Linear Algebra and Its Applications 4th Edition – Complete Solutions Manual & Instructor’s Guide by David C. Lay (All Chapters Included)

SOLUTION MANUAL
All Chapters Included

,SOLUTIONS MANUAL FOR ELEMENTARY LINEAR ALGEBRA 4TH EDITION




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


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,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) In this case, [5, —1]— [—4, 3] = [9, —4]. And so, the desired vector is [9, —4]. The distance between
√ √
the two points = [9, —4] = 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 = [—1, —1, 2, —3, —4] =
√ √
(—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).




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, 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 23 v (which
is 2
3
the length of v) to the vector for the initial point (—4, 7, 2):
28
[—4, 7, 2] + 32 v = [—4, 7, 2] + 23[14, —17, 9] = [—4, 7, 2] + 3
, — 34
3
,6 = 16
3
, — 133 , 8 ,
16 13
so the desired point is 3, — 3 , 8 .
(5) Let v represent the given vector. Then the desired unit vector u equals 1
v v.

1 1
1 3 5 6
(a) u = v v= [3, —5, 6] = √ 32+(—5)
[3,—5,6] 2+62 [3, —5, 6] = √ 70 , — √70 , √ 70 ;

√
u is shorter than v since u = 1 ≤ 70 = v .
1 1
1
(c) u = v = [0.6, —0.8] = √ [0.6, —0.8] = [0.6, —0.8]. Neither vector is
v [0.6,—0.8] (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 34[12, —16] = [9, —12].
(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 = — 43 instead. But
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○
angle with the positive x-axis. So, as shown in Figure 1.10 in the textbook, the minute
√
hand makes
the vector v = [ v cos θ, v sin θ] = [10 cos(210 ), 10 sin(210 )] = 10 — 2 , 10 — 12
○ ○ 3
=
√
[—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].


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