Elementary Linear Algebra, Students Solutions Manual (e-only)

300 Pages·2010·4.09 MB·English

by  Stephen Andrilli

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Dedication
To all the students who have used
the various editions of our book
over the years
Table of Contents
Preface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .iv
Chapter 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Chapter 3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
Chapter 4 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .115
Chapter 5 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .154
Chapter 6 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .184
Chapter 7 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .206
Chapter 8 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .225
Chapter 9 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .266
Appendix B Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .293
Appendix C Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .295
iii
Preface
This Student Solutions Manual is designed to
accompany the fourth edition of Elementary Linear
Algebra, by Andrilli and Hecker. It contains detailed
solutions for all of the exercises in the textbook marked
(cid:2) (cid:3)
with a star ( ) or a triangle ( ). In the triangle exercises,
the student is typically asked to prove a theorem that
appears in the textbook.
The solutions presented are generally self-contained,
except that a comment may appear at the beginning of an
exercise that applies to all the solutions for each part of the
exercise. For example, the solution to Exercise 1 in
Section 1.1 begins with a description of the general
strategy used to solve problems in each of parts (a) and (c).
Then the solution is given to each part separately, following
the previously stated strategy. Therefore, you should
always check for a comment at the heading of a problem
before jumping to the part in which you are interested.
We hope you find these solutions helpful. You can find
other useful information at the web site at which you
obtained this manual.
Stephen Andrilli
David Hecker
August 2009
iv
Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Section 1.1
Solutions to Selected Exercises
Chapter 1
Section 1.1
(1) In each part, to find the vector, subtract the corresponding coordinates of the initial vector from the
terminal vector.
(a) Inthiscase,[5,−1]−[−4,3]=[9,−4]. Andso,thedesiredvectoris[9,−4]. Thedistancebetween
(cid:2) √
the two points = (cid:2)[9,−4](cid:2)= 92+(−4)2 = 97.
(c) In this case, [0,−3,2,−1,−1]−[1,−2,0,2,3]=[−1,−1,2,−3,−4]. And so, the desired vector is
[−1,−1,2,−3,−4]. The distance between the two points = (cid:2)[−1,−1,2,−3,−4](cid:2)=
(cid:2) √
(−1)2+(−1)2+22+(−3)2+(−4)2 = 31
(2) In each part, the terminal point is found by adding the coordinates of the given vector to the corre-
sponding coordinates of the initial point (1,1,1).
(a) [1,1,1]+[2,3,1]=[3,4,2] (see Figure 1), so the terminal point = (3,4,2).
(c) [1,1,1]+[0,−3,−1]=[1,−2,0] (see Figure 2), so the terminal point = (1,−2,0).
Figure 1 Figure 2
(3) In each part, the initial point is found by subtracting the coordinates of the given vector from the
corresponding coordinates of the given terminal point.
(a) [6,−9]−[−1,4]=[7,−13], so the initial point is (7,−13).
(c) [2,−1,−1,5,4]−[3,−4,0,1,−2]=[−1,3,−1,4,6], so the initial point = (−1,3,−1,4,6).
1
Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Section 1.1
(4) (a) First, we find the vector v having the given initial and terminal point by subtracting:
[10,−10,11]−[−4,7,2]=[14,−17,9]=v. Next, the desired point is found by adding 2v (which
3
is 2 the length of v) to the vector for the initial point (−4,7,2):
3 (cid:3) (cid:4) (cid:3) (cid:4)
[−4,7,2]+ 2v=[−4,7,2]+ 2[14,−17,9]=[−4,7,2]+ 28,−34,6 = 16,−13,8 ,
3 (cid:5) 3 (cid:6) 3 3 3 3
so the desired point is 16,−13,8 .
3 3
(5) Let v represent the given vector. Then the desired unit vector u equals 1 v.
(cid:2)v(cid:2)
(cid:7) (cid:8)
(a) u= 1 v= 1 [3,−5,6]= √ 1 [3,−5,6]= √3 ,−√5 ,√6 ;
(cid:2)v(cid:2) (cid:2)[3,−5,6](cid:2) 32+(−5)2+62 70 70 70
√
u is shorter than v since (cid:2)u(cid:2)=1≤ 70=(cid:2)v(cid:2).
(c) u = 1 v = 1 [0.6,−0.8] = √ 1 [0.6,−0.8] = [0.6,−0.8]. Neither vector is
(cid:2)v(cid:2) (cid:2)[0.6,−0.8](cid:2) (0.6)2+(−0.8)2
longer because u=v.
(6) Two nonzero vectors are parallel if and only if one is a scalar multiple of the other.
(a) [12,−16] and [9,−12] are parallel because 3[12,−16]=[9,−12].
4
(c) [−2,3,1] and [6,−4,−3] are not parallel. To show why not, suppose they are. Then there would
be a c ∈ R such that c[−2,3,1] = [6,−4,−3]. Comparing first coordinates shows that −2c = 6,
or c=−3. However, comparing second coordinates shows that 3c=−4, or c=−4 instead. But
3
c cannot have both values.
(7) (a) 3[−2,4,5]=[3(−2),3(4),3(5)]=[−6,12,15]
(c) [−2,4,5]+[−1,0,3]=[(−2+(−1)),(4+0),(5+3)]=[−3,4,8]
(e) 4[−1,0,3] − 5[−2,4,5] = [4(−1),4(0),4(3)] − [5(−2),5(4),5(5)] = [−4,0,12] − [−10,20,25] =
[(−4−(−10)),(0−20),(12−25)]=[6,−20,−13]
(8) (a) x+y=[−1,5]+[2,−4]=[(−1+2),(5−4)]=[1,1],x−y=[−1,5]−[2,−4]=[(−1−2),(5−(−4))]=
[−3,9], y−x=[2,−4]−[−1,5]=[(2−(−1)),((−4)−5)]=[3,−9] (see Figure 3, next page)
(c) x+y=[2,5,−3]+[−1,3,−2]=[(2+(−1)),(5+3),((−3)+(−2))]=[1,8,−5],
x−y=[2,5,−3]−[−1,3,−2]=[(2−(−1)),(5−3),((−3)−(−2))]=[3,2,−1],
y−x = [−1,3,−2]−[2,5,−3] = [((−1)−2),(3−5),((−2)−(−3))] = [−3,−2,1] (see Figure 4,
next page)
(10) In each part, consider the center of the clock to be the origin.
(a) At 12 PM, the tip of the minute hand is at (0,10). At 12:15 PM, the tip of the minute hand is
at (10,0). To find the displacement vector, we subtract the vector for the initial point from the
vector for the terminal point, yielding [10,0]−[0,10]=[10,−10].
(b) At 12 PM, the tip of the minute hand is at (0,10). At 12:40 PM, the minute hand makes a 210◦
anglewiththepositivex-axis. So,asshowninFigure1.10inthetextbook(cid:7),th(cid:9)em√inu(cid:10)teha(cid:5)ndm(cid:6)(cid:8)akes
the vector v = [(cid:2)v(cid:2)cosθ,(cid:2)v(cid:2)sinθ] = [10cos(210◦),10sin(210◦)] = 10 − 3 ,10 −1 =
√ 2 2
[−5 3,−5]. To find the displacement vecto√r, we subtract the vect√or for the initial point from the
vector for the terminal point, yielding [−5 3,−5]−[0,10]=[−5 3,−15].
2
Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Section 1.1
Figure 3 Figure 4
(13) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2, v = [(cid:2)v(cid:2)cosθ,(cid:2)v(cid:2)sinθ], where θ is
the angle v makes with the positive x-axis. Now “southwest” corresponds to 225◦, “east” corresponds
to 0◦, and northwest corresponds to 135◦. Hence, if v1, v2, and v3 are the vectors corresponding to
the three given movements, then(cid:7) (cid:8)
√ √
v1 =[1cos(225◦),1sin(225◦)]= − (cid:7)22,−(cid:9)22 , (cid:10)v2 =[0.5(cid:8)cos(0◦),0.5sin(0◦)]=[0.5,0], and
√ √
v3 =[0.2cos(1(cid:7)35◦√),0.2√sin(cid:8)(135◦)]= (cid:7)0.2 (cid:9)− √22 (cid:10),0.2 √22 (cid:8). Hence, the√result of√all three putts is
v1+v2+v3 = − 22,− 22 +[0.5,0]+ 0.2 − 22 ,0.2 22 =[0.5−0.6 2,−0.4 2]≈[−0.3485,−0.5657].
(15) AsshowninFigure1.10inthetextbook,foranyvectorv∈R2,v=[(cid:2)v(cid:2)cosθ,(cid:2)v(cid:2)sinθ],whereθisthe
angle v makes with the positive x-axis. Now “northwestward” corresponds to 135◦, and “southward”
corresponds to 270◦. Hence, if v(cid:7)1 c(cid:9)orr√esp(cid:10)ond√s t(cid:8)o th(cid:3)e ro√wer,√and(cid:4) v2 corresponds to the current, then
v1 =[4cos(135◦),4sin(135◦)]= 4 − 22 ,4 22 = −2 2,2 2 and v2 =[3cos(270◦),3sin(270◦)]=
(cid:3) √ √ (cid:4) √ √
[0,−3]. Therefore,therower’snetvelocity=v1+v2 = −2 2,2 2 +[0,−3]=[−2 2,−3+2 2]. The
(cid:11) √ √ (cid:11) (cid:12) √ (cid:5) √ (cid:6) (cid:2) √
resultant speed =(cid:11)[−2 2,−3+2 2](cid:11)= (−2 2)2+ −3+2 2 2 = 25−12 2≈ 2.83 km/hr.
(17) If v1 corresponds to the rower, andv2 corresponds to the current, andv3 corresponds to the resultant
velocity, then v3 = v1+v2, or v1 = v3−v2. Now, as shown in Figure 1.10 in the textbook, for any
vector v ∈ R2, v = [(cid:2)v(cid:2)cosθ,(cid:2)v(cid:2)sinθ], where θ is the angle v makes with the positive x-axis. Also,
“westward” corresponds to 180◦(cid:7), a√nd “√nor(cid:8)thea(cid:3)s√twar√d”(cid:4)corresponds to 45◦. Thus,
v2 = [2cos(45◦),2sin(45◦)] = 2 22,2 22 = 2, 2 and v3 = [8cos(180◦),8sin(180◦)] = [−8,0].
(cid:3)√ √ (cid:4) √ √
Hence, v1 =v3−v2 =[−8,0]− 2, 2 =[−8− 2,− 2].
(18) Let f and a represent the resultant force and the resultant acceleration, respectively. Then f =
R R R
3
Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Section 1.1
(cid:3) (cid:4)
f1+f2+f3. Now, f1 =4(cid:2)[[33,,−−1122,,44]](cid:2) = √32+(−412)2+42 [3,−12,4]= 143[3,−12,4]= 1123,−4183,1163 ,
(cid:3) (cid:4)
f2 =2(cid:2)[[00,,−−44,,−−33]](cid:2) = √02+(−42)2+(−3)2 [0,−4,−3]= 25[0,−4,−3]= 0,−85,−65 , and
(cid:3) (cid:4) (cid:3) (cid:4)
f(cid:3)3 =6(cid:2)[[00,,00,,11](cid:2) =(cid:4) 61[0,0,1]=[0,0,6]. Therefore,fR =f1+f2+f3 = 1123,−4183,1163 + 0,−85,−65 +[0,0,6]=
12,−344,392 . Finally, f = ma . Since m = 20, acceleration = a = 1f = 1 [12,−344,392] ≈
13 65 65 R R R m R 20 13 65 65
[0.0462,−0.2646,0.3015].
(21) As shown in Figure 1.10 in the textbook, for any vector v ∈ R2, v = [(cid:2)v(cid:2)cosθ,(cid:2)v(cid:2)sinθ], where θ is
the angle v makes with the positive x-axis. Thus, if we let r =(cid:2)a(cid:2) and s=(cid:2)b(cid:2), then
(cid:7) (cid:8) (cid:7) (cid:8)
√ √ √
a = [rcos(135◦),rsin(135◦)] = −r 2,r 2 and b = [scos(60◦),ssin(60◦)] = s1,s 3 . Now, since
2 2 (cid:7) (cid:8) (cid:7) (cid:8) 2 2
√ √ √
the weight is not moving, a+b+g=0. That is, −r 2,r 2 + s1,s 3 +[0,−mg]=[0,0]. Using
2 2 2 2
√ √
the first coordinates gives us −r 2 +s1 =0. Hence, s= 2r. Using the second coordinates gives us
√ √ 2 2 √ √ √ √
r 2 +s 3 −mg =0. Substituting in s= 2r yields r 2 + 2r 3 −mg =0. Solving for r produces
2 2 √ 2 2
r = √ 2mg√ , and s = 2r = 2m√g . Plugging these values in to the formulas for a and b gives
2(1+ 3) 1+ 3
√
a=[−m√g , m√g ] and b=[ m√g ,mg√3].
1+ 3 1+ 3 1+ 3 1+ 3
(cid:2) (cid:2)
(23) Let x(cid:2)=[x1,x2,...,xn]. Then (cid:2)cx(cid:2)= (cx1)2+···+(cxn)2 = c2(x21+···+x2n)
=|c| x2+···+x2 =|c|(cid:2)x(cid:2).
1 n
(24) In each part, let x=[x1,x2,...,xn], y=[y1,y2,...,yn], and c∈R.
(b) x+(−x) = [x1,...,xn]+[−x1,...,−xn] = [(x1 +(−x1)),...,(xn +(−xn))] = [0,...,0] = 0.
Also, (−x)+x=x+(−x) (by part (1) of Theorem 1.3) =0, by the above.
(c) c(x+y)=c([(x1+y1),...,(xn+yn)])=[c(x1+y1),...,c(xn+yn)]=[(cx1+cy1),...,(cxn+cyn)]=
[cx1,...,cxn]+[cy1,...,cyn]=cx+cy.
(25) If c = 0, we are done. Otherwise, cx = 0 ⇒ (1)(cx) = 1(0) ⇒ (1 ·c)x = 0 (by part (7) of Theorem
c c c
1.3) ⇒x=0. Thus, either c=0 or x=0.
(cid:2)
(27) (a) False. Thelengthof[a1,a2,a3]is a21+a22+a23. Theformulagivenintheproblemismissingthe
square root. So, for example, the length of [0,3,4] is actually 5, not 02+32+42 =25.
(b) True. This is easily proved using parts (1) and (2) of Theorem 1.3.
(c) True. [2,0,−3]=2[1,0,0]+(−3)[0,0,1].
(d) False. Two nonzero vectors are parallel if and only if one is a scalar multiple of the other. To
showwhy[3,−5,2]and[6,−10,5]arenotparallel,supposetheyare. Thentherewouldbeac∈R
such that c[3,−5,2] = [6,−10,5]. Comparing first coordinates shows that 3c = 6, or c = 2. But
this value of c must work in all coordinates. However, it does not work in the third coordinate,
since (2)(2)(cid:8)=5.
(e) True. Multiply both sides of dx=0 by 1.
d
(f) False. Parallel vectors can be in opposite directions. For example, [1,0] and [−2,0] are parallel
but are not in the same direction.
(g) False. The properties of vectors in Theorem 1.3 are independent of the location of the initial
points of the vectors.
4
Student Manual for Andrilli and Hecker - Elementary Linear Algebra, 4th edition Section 1.2
Section 1.2
x·y
(1) In each part, we use the formula cosθ = .
(cid:2)x(cid:2)(cid:2)y(cid:2)
(a) cosθ = x·y = [−4,3]·[6,−1] = √ (−4)(6)+√(3)(−1) = −√27 . And so, using a calculator yields
(cid:2)x(cid:2)(cid:2)y(cid:2) (cid:2)[−4,3](cid:2)(cid:2)[6,−1](cid:2) (−4)2+32 62+(−1)2 5 37
θ = arccos(− √27 ), or about 152.6◦, or 2.66 radians.
5 37
(c) cosθ = x·y = [7,−4,2]·[−6,−10,1] = √ (7)(−6)+(−√4)(−10)+(2)(1) = √ 0√ = 0. And so,
(cid:2)x(cid:2)(cid:2)y(cid:2) (cid:2)[7,−4,2](cid:2)(cid:2)[−6,−10,1](cid:2) 72+(−4)2+22 (−6)2+(−10)2+12 69 137
θ = arccos(0), which is 90◦, or π radians.
2
(4) When a force f moves an object by a displacement d, the work performed is f ·d.
(b) First, we need to compute f. A unit vector in the direction of f is u = 1 [−2,4,5] =
(cid:2)[−2,4,5](cid:2)
√ 1 [−2,4,5] = √1 [−2,4,5]. Thus, since (cid:2)f(cid:2) = 26, f = 26u = 2√6 [−2,4,5] =
(cid:7) (−2)2+42+52 (cid:8) 3 5 3 5
√ √ √
−52 5,104 5,26 5 , where we have rationalized the denominators.
15 15 3
Next, we need to compute d. A unit vector in the direction of d is v = 1 [−1,2,2] =
(cid:3) (cid:4) (cid:3) (cid:2)[−1,2,2(cid:4)](cid:2)
√ 1 [−1,2,2]= −1,2,2 . Thus, since (cid:2)d(cid:2) =10, d=10v= −10,20,20 .
(−1)2+22+22 3 3 3 3 3 3
Hence, the work performed is
(cid:7) √ √ √ (cid:8) (cid:3) (cid:4) (cid:9) √ (cid:10)(cid:5) (cid:6) (cid:9) √ (cid:10)(cid:5) (cid:6) (cid:9) √ (cid:10)(cid:5) (cid:6)
f · d = −52 5,104 5,26 5 · −10,20,20 = −52 5 −10 + 104 5 20 + 26 5 20
15 15 3 3 3 3 15 3 15 3 3 3
(cid:5) √ √ √ (cid:6) √
= 1 104 5+416 5+520 5 = 1040 5, or 258.4 joules.
9 9
(6) In all parts, let x=[x1,x2,...,xn], y=[y1,y2,...,yn], and z=[z1,z2,...,zn].
Part (1): x·y=[x1,x2,...,xn]·[y1,y2,...,yn]=x1y1+···+xnyn =y1x1+···+ynxn =
[y1,y2,...,yn]·[x1,x2,...,xn]=y·x
Part(2): x·x=[x1,x2,...,xn]·[x1,x2,...,xn]=x1x1+···+xnxn =x21+···+x2n. Nowx21+···+x2n
is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and so
(cid:9)(cid:2) (cid:10)2
its square root is defined. Thus, 0≤x·x=x2+···+x2 = x2+···+x2 =(cid:2)x(cid:2)2.
1 n 1 n
Part (3): Suppose x·x = 0. From part (2), 0 = x2+···+x2 ≥ x2, for each i, since the sum of the
1 n i
remaining squares (without x2) is nonnegative. Hence, 0 ≥ x2 for each i. But x2 ≥ 0, because it is a
i i i
square. Hence, each x =0. Therefore, x=0.
i
Next, suppose x=0. Then x·x=[0,...,0]·[0,...,0]=(0)(0)+···+(0)(0)=0.
Part (4): c(x·y)=c([x1,x2,...,xn]·[y1,y2,...,yn])=c(x1y1+···+xnyn) =cx1y1+···+cxnyn =
[cx1,cx2,...,cxn]·[y1,y2,...,yn]=(cx)·y.
Next, c(x·y)=c(y·x) (by part (1)) =(cy)·x (by the above) =x·(cy), by part (1).
Part (6): (x+y)·z=([x1,x2,...,xn]+[y1,y2,...,yn])·[z1,z2,...,zn]
=[x1+y1,x2+y2,...,xn+yn]·[z1,z2,...,zn]=(x1+y1)z1+(x2+y2)z2+···+(xn+yn)zn
=(x1z1+x2z2+···+xnzn)+(y1z1+y2z2+···+ynzn).
Also, (x·z)+(y·z)=([x1,x2,...,xn]·[z1,z2,...,zn])+([y1,y2,...,yn]·[z1,z2,...,zn])
=(x1z1+x2z2+···+xnzn)+(y1z1+y2z2+···+ynzn). Hence, (x+y)·z=(x·z)+(y·z).
(7) No. Consider x = [1,0], y = [0,1], and z = [1,1]. Then x·z = [1,0]·[1,1] = (1)(1)+(0)(1) = 1 and
y·z=[0,1]·[1,1]=(0)(1)+(1)(1)=1, so x·z=y·z. But x(cid:8)=y.
5