
(c) c(x+y)=c([(x +y );:::;(x +y )])=[c(x +y );:::;c(x +y )]=
1 1 n n 1 1 n n
[(cx +cy );:::;(cx +cy )]=[cx ;:::;cx ]+[cy ;:::;cy ]=cx+cy
1 1 n n 1 n 1 n
(d) (cd)x=[((cd)x );:::;((cd)x )]=[(c(dx ));:::;(c(dx ))]=
1 n 1 n
c[(dx );:::;(dx )] =c(d[x ;:::;x ])=c(dx)
1 n 1 n
(25) If c = 0, done. Otherwise, (1)(cx) = 1(0) (1 c)x = 0 (by part (7) of
c c ) c (cid:1)
Theorem 1.3) x=0: Thus either c=0 or x=0.
)
(26) Let x=[x ;x ;:::;x ]. Then [0;0;:::;0] = c x c x = [(c c )x ;
1 2 n 1 2 1 2 1
(cid:0) (cid:0)
:::;(c c )x ]. Since c c =0, we have x =x = =x =0.
1 2 n 1 2 1 2 n
(cid:0) (cid:0) 6 (cid:1)(cid:1)(cid:1)
(27) (a) F (b) T (c) T (d) F (e) T (f) F (g) F
Section 1.2
(1) (a) arccos( 27 ), or about 152:6 , or 2.66 radians
(cid:0)5p37 (cid:14)
(b) arccos( 13 ), or about 63:7 , or 1.11 radians
p13p66 (cid:14)
(c) arccos(0), which is 90 , or (cid:25) radians
(cid:14) 2
(d) arccos( 1), which is 180 , or (cid:25) radians (since x= 2y)
(cid:14)
(cid:0) (cid:0)
(2) The vector from A to A is [2; 7; 3], and the vector from A to A is
1 2 1 3
(cid:0) (cid:0)
[5;4; 6]. These vectors are orthogonal.
(cid:0)
(3) (a) [a;b] [ b;a]=a( b)+ba=0: Similarly, [a; b] [b;a]=0:
(cid:1) (cid:0) (cid:0) (cid:0) (cid:1)
(b) A vector in the direction of the line ax+by+c=0 is [b; a], while
(cid:0)
a vector in the direction of bx ay+d=0 is [a;b].
(cid:0)
(4) (a) 12 joules (c) 124 joules
(cid:0)
(b) 1040p5, or 258:4 joules
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(5) Note that yz is a scalar, so x(yz) is not de(cid:133)ned.
(cid:1) (cid:1) (cid:1)
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Andrilli/Hecker - Answers to Exercises Section 1.2
(6) In all parts, let x=[x ;x ;:::;x ]; y=[y ;y ;:::;y ]; and
1 2 n 1 2 n
z=[z ;z ;:::;z ]:
1 2 n
Part (1): x y=[x ;x ;:::;x ] [y ;y ;:::;y ]=x y + +x y =
1 2 n 1 2 n 1 1 n n
(cid:1) (cid:1) (cid:1)(cid:1)(cid:1)
y x + +y x =[y ;y ;:::;y ] [x ;x ;:::;x ]=y x
1 1 n n 1 2 n 1 2 n
(cid:1)(cid:1)(cid:1) (cid:1) (cid:1)
Part (2): x x = [x ;x ;:::;x ] [x ;x ;:::;x ] = x x + +x x =
1 2 n 1 2 n 1 1 n n
(cid:1) (cid:1) (cid:1)(cid:1)(cid:1)
x2+ +x2. Nowx2+ +x2 isasumofsquares,eachofwhichmustbe
1 (cid:1)(cid:1)(cid:1) n 1 (cid:1)(cid:1)(cid:1) n
nonnegative. Hence, the sum is also nonnegative, and so its square root
2
is de(cid:133)ned. Thus, 0 x x=x2+ +x2 = x2+ +x2 = x 2.
(cid:20) (cid:1) 1 (cid:1)(cid:1)(cid:1) n 1 (cid:1)(cid:1)(cid:1) n k k
Part(3): Supposex x=0. Frompart(2), 0=(cid:16)px x=x2+ (cid:17)+x2 x2,
(cid:1) (cid:1) 1 (cid:1)(cid:1)(cid:1) n (cid:21) i
for each i, since the sum of the remaining squares (without x2), which is
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nonnegative. Hence, 0 x2 for each i. But x2 0, because it is a square.
(cid:21) i i (cid:21)
Hence each x =0. Therefore, x=0.
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Next, suppose x=0. Then x x=[0;:::;0] [0;:::;0]
(cid:1) (cid:1)
=(0)(0)+ +(0)(0)=0.
(cid:1)(cid:1)(cid:1)
Part (4): c(x y)=c([x ;x ;:::;x ] [y ;y ;:::;y ])
1 2 n 1 2 n
(cid:1) (cid:1)
=c(x y + +x y ) =cx y + +cx y =
1 1 n n 1 1 n n
(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)
[cx ;cx ;:::;cx ] [y ;y ;:::;y ]=(cx) y.
1 2 n 1 2 n
(cid:1) (cid:1)
Next, c(x y)=c(y x) (by part (1)) =(cy) x (by the above) =x (cy),
(cid:1) (cid:1) (cid:1) (cid:1)
by part (1):
Part (6): (x+y) z=([x ;x ;:::;x ]+[y ;y ;:::;y ]) [z ;z ;:::;z ]
1 2 n 1 2 n 1 2 n
(cid:1) (cid:1)
=[x +y ;x +y ;:::;x +y ] [z ;z ;:::;z ]
1 1 2 2 n n 1 2 n
(cid:1)
=(x +y )z +(x +y )z + +(x +y )z
1 1 1 2 2 2 n n n
(cid:1)(cid:1)(cid:1)
=(x z +x z + +x z )+(y z +y z + +y z ):
1 1 2 2 n n 1 1 2 2 n n
(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)
Also, (x z)+(y z)=([x ;x ;:::;x ] [z ;z ;:::;z ])
1 2 n 1 2 n
(cid:1) (cid:1) (cid:1)
+([y ;y ;:::;y ] [z ;z ;:::;z ])
1 2 n 1 2 n
(cid:1)
=(x z +x z + +x z )+(y z +y z + +y z ).
1 1 2 2 n n 1 1 2 2 n n
(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)
Hence, (x+y) z=(x z)+(y z).
(cid:1) (cid:1) (cid:1)
(7) No; consider x=[1;0], y= [0;1], and z= [1;1].
(8) A method similar to the (cid:133)rst part of the proof of Theorem 1.6 in the
textbook yields:
a b 2 0 (a a) (b a) (a b)+(b b) 0
k (cid:0) k (cid:21) ) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:1) (cid:21)
1 2(a b)+1 0 a b 1.
) (cid:0) (cid:1) (cid:21) ) (cid:1) (cid:20)
(9) Note that (x+y)(x y) = (xx)+(yx) (xy) (yy) = x 2 y 2.
(cid:1) (cid:0) (cid:1) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) k k (cid:0)k k
(10) Note that x+y 2 = x 2+2(xy)+ y 2, while
k k k k (cid:1) k k
x y 2 = x 2 2(xy)+ y 2.
k (cid:0) k k k (cid:0) (cid:1) k k
(11) (a) This follows immediately from the solution to Exercise 10 above.
(b) Note that x+y+z 2 = (x+y)+z 2
k k k k
= x+y 2 + 2((x+y)z)+ z 2
k k (cid:1) k k
= x 2+2(xy)+ y 2+2(xz)+2(yz)+ z 2.
k k (cid:1) k k (cid:1) (cid:1) k k
(c) This follows immediately from the solution to Exercise 10 above.
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