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Students’ Solutions Manual

Linear Algebra

This manual contains solutions to odd-numbered exercises from the book Linear Algebra by MiroslavLovric, published by Nelson Publishing.

Keep in mind that the solutions provided represent one way of answering a question or solving anexercise. In many cases there are alternatives, so make sure that you don’t dismiss your solution justbecause it does not look like the solution in this manual.

This solutions manual is not meant to be read! Think, try to solve an exercise on your own, investigatedifferent approaches, experiment, see how far you get. If you get stuck and don’t know how to proceed,try to understand why you are having difficulties before looking up the solution in this manual. If youjust read a solution you might fail to recognize the hard part(s); even worse, you might completelymiss the point of the exercise.

I accept full responsibility for errors in this text and will be grateful to anybody who brings them tomy attention. Your comments and suggestions will be greatly appreciated.

Miroslav Lovric September 2011Department of Mathematics and StatisticsMcMaster Universitye-mail: [email protected]

Section 1 [Solutions] L1-1

Section 1 Identifying Location in a Plane and in Space

1. The distance from a point P = (x, y) to the origin is given by

d(P, 0) =√

(x − 0)2 + (y − 0)2 =√

x2 + y2

The distance from the point A = (1, 3) to the origin is d(A, 0) =√

(1)2 + (3)2 =√

10. Likewise,d(B, 0) =

√(0)2 + (4)2 =

√16 = 4 and d(C, 0) =

√(2)2 + (2)2 =

√8. Thus, C is closest to the

origin.

3. From

d((3,−1), (a, 1)) =√

(3 − a)2 + (−1 − 1)2 =√

a2 − 6a + 9 + 4 = 10

we get a2 − 6a + 13 = 100 and a2 − 6a − 87 = 0. Thus,

a =6 ±

√36 + 4 · 87

2=

6 ± 2√

9 + 872

i.e., a = 3 ±√

96 = 3 ±√

16 · 6 = 3 ± 4√

6.

5. The distance is

d =√

(4 − 1)2 + (5 − 2)2 + (6 − 3)2 =√

9 · 3 = 3√

3

7. From

d((3,−1, 2), (9, 0,−4)) =√

(6)2 + (1)2 + (−6)2 =√

73

d((0, 0, 6), (9, 0,−4)) =√

(9)2 + (0)2 + (−10)2 =√

181

we see that (3,−1, 2) is closer to (9, 0,−4).

9. Note that π/2 ≈ 1.57, π/4 ≈ 0.79, so the angle of 1 radian is between π/4 and π/2, closer to theangle of π/4. All three points lie on a circle of radius 1 centred at the origin. See the figure below.

0

(1,π/4)

(1,π/2)(1,1)

x1

y

11. The Cartesian coordinates are (10 cosπ, 10 sinπ) = (−10, 0).

13. The Cartesian coordinates are (10 cos(3π/2), 10 sin(3π/2)) = (0,−10).

15. The Cartesian coordinates are (4 cos 4, 4 sin 4) ≈ (−2.615,−3.027).

17. Let A = (4, π/4) and B = (1, 3π/4) (in polar coordinates). Their Cartesian coordinates areA = (4 cos(π/4), 4 sin(π/4)) = (2

√2, 2

√2) and B = (cos(3π/4), sin(3π/4)) = (−

√2/2,

√2/2), and

their distance is

d(A, B) =

√√√√(2√

2 +√

22

)2

+

(2√

2 −√

22

)2

L1-2 Linear Algebra [Solutions]

=

√√√√(8 + 2(2

√2)

√2

2+

12

)+

(8 − 2(2

√2)

√2

2+

12

)=

√17

Note: in Exercise 42 we derive the distance formula in polar coordinates (in which case no conversionto Cartesian coordinates is needed). Using that formula [with A = (r1 = 4, θ1 = π/4) and B = (r2 =1, θ2 = 3π/4)], we obtain

d(A, B) =√

r21 + r2

2 − 2r1r2 cos(θ1 − θ2) =√

(4)2 + (1)2 − 2(4)(1) cos(−π/2) =√

17

since cos(−π/2) = 0.

19. We compute r =√

(−5)2 + (5)2 =√

50 = 5√

2 and tan θ = 5/(−5) = −1. We get θ = −π/4+kπ,and since the given point lies in the second quadrant, θ = −π/4 + π = 3π/4. The polar coordinatesof (−5, 5) are (5

√2, 3π/4).


(1)2 + (√

3)2 =√

4 = 2 and tan θ =√

3. We get θ = π/3 + πk, and since the

given point lies in the first quadrant, θ = π/3. The polar coordinates of (1,√

3) are (2, π/3).


(−2)2 + (5)2 =√

29 and tan θ = −5/2. Using a calculator we get θ ≈−1.190+πk, and since the given point lies in the second quadrant, θ ≈ −1.190+π ≈ 1.951. The polarcoordinates of (−2, 5) are (approximately) (

√29, 1.951).

25. All three points have the second coordinate equal to 2. They belong to the curve θ = 2, which isa line through the origin which makes an angle of 2 radians with respect to the positive x-axis.

27. We get θ = −π/3 + πk; since θ is in the second quadrant, θ = −π/3 + π = 2π/3.

29. We get θ = −π/4 + πk; since θ is in the fourth quadrant, θ = −π/4 + 2π = 7π/4.

31. We get θ = −π/4 + πk; since θ is in the second quadrant, θ = −π/4 + π = 3π/4.

33. We get r =√

1 + 1 =√

2, and tan θ = 1/(−1) = −1, i.e., θ = −π/4 + π = 3π/4. The polarcoordinates are (

√2, 3π/4).

35. We get r =√

1 + 3 = 2, and tan θ =√

3, i.e., θ = π/3 + π = 4π/3. The polar coordinates are(2, 4π/3).

37. We get r =√

3 + 1 = 2, and tan θ = 1/√

3, i.e., θ = π/6 + π = 7π/6. The polar coordinates are(2, 7π/6).

39. The midpoint between (−2, 3) and (3, 5) is ((−2 + 3)/2, (3 + 5)/2) = (1/2, 4). Thus, a = 1/2.

41. Let A = (a1, a2), B = (b1, b2), and M = ((a1 + b1)/2, (a2 + b2)/2). We show that d2(A, M) =d2(M, B), which implies that d(A, M) = d(M, B). Computing the common denominator within eachset of parentheses, we get

d2(A, M) =(

a1 −a1 + b1

2

)2

+(

a2 −a2 + b2

2

)2

=(

a1 − b1

2

)2

+(

a2 − b2

2

)2

d2(M, B) =(

a1 + b1

2− b1

)2

+(

a2 + b2

2− b2

)2

=(

a1 − b1

2

)2

+(

a2 − b2

2

)2


43. What can we say about integers a and b if they satisfy a2 +b2 < 52? Clearly, a2 < 52 and b2 < 52;so, a and b are some numbers satisfying −5 < a < 5 and −5 < b < 5. Not all combinations of suchnumbers work: for instance, a = 4 and b = −2 satisfy a2 + b2 = 16 + 4 = 20 < 25; however, if a = −4and b = 3 then a2 + b2 = 16 + 9 = 25 (and not smaller than 25).

Let (r, f) represent the current year’s population of rabbits and foxes. Then

d((r, f), (107, 12)) =√

(r − 107)2 + (f − 12)2 < 5

i.e., (r − 107)2 + (f − 12)2 < 52. Take a = r − 107 and b = f − 12. Thus, we can conclude that themost either population changes is by ±4 members. For instance, there could be 4 more rabbits and 2fewer foxes. However, 4 fewer rabbits and 3 extra foxes cannot happen.

We argue the remaining cases analogously. f the distance is less than 50, then the most either foxesor rabbits population can change is by ±49. In this case, large changes in the numbers are possible.For instance, it could happen that foxes completely disappear.

If the distance is larger than 100, then the changes in the population numbers (of either foxes, orrabbits, or both foxes and rabbits) are large.


Section 2 Vectors

1. Let A = (10, 3) and C = (−4,−7). The representative directed line segment of v = [3,−1] whichstarts at A is −−→AB, where B = (10+3, 3− 1) = (13, 2). The representative directed line segment whichstarts at C is −−→

CD, where D = (−4 + 3,−7 − 1) = (−1,−8).

3. The vector v is given by v = [3 − (−4),−2 − (−1)] = [7,−1]. The representative directed linesegment of v that starts at C = (0, 1) is −−→

CD, where D = (0 + 7, 1 − 1) = (7, 0).

5. We get v = [‖v‖ cos(3π/4), ‖v‖ sin(3π/4)] = [15(−√

2/2), 15(√

2/2)] = [−15√

2/2, 15√

2/2].

7. Using ‖tv‖ = |t| ‖v‖ with t = −4‖v‖, we get

‖w‖ =∥∥ (−4‖v‖v)

∥∥ =∣∣ − 4‖v‖

∣∣ ‖v‖ = 4‖v‖ ‖v‖ = 4‖v‖2 = 4(3)2 = 36

Alternatively, simplify w first: w = −4‖v‖v = −12‖v‖; thus, ‖w‖ = | − 12| ‖v‖ = 12 · 3 = 36.

9. We are looking for a multiple w of v = [−1, 4, 1] whose length is 10. We know that the vectorv/‖v‖ is a unit vector parallel to v. Thus, w = 10 v

‖v‖ . Since ‖v‖ =√

1 + 16 + 1 =√

18 = 3√

2, itfollows that w = 10

3√

2[−1, 4, 1].

11. We compute 3a− b + j = 3[3,−2] − [−1, 1] + [0, 1] = [9,−6] + [1,−1] + [0, 1] = [10,−6].

13. From a + i = [3,−2] + [1, 0] = [4,−2], we get ‖a + i‖ =√

16 + 4 =√

20 = 2√

5.

15. We get r =√

2 and tan θ = −1; since (−1, 1) is in the second quadrant, θ = 3π/4. Thus,b = [

√2 cos(3π/4),

√2 sin(3π/4)].

17. Since ‖b‖ =√

2, the unit vector in the direction of b is 1√2b = 1√

2[−1, 1].

19. From d = αb + βc we get[04

]= α

[−11

]+ β

[−10

]and

[04

]=

[−α − β

α

]

Thus −α − β = 0 and α = 4; it follows that β = −4.

21. From a + c = [3,−2] + [−1, 0] = [2,−2] we compute ‖a + c‖ =√

4 + 4 =√

8. On the other hand,‖a‖ + ‖c‖ =

√9 + 4 +

√1 + 0 =

√13 + 1.

23. From [5, y] = x[4, 1] = [4x, x] we conclude that 5 = 4x and y = x. Thus, x = y = 5/4.

25. Using properties of vector operations,

3(2a− x − b) = b − 2(a + 7x)6a − 3x− 3b = b − 2a − 14x

−3x + 14x = b − 2a − 6a + 3b11x = 4b − 8a

x =4b − 8a

11=

111

(4b − 8a)


27. We compute

−2a + 3j − 4b = −2

⎡⎣ 4

0−2

⎤⎦ + 3

⎡⎣ 0

10

⎤⎦ − 4

⎡⎣−1

60

⎤⎦ =

⎡⎣−8

04

⎤⎦ +

⎡⎣ 0

30

⎤⎦ +

⎡⎣ 4−24

0

⎤⎦ =

⎡⎣ −4−21

4

⎤⎦

29. Because ‖b‖ =√

1 + 36 + 0 =√

37, the unit vector in the direction of b is given by

u =1

‖b‖ b =1√37

⎡⎣−1

60

⎤⎦ =

⎡⎣−1/

√37

6/√

370

⎤⎦

31. From a = αc + βi + γk we get⎡⎣ 4

0−2

⎤⎦ = α

⎡⎣ 2

7−4

⎤⎦ + β

⎡⎣ 1

00

⎤⎦ + γ

⎡⎣ 0

01

⎤⎦ and

⎡⎣ 4

0−2

⎤⎦ =

⎡⎣ 2α + β

7α

−4α + γ

⎤⎦

and so 2α + β = 4, 7α = 0, and −4α + γ = −2. It follows that α = 0, β = 4 and γ = −2.

Note that we could have guessed the values: a linear combination of i and k always gives 0 in thesecond component. Since the given vector has 0 as a second component, α must be zero.

33. Since a + b = [3, 6,−2], we get ‖a + b‖ =√

9 + 36 + 4 =√

49 = 7. On the other hand

‖a‖ + ‖b‖ =√

16 + 0 + 4 +√

1 + 36 + 0 =√

20 +√

37

Thus

‖a‖ + ‖b‖ =√

20 +√

37 > 4 + 6 = 10 > 7 = ‖a + b‖

35. Writing w = [v1/√

v21 + v2

2 , v2/√

v21 + v2

2 ] = (1/√

v21 + v2

2)[v1, v2], we compute

‖w‖ =1√

v21 + v2

2

‖[v1, v2]‖ =1√

v21 + v2

2

√v21 + v2

2 = 1

37. Let v = [v1, v2, v3] and w = [w1, w2, w3]. Then −v = [−v1,−v2,−v3] and

v + (−v) = [v1, v2, v3] + [−v1,−v2,−v3] = [v1 − v1, v2 − v2, v3 − v3] = [0, 0, 0] = 0

That’s (d).Let t be a real number. The two vectors

t(v + w) = t([v1, v2, v3] + [w1, w2, w3]) = t[v1 + w1, v2 + w2, v3 + w3]= [t(v1 + w1), t(v2 + w2), t(v3 + w3)]

and

tv + tw = t[v1, v2, v3] + t[w1, w2, w3] = [tv1, tv2, tv3] + [tw1, tw2, tw3]= [tv1 + tw1, tv2 + tw2, tv3 + tw3]

are equal due to the distributivity property of the real numbers. This completes the proof of (e).To prove (h), we compute

0v = 0[v1, v2, v3] = [0 · v1, 0 · v2, 0 · v3] = [0, 0, 0] = 0

and

1v = 1[v1, v2, v3] = [1 · v1, 1 · v2, 1 · v3] = [v1, v2, v3] = v


39. See the figure below.

y

x

w

v

0

w+zv+w

(v+w )+z

z

v+(w+z )

w

z

y

x

w

v

0

z

z

w+z


Section 3 The Dot Product

1. From v ·w = ‖v‖ ‖w‖ cos θ < 0 we get that cos θ < 0. Thus, the angle θ between the vectors whosedot product is negative satisfies π/2 < θ ≤ π.

3. We cannot add a real number v · w to a vector w.

5. The norm of a vector ‖v‖ is a real number, and so is the dot product v · v. Thus, ‖v‖(v · v) is areal number.

7. The dots in v ·w and w ·v represent the dot product, and thus both expressions in the parenthesesare real numbers. The dot between the parentheses is the multiplication of real numbers. Therefore,the expression (v · w) · (w · v) is a real number.

9. [−2 1 ] · [ 2 −1 ] = (−2)(2) + (1)(−1) = −5.

11.

⎡⎣ 4

80

⎤⎦ ·

⎡⎣ 0

03

⎤⎦ = (4)(0) + (8)(0) + (0)(3) = 0.

13. (3j + 2k) · (5i + 4j − 3k) = (0)(5) + (3)(4) + (2)(−3) = 6.

15. Let [x, y] be a vector in R2. The orthogonality assumption implies that [x, y] · [2,−5] = 0, i.e.,

2x − 5y = 0. Thus, all vectors orthogonal to the vector v = [2,−5] lie on the line y = 2x/5 (of slope2/5 going through the origin).

17. Let [x, y, z] be a vector in R3. The orthogonality assumption implies that [x, y, z] · [1,−5, 6] = 0,

i.e., x − 5y + 6z = 0. Thus, all vectors orthogonal to the vector v = [1,−5, 6] lie in the planex − 5y + 6z = 0.

19. From v · w = [ 1 2 0 ] · [ 0 3 1 ] = 6, ‖v‖ =√

5, and ‖w‖ =√

10, we get

cos θ =v · w

‖v‖ ‖w‖ =6√

5√

10≈ 0.849

Using a calculator, θ ≈ 0.557 radians.

21. From v · w = (2i + 4k) · (−i− 2j + 5k) = 18, ‖v‖ =√

20, and ‖w‖ =√

30, we get

cos θ =v · w

‖v‖ ‖w‖ =18√

20√

30≈ 0.735

Using a calculator, θ ≈ 0.745 radians.

23. Let A = (1, 1, 5), B = (3, 2, 7), and C = (3,−3, 5). Form the vectors (directed line segments):v = vector from A to B = [2, 1, 2]; w = vector from A to C = [2,−4, 0]; z = vector from B to C= [0,−5,−2]. From v · w = (2)(2) + (1)(−4) + (2)(0) = 0 we conclude that the sides AB and AC inthe triangle ABC are perpendicular.

Alternatively, we calculate the lengths of the sides: ‖v‖ =√

9, ‖w‖ =√

20, and ‖z‖ =√

29. Since‖z‖2 = ‖v‖2 + ‖w‖2, the triangle ABC is a right-angled triangle (by the Pythagorean Theorem).


25. (a) By the definition of the dot product, v ·w = v1w1 + v2w2 and w · v = w1v1 + w2v2. Becausethe multiplication of real numbers is commutative, we conclude that v · w = w · v.

(b) Using the distributivity of the multiplication of real numbers, we get

v · (w + z) = [v1, v2] · [w1 + z1, w2 + z2]= v1(w1 + z1) + v2(w2 + z2)= v1w1 + v1z1 + v2w2 + v2z2

= (v1w1 + v2w2) + (v1z1 + v2z2)= v · w + v · z

(b) Using the properties of real numbers, we get

(tv) · w = [tv1, tv2] · [w1, w2]= (tv1)w1 + (tv2)w2

= t(v1w1) + t(v2w2)= t(v1w1 + v2w2)= t(v · w)

27. (a) It is assumed that v · w = w · v = 0. Using ‖a‖2 = a · a, we compute

‖v + w‖2 = (v + w) · (v + w)= v · v + v · w + w · v + w · w= v · v + w · w= ‖v‖2 + ‖w‖2

(b) Using the equality of the norm squared and the dot product as in (a),

‖v + w‖2 = ‖v‖2 + ‖w‖2

(v + w) · (v + w) = ‖v‖2 + ‖w‖2

v · v + v · w + w · v + w · w = ‖v‖2 + ‖w‖2

‖v‖2 + 2v · w + ‖w‖2 = ‖v‖2 + ‖w‖2

2v · w = 0

Thus, v · w = 0, i.e., v and w are orthogonal.


Section 4 Equations of Lines and Planes

1. Recall that a line whose direction is given by the vector [v1, v2] has the slope of v2/v1, if v1 �= 0.Thus, [a, 4a], where a �= 0, is a direction vector of a line of slope 4.

Alternatively, a line of slope 4 has the equation y = 4x+ b, where b is any real number. Let x = t;then y = 4t + b, and the vector equation of the line is[

x

y

]=

[t

4t + b

]=

[0b

]+ t

[14

]Thus, the direction vector is [1, 4].

3. The slope is “rise over run”, i.e., v2/v1, if v1 �= 0. If v1 = 0, the line is vertical and its slope is nota real number.

5. The y-axis is characterized by x = 0 and z = 0; its parametric equations are⎧⎨⎩

x = 0

y = t

z = 0where t is a real number.

7. The xz-plane is characterized by y = 0; its parametric equations are⎧⎨⎩

x = s

y = 0

z = twhere s, t ∈R.

9. Let y = t (to avoid fractions). Then x = 5t − 9, and the vector equation is[x

y

]=

[5t − 9

t

]=

[−90

]+ t

[51

]The parametric equations are {

x = 5t − 9

y = t

where t is a real number.

11. As a direction vector v, we take the vector from (4, 0) to (0,−2); thus, v = [−4,−2]. A vectorequation of the line is (we pick (4, 0) for a point on the line):[

x

y

]=

[40

]+ t

[−4−2

]where t ∈R. In parametric form, {

x = 4 − 4t

y = −2t

To obtain an implicit form, we eliminate t: substituting t = −y/2 into the first equation, we getx = 4 + 2y, i.e., x − 2y = 4.

13. As a direction vector v, we take the vector from (0, 5, 0) to (−1, 8, 2); thus, v = [−1, 3, 2]. Avector equation of the line is (we pick (0, 5, 0) for a point on the line):⎡

⎣x

y

z

⎤⎦ =

⎡⎣ 0

50

⎤⎦ + t

⎡⎣−1

32

⎤⎦


where t ∈R. In parametric form, ⎧⎨⎩

x = −t

y = 5 + 3t

z = 2twhere t ∈R.

15. The vector equation is ⎡⎣x

y

z

⎤⎦ =

⎡⎣ 1

27

⎤⎦ + t

⎡⎣−2

04

⎤⎦

and the parametric equations are ⎧⎪⎨⎪⎩

x = 1 − 2t

y = 2

z = 7 + 4t

where t ∈R.

17. The normal form is [1, 1] · [x − (−4), y − 6] = 0, or (x + 4) + (y − 6) = 0, i.e., x + y − 2 = 0. Toobtain parametric equations, we introduce a parameter: let x = t; then y = −x + 2 = −t + 2. Thus,{

x = t

y = −t + 2where t ∈R.

19. The line 2x − 5y − 4 = 0, or y = 2x/5 − 4/5, has slope 2/5. The slope of a perpendicular line is−5/2, so we can take [2,−5] as its direction vector. Thus, the equations[

x

y

]=

[3

−1

]+ t

[2

−5

]and {

x = 3 + 2t

y = −1 − 5t

where t ∈R, represent the line that passes through (3,−1) and is perpendicular to 2x − 5y − 4 = 0.

21. The vector equation is ⎡⎣x

y

z

⎤⎦ =

⎡⎣ 3

4−1

⎤⎦ + s

⎡⎣ 0

14

⎤⎦ + t

⎡⎣ 0

02

⎤⎦


x = 3

y = 4 + s

z = −1 + 4s + 2t

where s, t ∈R.

23. Designate A = (1, 3, 4) to be the point which we will use in the equations. Let u be the vectorfrom A to B = (3, 9,−6), i.e., u = [2, 6,−10]; let v be the vector from A to C = (1, 0, 0), i.e.,v = [0,−3,−4]. The vectors u and v (obviously non-zero) are not multiples of each other, and henceare not parallel. The vector form of the equation of the plane is⎡

⎣x

y

z

⎤⎦ =

⎡⎣ 1

34

⎤⎦ + s

⎡⎣ 2

6−10

⎤⎦ + t

⎡⎣ 0−3−4

⎤⎦



x = 1 + 2s

y = 3 + 6s − 3t

z = 4 − 10s − 4t

where s, t ∈R.

25. From ⎡⎣ 3

01

⎤⎦ ·

⎡⎣x − 4

y − 2z + 5

⎤⎦ = 0

we get 3(x− 4) + 1(z + 5) = 0, and 3x + z − 7 = 0. To obtain parametric form, we use one parameterfor the missing variable y, since it can take on any value: y = s. When x = t, 3t + z − 7 = 0 andz = −3t + 7. Thus, the parametric equations of the given plane are⎧⎨

⎩x = t

y = s

z = 7 − 3t

where s, t ∈R.

27. The normal vector to the given plane, n = [1, 0, 4], is also a normal vector to any plane parallel toit. Thus, a vector equation of the plane passing through (2, 1, 3) parallel to the plane x+4z−1 = 0 is⎡

⎣ 104

⎤⎦ ·

⎡⎣x − 2

y − 1z − 3

⎤⎦ = 0

We get 1(x − 2) + 4(z − 3) = 0, and x + 4z − 14 = 0. In parametric form,⎧⎨⎩

x = 14 − 4t

y = s


29. We need a point in the plane and two vectors parallel to it. The two lines intersect at (0, 2,−6);the direction vectors of the two lines [0, 4, 2] and [1,−1, 2] are parallel to the plane. Thus, the equationof the plane is ⎡

⎣x

y

z

⎤⎦ =

⎡⎣ 0

2−6

⎤⎦ + s

⎡⎣ 0

42

⎤⎦ + t

⎡⎣ 1−1

2

⎤⎦

and the parametric equations are ⎧⎨⎩

x = t

y = 2 + 4s − t

z = −6 + 2s + 2t

where s, t ∈R.

31. We need to declare two variables as parameters. To avoid fractions, we set x = s and z = t; theny = 3x − 2z + 4 = 3s − 2t + 4. The parametric equations are⎧⎨

⎩x = s

y = 4 + 3s − 2t



33. The equations are ⎧⎪⎨⎪⎩

x = −s + 3t

y = 4 + 2s − 7t

z = 2 − 2s + 2t

Multiplying the first equation by 2 and adding to the second equation, we get 2x + y = 4 − t andt = 4 − 2x − y. Substituting this expression for t into the first equation gives x = −s + 3(4 − 2x − y)and s = 12 − 7x − 3y. Finally, using the third equation,

z = 2 − 2s + 2t = 2 − 2(12 − 7x − 3y) + 2(4 − 2x − y) = −14 + 10x + 4y

or 10x + 4y − z − 14 = 0.

35. The normal vectors to the two planes, n1 = [1,−2, 8] and n1 = [2,−4, 8], are not parallel. Thus,the given planes are not parallel.

37. We need to reduce the two equations to one parameter. Let z = t; from the first equation,x = −3z +1 = −3t+1. Substituting x and z into 2x+y−3z−6 = 0 we get 2(−3t+1)+y−3t−6 = 0and y = 9t + 4. So, the parametric equations of the line are⎧⎨

⎩x = 1 − 3t

y = 4 + 9t

z = twhere t ∈R.

39. The normal vector to the plane, n = [1,−3, 1], is parallel to the line we are looking for (and thuscan be taken as its direction vector). The vector equation of the line is⎡

⎣x

y

z

⎤⎦ =

⎡⎣ 0

82

⎤⎦ + t

⎡⎣ 1−3

1

⎤⎦

and the parametric equations are ⎧⎨⎩

x = t

y = 8 − 3t

z = 2 + t

where t ∈R.

41. The direction vector of the line, v = [4, 1,−1], is parallel to the normal vector of the plane,n = [−8,−2, 2]. Thus, the vector and the plane are perpendicular.


Section 5 Systems of Linear Equations

1. Substituting, we get (3) − (4) + 3(a) = 11. Thus, 3a = 12 and a = 4.

3. The lines x + y = 2 (i.e., y = −x + 2) and 2x − 2y = 5 (i.e., y = x − 5/2) do not have equalslopes. Since they are not parallel, they intersect at a single point. Thus, the given system has aunique solution.

To find the solution, compute y from the first equation (y = −x + 2) and substitute into thesecond equation: 2x − 2(−x + 2) = 5; it follows that 4x = 9 and x = 9/4. The unique solution isx = 9/4, and y = −9/4 + 2 = −1/4.

5. The solution set is the line 2y = 4x, i.e., y = 2x (so there are infinitely many solutions). Toparametrize the solution set, let x = t; then y = 2t, and so the solution is{

x = t

y = 2t

where t ∈R.

7. The solution set is a plane in R3. To parametrize it, let x = s and y = t. Then z = 4 − 2x + y =

4 − 2s + t, and the solution is ⎧⎨⎩

x = s

y = t

z = 4 − 2s + t

where s, t ∈R.

9. The two equations represent the lines y = −3x + 4 and y = x. They are not parallel, and thereforeintersect at a single point; thus, the solution set of the given system contains one element (one point).Substituting y = x into 3x + y = 4 we get 4x = 4, and x = 1. Thus, the unique solution of the systemis x = 1, y = 1.

11. Multiplying the second equation by 2 we obtain the first equation. Thus, the lines represented bythe two equations overlap, i.e., are the same line. The system has infinitely many solutions, and thesolution set is the line 5x − 2y = 1. Take x = t; then 5t − 2y = 1 and y = 5t/2 − 1/2. The solution,written in parametric form, is { x = t

y = 52 t − 1

2

where t ∈R.

13. The two equations represent the planes with normal vectors [2,−1,−1] and [3, 0, 1]. Knowing thatthe two normal vectors are not parallel, we conclude that the planes 2x − y − z = 4 and 3x + z = −2are not parallel, and thus intersect along a line. To find the parametric equations of the line, let x = t.From the second equation, 3t + z = −2 and z = −3t− 2. Substituting x and z into 2x− y − z = 4 weget 2t − y − (−3t − 2) = 4 and y = 5t − 2. Thus,⎧⎨

⎩x = t

y = 5t − 2

z = −3t − 2where t ∈R.


15. First we compute x:

Ax + By = P (R1)Cx + Dy = Q (R2)

ADx + BDy = DP (R3 ← DR1)−BCx − BDy = −BQ (R4 ← (−B)R2)

(AD − BC)x = DP − BQ (R5 ← R3 + R4)

x =DP − BQ

AD − BC(R6 ← R5/(AD − BC))

In the same way, we compute y:

Ax + By = P (R1)Cx + Dy = Q (R2)

ACx + BCy = CP (R3 ← CR1)−ACx − ADy = −AQ (R4 ← (−A)R2)

(−AD + BC)y = CP − AQ (R5 ← R3 + R4)

y =−(AQ − CP )−(AD − BC)

(R6 ← R5/(AD − BC))

y =AQ − CP

AD − BC

17. The equations R1 and R4 below represent the system in upper-triangular form.

2x + 10y = 6 (R1)−x + y = 9 (R2)

12y = 24 (R3 ← 2R2 + R1)

2x + 10y = 6 (R1)y = 2 (R4 ← R3/12)

Substituting y = 2 into R1 we get 2x = −14 and x = −7. So, the (unique) solution is x = −7, y = 2.

19. The equations R1 and R3 (or R2 and R3) below represent the system in upper-triangular form.

7x − 2y = 2 (R1)7x − 3y = 2 (R2)

y = 0 (R3 ← R1 − R2)

Substituting y = 0 into R1 we get x = 2/7. So, the (unique) solution is x = 2/7, y = 0.

21. The equations R1 and R4 below represent the system in upper-triangular form.

12x + y = 6 (R1)2x + 7y = 1 (R2)

−41y = 0 (R3 ← R1 − 6R2)

y = 0 (R4 ← R3/(−41))

Substituting y = 0 into R1 we get x = 1/2. So, the (unique) solution is x = 1/2, y = 0.

23. We start be eliminating the x terms from the second and the third equations:

x + y + 2z = 7 (R1)3x − y + 4z = 19 (R2)


−2x − y + z = 0 (R3)

x + y + 2z = 7 (R1)−4y − 2z = −2 (R4 ← −3R1 + R2)

y + 5z = 14 (R5 ← 2R1 + R3)

x + y + 2z = 7 (R1)−4y − 2z = −2 (R4)

18z = 54 (R6 ← R4 + 4R5)

Thus, z = 3. From R4 we compute −4y = 4 and y = −1. The equation R1 reads x + (−1) + 2(3) = 7and so x = 2.

25. We compute

7x − y − z = 0 (R1)y + 6z = 37 (R2)

3x + 2y = 5 (R3)

7x − y − z = 0 (R1)y + 6z = 37 (R2)

177

y +37z = 5 (R4 ← R3 − 3R1/7)

17y + 3z = 35 (R5 ← 7R4)

7x − y − z = 0 (R1)y + 6z = 37 (R2)−33y = −33 (R6 ← R2 − 2R5)

(Note that this is upper-triangular form when the variables are ordered x, z, y.) From the lastequation, y = 1. Using R2 we compute z = 6. The equation R1 reads 7x − 7 = 0 and so x = 1.

27. We compute

x − y + z = 0 (R1)3x + z = 5 (R2)

5x − 2y + 3z = 5 (R3)

x − y + z = 0 (R1)3y − 2z = 5 (R4 ← R2 − 3R1)3y − 2z = 5 (R5 ← R3 − 5R1)

The equations R1 and R4 (or R5) represent the system in upper-triangular form. We need oneparameter: let z = t, where t is a real number. Then (from R4) 3y − 2t = 5 and y = 2t/3 + 5/3.Substituting z and y into R1, we get x − (2t/3 + 5/3) + t = 0 and x = −t/3 + 5/3.

29. Multiplying the first equation by 2 we get 2x − 4y + 2z = 8. Since the third equation reads2x − 4y + 2z = 5, we conclude that the given system is inconsistent (i.e., has no solutions).

31. We reduce the given system to upper-triangular form:

Ax + By = P (R1)Cx + Dy = Q (R2)

−CB

Ay + Dy = −C

AP + Q (R3 ← R2 − CR1/A)

(AD − BC)y = −CP + QA (R4 ← AR3)


By assumption, AD − BC = 0, and the upper-triangular form is:

Ax + By = P (R1)0y = −CP + QA (R4)

If −CP + QA �= 0, the system has no solutions (that’s the answer to (b); note that −CP + QA �= 0can be written as P/Q �= A/C).

If −CP + QA = 0 (i.e., P/Q = A/C), then there are infinitely many solutions (part (a)): takey = t, where t ∈R. Then from Ax + Bt = P we get x = −Bt/A + P/A.

33. Rewrite the system as y = ax− 2 and y = x− 3. If a �= 1 then the two lines have different slopesand intersect at a single point (unique solution). If a = 1, the system y = x− 2 and y = x− 3 has nosolutions (subtracting the second equation from the first, we get 0 = 1). There are no values of a forwhich the given system has infinitely many solutions.

35. First we compute x:

Ax + By = 0 (R1)Cx + Dy = 0 (R2)

ADx + BDy = 0 (R3 ← DR1)−BCx − BDy = 0 (R4 ← (−B)R2)

(AD − BC)x = 0 (R5 ← R3 + R4)0x = 0 (R6)

The equations R1 (or R2) and R6 represent the given system in upper triangular form. From R6 weconclude that x can be any real number: x = t, where t ∈R. From R1, y = −Ax/B = −At/B, ifB �= 0.

If B = 0 then AD−BC = 0 implies that AD = 0 and D = 0 (since A �= 0), and the system readsAx = 0 and Cx = 0. In this case, the solution is x = 0 and y = t, t ∈R.


Section 6 Gaussian Elimination

1. We need to repeat the same equation three times; for instance, the system 2x+3y = 4, −2x−3y =−4, and 6x + 9y = 12 has infinitely many solutions.

3. The leading entry in the first row is not to the left of the leading entry in the second row (or, thereshould be a 0 in the second row below the leading 1 in the first row). To obtain the matrix in rowechelon form: [

1 11 0

]→

[1 10 −1

]((−1)R1 + R2 → R2)

5. The leading entry in the second row is not to the left of the leading entry in the third row. Toobtain the matrix in row echelon form:⎡

⎣ 1 2 30 0 10 0 4

⎤⎦ →

⎡⎣ 1 2 3

0 0 10 0 0

⎤⎦ ((−4)R2 + R3 → R3)

7. We need one step only: [1 23 4

]→

[1 20 −2

]((−3)R1 + R2 → R2)

9. We need to work on the third row:⎡⎣ 1 0 −3

0 4 0−3 0 1

⎤⎦ →

⎡⎣ 1 0 −3

0 4 00 0 −8

⎤⎦ (3R1 + R3 → R3)

11. We need 0 below the leading entry −2 in the first row:[−2 1 1

8 0 0

]→

[−2 1 1

0 4 4

](4R1 + R2 → R2)

13. We start by introducing 0 in the second row and the first column:⎡⎣ 1 3 −2

2 4 00 4 1

⎤⎦ →

⎡⎣ 1 3 −2

0 −2 40 4 1

⎤⎦ ((−2)R1 + R2 → R2)

→

⎡⎣ 1 3 −2

0 −2 40 0 9

⎤⎦ (2R2 + R3 → R3)

15. We solve the system of equations x − 3y + z = 2 and 2x + 3y − z = 5 by working with theaugmented matrix: [

1 −3 12 3 −1

∣∣∣∣ 25

](R1)(R2)[

1 −3 10 9 −3

∣∣∣∣ 21

](R1)(R3 ← −2R1 + R2)


Let z = t. From 9y − 3z = 1, i.e., 9y − 3t = 1 we compute y = t/3 + 1/9. By back substitution,

x − 3y + z = 2

x − 3(

t

3+

19

)+ t = 2

x =219

=73

Thus, the set of points that belong to both planes is the line⎡⎣ x

y

z

⎤⎦ =

⎡⎣ 7/3

t/3 + 1/9t

⎤⎦ =

⎡⎣ 7/3

1/90

⎤⎦ + t

⎡⎣ 0

1/31

⎤⎦

where t ∈R.

17. The matrix represents a system of three equations in three variables. The zeros in the last rowimply that the system is consistent. To solve, we need to take one of the two variables appearing inthe second equation as a parameter. Thus, the system has infinitely many solutions.

19. The matrix represents a system of three equations in four variables. Looking at the last row wesee that the system is consistent. There are four variables but only three equations, so the system hasinfinitely many solutions. (Note that the solution for the fourth variable (last row) is unique (equalto 5) and the solution for the third variable is unique (second row); thus, the parameter appears inthe expressions for the first and the second variables.)

21. The matrix represents a system of three equations in three variables. Looking at the last row wesee that the system is consistent. Using back substitution, we get a unique solution.

23. We need to find real numbers a, b, and c so that⎡⎣ 8−4

8

⎤⎦ = a

⎡⎣ 3

01

⎤⎦ + b

⎡⎣ 1

2−1

⎤⎦ + c

⎡⎣ 4

12

⎤⎦

Thus, we need to solve the system

3a + b + 4c = 82b + c = −4

a − b + 2c = 8

for a, b, and c. Using augmented matrix,⎡⎣ 3 1 4

0 2 11 −1 2

∣∣∣∣∣∣8

−48

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 0 4 −20 2 11 −1 2

∣∣∣∣∣∣−16−4

8

⎤⎦ (R4 ← −3R3 + R1)

(R2)(R3)⎡

⎣ 1 −1 20 2 10 4 −2

∣∣∣∣∣∣8

−4−16

⎤⎦ (R3)

(R2)(R4)⎡

⎣ 1 −1 20 2 10 0 −4

∣∣∣∣∣∣8

−4−8

⎤⎦ (R3)

(R2)(R5 ← (−2)R2 + R4)


Thus, −4c = −8 and c = 2. from 2b + c = −4 we get b = −3, and then a − b + 2c = 8 it implies thata − (−3) + 2(2) = 8, i.e., a = 1.

25. To simplify working with the augmented matrix, we switch the first two equations:⎡⎣ 1 1 2

3 −1 4−2 −1 1

∣∣∣∣∣∣7

190

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 1 20 −4 −20 1 5

∣∣∣∣∣∣7

−214

⎤⎦ (R1)

(R4 ← −3R1 + R2)(R5 ← 2R1 + R3)⎡

⎣ 1 1 20 1 50 −4 −2

∣∣∣∣∣∣7

14−2

⎤⎦ (R1)

(R5)(R4)⎡

⎣ 1 1 20 1 50 0 18

∣∣∣∣∣∣7

1454

⎤⎦ (R1)

(R5)(R6 ← 4R5 + R4)

Thus, 18z = 54 and z = 3. By back substitution, y + 5z = 14 and y = −1; from x + y + 2z = 7 we getx = 2.

27. We work with the augmented matrix:⎡⎣ 1 1 −2

2 0 13 1 −1

∣∣∣∣∣∣404

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 1 −20 −2 50 −2 5

∣∣∣∣∣∣4

−8−8

⎤⎦ (R1)

(R4 ← −2R1 + R2)(R5 ← −3R1 + R3)⎡

⎣ 1 1 −20 −2 50 0 0

∣∣∣∣∣∣4

−80

⎤⎦ (R1)

(R4)(R6 ← −R4 + R5)

The system has infinitely many solutions. Let z = t be a parameter. From −2y + 5z = −8 we get−2y + 5t = −8 and y = 5t/2 + 4. Finally, from x + y − 2z = 4 it follows that x + (5t/2 + 4) − 2t = 4and x = −t/2.

29. We work with the augmented matrix (to start, we switch the second and the third equation):⎡⎣ 7 −1 −1

3 2 00 1 6

∣∣∣∣∣∣05

37

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 7 −1 −10 17/7 3/70 1 6

∣∣∣∣∣∣05

37

⎤⎦ (R1)

(R4 ← −3R1/7 + R2)(R3)⎡

⎣ 7 −1 −10 17 30 1 6

∣∣∣∣∣∣0

3537

⎤⎦ (R1)

(R5 ← 7R4)(R3)⎡

⎣ 7 −1 −10 1 60 17 3

∣∣∣∣∣∣0

3735

⎤⎦ (R1)

(R3)(R5)


⎡⎣ 7 −1 −1

0 1 60 0 −99

∣∣∣∣∣∣0

37−594

⎤⎦ (R1)

(R3)(R6 ← −17R3 + R5)

From −99z = −594 it follows that z = 6. Using back substitution: from y + 6z = 37 we get y = 1,and 7x − y − z = 0 yields x = 1.

31. We work with the augmented matrix:⎡⎣ 1 −1 1

3 0 15 −2 3

∣∣∣∣∣∣055

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 −1 10 3 −20 3 −2

∣∣∣∣∣∣055

⎤⎦ (R1)

(R4 ← −3R1 + R2)(R5 ← −5R1 + R3)⎡

⎣ 1 1 −20 3 −20 0 0

∣∣∣∣∣∣050

⎤⎦ (R1)

(R4)(R6 ← −R4 + R5)

The system has infinitely many solutions. Let z = t be a parameter. From 3y − 2z = 5 we get3y − 2t = 5 and y = 2t/3 + 5/3. Finally, from x − y + z = 0 it follows that x − (2t/3 + 5/3) + t = 0and x = −t/3 + 5/3.

33. We row reduce the augmented matrix:[A B

C D

∣∣∣∣ P

Q

](R1)(R2)[

A B

0 −CB/A + D

∣∣∣∣ P

−CP/A + Q

](R1)(R3 ← −CR1/A + R2)[

A B

0 AD − BC

∣∣∣∣ P

AQ − CP

](R1)(R4 ← AR3)

Thus, (AD − BC)y = AQ − CP and y = (AQ − CP )/(AD − BC). Using Ax + By = P, we get

Ax = P − BAQ − CP

AD − BC

Ax =P (AD − BC) − B(AQ − CP )

AD − BC

Ax =PAD − ABQ

AD − BC=

A(DP − BQ)AD − BC

x =DP − BQ

AD − BC

35. We row reduce the augmented matrix:⎡⎣ 3 1

4 35 −1

∣∣∣∣∣∣55

11

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 3 10 5/30 −8/3

∣∣∣∣∣∣5

−5/38/3

⎤⎦ (R1)

(R4 ← −4R1/3 + R2)(R5 ← −5R1/3 + R3)


⎡⎣ 3 1

0 10 −1

∣∣∣∣∣∣5

−11

⎤⎦ (R1)

(R6 ← R4/(5/3)(R7 ← R5/(8/3))⎡

⎣ 3 10 10 0

∣∣∣∣∣∣5

−10

⎤⎦ (R1)

(R6)(R8 ← R6 + R7)

Thus y = −1. From 3x + y = 5 it follows that x = 2.

37. We need to solve the system ⎡⎣ 6 − t

−2 + t

5 − t

⎤⎦ =

⎡⎣ 3s

2 − 2s

1 + 2s

⎤⎦

i.e.,

3s + t = 62s + t = 42s + t = 4

for s and t. By subtracting the second equation from the first equation we get s = 2. From eitherequation it follows that t = 0. Thus, the two lines intersect at a single point (6,−2, 5) (we obtainedthis point by substituting either s = 2 or t = 0 into the corresponding equation of the line).


Section 8 Matrices

1. The product is a 1 × 1 matrix (i.e., a matrix with a single entry), which can be identified with areal number.

3. For example, if

A =[

0 20 0

], then A2 =

[0 20 0

] [0 20 0

]=

[0 00 0

]There are many other matrices with the same property, such as:[

0 a

0 0

],

[0 0a 0

], and

[a a

−a −a

],

where a is a non-zero real number.

5. Recall that a matrix D is called symmetric if Dt = D; replacing D by A + At, we have to showthat (A + At)t = A + At. Let

A =

⎡⎣ a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎦

Then

A + At =

⎡⎣ a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎦ +

⎡⎣ a11 a21 a31

a12 a22 a32

a13 a23 a33

⎤⎦ =

⎡⎣ 2a11 a12 + a21 a13 + a31

a21 + a12 2a22 a23 + a32

a31 + a13 a32 + a23 2a33

⎤⎦

Switching the rows and columns of A + At will produce the same matrix. Thus, (A + At)t = A + At.

7. The matrix operations involved are defined. We compute

7A − 2B + I2 = 7[

1 20 4

]− 2

[4 00 −2

]+

[1 00 1

]=

[7 140 28

]+

[−8 00 4

]+

[1 00 1

]=

[0 140 33

]

9. The matrix operations involved are defined. We compute

3E − 4F t = 3

⎡⎣ 1 2

3 02 5

⎤⎦ − 4

[1 0 01 2 3

]t

=

⎡⎣ 3 6

9 06 15

⎤⎦ − 4

⎡⎣ 1 1

0 20 3

⎤⎦ =

⎡⎣−1 2

9 −86 3

⎤⎦

11. A is a 2× 2 matrix, and E is a 3× 2 matrix. The product AE is not defined. (Note: the productEA is defined, and is a 3 × 2 matrix.)

13. E is a 3 × 2 matrix, and so the power E2 = E · E is not defined. Thus, D − E2 is not defined.

15. The product EF is defined, and is a 3 × 3 matrix. Thus, D − EF is defined. We compute:

D − EF =

⎡⎣ 0 3 2

0 0 60 0 0

⎤⎦ −

⎡⎣ 1 2

3 02 5

⎤⎦ ·

[1 0 01 2 3

]=

⎡⎣ 0 3 2

0 0 60 0 0

⎤⎦ −

⎡⎣ 3 4 6

3 0 07 10 15

⎤⎦ =

⎡⎣−3 −1 −4−3 0 6−7 −10 −15

⎤⎦


17. Since D is a 3 × 3 matrix, any power of D is defined (and is a 3 × 3 matrix). From

D2 = D · D =

⎡⎣ 0 3 2

0 0 60 0 0

⎤⎦ ·

⎡⎣ 0 3 2

0 0 60 0 0

⎤⎦ =

⎡⎣ 0 0 18

0 0 00 0 0

⎤⎦

we get

D3 = D2 · D =

⎡⎣ 0 0 18

0 0 00 0 0

⎤⎦ ·

⎡⎣ 0 3 2

0 0 60 0 0

⎤⎦ =

⎡⎣ 0 0 0

0 0 00 0 0

⎤⎦

19. Since B is a 2 × 2 matrix, G is a 2 × 1 matrix, and Gt is a 1 × 2 matrix, the product BGGt isdefined, and is a 2 × 2 matrix. From

GGt =[

56

]· [ 5 6 ] =

[25 3030 36

]we obtain

BGGt = B(GGt) =[

4 00 −2

]·[

25 3030 36

]=

[100 120−60 −72

]

21. Let

A =[

a11 a12

a21 a22

]Then

AS =[

a11 a12

a21 a22

]·[

s 00 s

]=

[sa11 sa12

sa21 sa22

]and

SA =[

s 00 s

]·[

a11 a12

a21 a22

]=

[sa11 sa12

sa21 sa22

]Thus, AS = SA. To multiply a matrix by a scalar matrix (whose non-zero entry is s), we multiplyeach entry in the matrix by s.

23. Let

A =

⎡⎣ a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎦

We compute:

AD =

⎡⎣ a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎦ ·

⎡⎣ d1 0 0

0 d2 00 0 d3

⎤⎦ =

⎡⎣ d1a11 d2a12 d3a13

d1a21 d2a22 d3a23

d1a31 d2a32 d3a33

⎤⎦

To compute AD, we multiply the first column of A by d1, the second column of A by d2, and thethird column of A by d3. In reversed order:

DA =

⎡⎣ d1 0 0

0 d2 00 0 d3

⎤⎦ ·

⎡⎣ a11 a12 a13

a21 a22 a23

a31 a32 a33

⎤⎦ =

⎡⎣ d1a11 d1a12 d1a13

d2a21 d2a22 d2a23

d3a31 d3a32 d3a33

⎤⎦

To compute DA, we multiply the first row of A by d1, the second row of A by d2, and the third rowof A by d3. Clearly, AD �= DA.


25. Using trig identities sin 2θ = 2 sin θ cos θ and cos 2θ = cos2 θ − sin2 θ, we get

A2 =[

cos θ − sin θ

sin θ cos θ

]·[

cos θ − sin θ

sin θ cos θ

]

=[

cos2 θ − sin2 θ − sin θ cos θ − sin θ cos θ

sin θ cos θ + sin θ cos θ − sin2 θ + cos2 θ

]=

[cos 2θ − sin 2θ

sin 2θ cos 2θ

]

27. From 4(X + D) = BC we get 4X + 4D = BC and

X =14(BC − 4D)

=14

([−2 00 −2

] [8 15 1

]− 4

[0 02 0

])

=14

([−16 −2−10 −2

]−

[0 08 0

])=

14

[−16 −2−18 −2

]

=[ −4 −1/2−9/2 −1/2

]

29. From 5X − A = X − 5C we get 4X = A − 5C and

X =14(A − 5C) =

14

([−1 20 4

]− 5

[8 15 1

])=

14

[−41 −3−25 −1

]=

[−41/4 −3/4−25/4 −1/4

]

31. Let

A =[

a11 a12

a21 a22

]and B =

[b11 b12

b21 b22

]Then

(αA)B =(

α

[a11 a12

a21 a22

]) [b11 b12

b21 b22

]=

[αa11 αa12

αa21 αa22

] [b11 b12

b21 b22

]

=[

(αa11)b11 + (αa12)b21 (αa11)b12 + (αa12)b22

(αa21)b11 + (αa22)b21 (αa21)b12 + (αa22)b22

]and

α(AB) = α

([a11 a12

a21 a22

] [b11 b12

b21 b22

])

= α

[a11b11 + a12b21 a11b12 + a12b22

a21b11 + a22b21 a21b12 + a22b22

]

=[

α(a11b11 + a12b21) α(a11b12 + a12b22)α(a21b11 + a22b21) α(a21b12 + a22b22)

]

=[

α(a11b11) + α(a12b21) α(a11b12) + α(a12b22)α(a21b11) + α(a22b21) α(a21b12) + α(a22b22)

]Because the multiplication of real numbers is associative (i.e., (xy)z = x(yz) for real numbers x, y,and z), it follows that (αA)B = α(AB).

33. Using the definition,

det(A) =∣∣∣∣ 1 −2−3 4

∣∣∣∣ = (1)(4) − (−3)(−2) = −2


35. Using the definition,

det(A) =

∣∣∣∣∣∣1 2 32 3 43 5 7

∣∣∣∣∣∣ = 1∣∣∣∣ 3 45 7

∣∣∣∣ − 2∣∣∣∣ 2 43 7

∣∣∣∣ + 3∣∣∣∣ 2 33 5

∣∣∣∣= 1[3 · 7 − 5 · 4] − 2[2 · 7 − 3 · 4] + 3[2 · 5 − 3 · 3] = 1(1) − 2(2) + 3(1) = 0

37. Let

A =[

a11 a12

a21 a22

]; then cA =

[ca11 ca12

ca21 ca22

]and

det(cA) =∣∣∣∣ ca11 ca12

ca21 ca22

∣∣∣∣ = ca11 · ca22 − ca21 · ca12 = c2 (a11 · a22 − a21 · a12) = c2det(A)

39. We compute

AB =[−1 3

1 4

] [2 31 1

]=

[1 06 7

]and

BA =[

2 31 1

] [−1 31 4

]=

[1 180 7

]Thus, AB �= BA. However,

det(AB) =∣∣∣∣ 1 06 7

∣∣∣∣ = 1 · 7 − 6 · 0 = 7

and

det(BA) =∣∣∣∣ 1 180 7

∣∣∣∣ = 1 · 7 − 0 · 18 = 7


Section 9 Matrices and Linear Systems

1. Yes. Think of ABC = CAB = I as (AB)C = C(AB) = I; i.e., the matrix AB multipliedby the matrix C in either order gives the identity matrix. This means that C−1 = AB (as well as(AB)−1 = C).

3. Think of A3 = I as A2 · A = I and A · A2 = I; i.e., A multiplied by A2 in either order gives theidentity matrix. Thus, A−1 = A2.

5. The determinant of the given matrix is∣∣∣∣ a 2 − a

a 1

∣∣∣∣ = a − a(2 − a) = a[1 − (2 − a)] = a(−1 + a)

Now a(−1 + a) = 0 implies that a = 0 or a = 1. Thus, the matrix is invertible if a �= 0 and a �= 1.

7. Since the determinant ∣∣∣∣ 0 12 3

∣∣∣∣ = −2 �= 0

the matrix is invertible. Using the formula given in Theorem 8,[0 12 3

]−1

=1−2

[3 −1

−2 0

]=

[−3/2 1/21 0

]

9. The determinant of the matrix is −24 + 24 = 0, and the matrix does not have an inverse.

11. Since the determinant∣∣∣∣ 0.1 0.20.4 −0.2

∣∣∣∣ = (0.1)(−0.2) − (0.4)(0.2) = −0.1 �= 0

the matrix is invertible. Using the formula given in Theorem 8,[0.1 0.20.4 −0.2

]−1

=1

−0.1

[−0.2 −0.2−0.4 0.1

]=

[2 24 −1

]

13. The determinant of the given matrix is∣∣∣∣∣∣1 −2 30 1 00 1 0

∣∣∣∣∣∣ = 1∣∣∣∣ 1 01 0

∣∣∣∣ − (−2)∣∣∣∣ 0 00 0

∣∣∣∣ + 3∣∣∣∣ 0 10 1

∣∣∣∣ = 0

We conclude that the matrix is non-invertible (i.e., singular).

15. We use Algorithm 2:⎡⎣ 1 −2 3

4 1 00 1 1

∣∣∣∣∣∣1 0 00 1 00 0 1

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 −2 30 9 −120 1 1

∣∣∣∣∣∣1 0 0

−4 1 00 0 1

⎤⎦ (R1)

(R4 ← −4R1 + R2)(R3)⎡

⎣ 1 0 50 0 −210 1 1

∣∣∣∣∣∣1 0 2

−4 1 −90 0 1

⎤⎦ (R5 ← 2R3 + R1)

(R6 ← −9R3 + R4)(R3)


⎡⎣ 1 0 5

0 1 10 0 1

∣∣∣∣∣∣1 0 20 0 1

4/21 −1/21 9/21

⎤⎦ (R5)

(R3)(R7 ← R6/(−21))⎡

⎣ 1 0 00 1 00 0 1

∣∣∣∣∣∣1/21 5/21 −3/21

−4/21 1/21 12/214/21 −1/21 9/21

⎤⎦ (R8 ← −5R7 + R5)

(R9 ← −R7 + R3)(R7)

Thus, ⎡⎣ 1 −2 3

4 1 00 1 1

⎤⎦−1

=

⎡⎣ 1/21 5/21 −3/21−4/21 1/21 12/21

4/21 −1/21 9/21

⎤⎦

17. Rewrite the two equations in the form y = 2x/6 − 13/6 and y = −3x/5 + 7/5. Since the twolines have different slopes, they intersect at a single point; therefore, the given system has a uniquesolution. Write the system as Ax = b, where

A =[

2 −63 5

], x =

[x

y

]and b =

[137

]The determinant of A is detA = 10 − (−18) = 28, and thus (Theorem 8)

A−1 =128

[5 6

−3 2

]The solution is

x = A−1b =128

[5 6

−3 2

]·[

137

]=

128

[107−25

]i.e., x = 107/28 and y = −25/28.

19. The matrix of the system is

A =

⎡⎣ 1 3 −1−1 0 1

2 2 1

⎤⎦

Since the determinant∣∣∣∣∣∣1 3 −1

−1 0 12 2 1

∣∣∣∣∣∣ = 1∣∣∣∣ 0 12 1

∣∣∣∣ − 3∣∣∣∣−1 1

2 1

∣∣∣∣ + (−1)∣∣∣∣−1 0

2 2

∣∣∣∣ = 1(−2) − 3(−3) − 1(−2) = 9

is non-zero, the corresponding system has a unique solution; it is given by x = A−1b where

x =

⎡⎣x

y

z

⎤⎦ and b =

⎡⎣ 2

31

⎤⎦

Next, we find A−1: ⎡⎣ 1 3 −1−1 0 1

2 2 1

∣∣∣∣∣∣1 0 00 1 00 0 1

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 3 −10 3 00 −4 3

∣∣∣∣∣∣1 0 01 1 0

−2 0 1

⎤⎦ (R1)

(R4 ← R1 + R2)(R5 ← −2R1 + R3)⎡

⎣ 1 3 −10 1 00 −4 3

∣∣∣∣∣∣1 0 0

1/3 1/3 0−2 0 1

⎤⎦ (R1)

(R6 ← R4/3)(R5)


⎡⎣ 1 0 −1

0 1 00 0 3

∣∣∣∣∣∣0 −1 0

1/3 1/3 0−2/3 4/3 1

⎤⎦ (R7 ← −3R6 + R1)

(R6)(R8 ← 4R6 + R5)⎡

⎣ 1 0 −10 1 00 0 1

∣∣∣∣∣∣0 −1 0

1/3 1/3 0−2/9 4/9 3/9

⎤⎦ (R7)

(R6)(R9 ← R8/3)⎡

⎣ 1 0 00 1 00 0 1

∣∣∣∣∣∣−2/9 −5/9 3/9

1/3 1/3 0−2/9 4/9 3/9

⎤⎦ (R10 ← R9 + R7)

(R6)(R9)

We are done. The inverse is

A−1 =

⎡⎣ 1 3 −1−1 0 1

2 2 1

⎤⎦−1

=19

⎡⎣−2 −5 3

3 3 0−2 4 3

⎤⎦

and ⎡⎣ x

y

z

⎤⎦ = A−1b =

19

⎡⎣−2 −5 3

3 3 0−2 4 3

⎤⎦

⎡⎣ 2

31

⎤⎦ =

19

⎡⎣−16

1511

⎤⎦

Thus, x = −16/9, y = 15/9, and z = 11/9.

21. If α = 0, then A = αI is a zero matrix, and does not have an inverse. If α �= 0, then we claimthat A−1 = (1/α)I. To prove:

A · A−1 = (αI)(

1α

I

)= α

1α

I · I = 1I = I

A−1 · A =(

1α

I

)(αI) =

1α

αI · I = 1I = I

We are done.

23. A 3 × 3 diagonal matrix is of the form

A =

⎡⎣ d1 0 0

0 d2 00 0 d3

⎤⎦

where d1, d2, and d3 are real numbers. The determinant of A is∣∣∣∣∣∣d1 0 00 d2 00 0 d3

∣∣∣∣∣∣ = d1

∣∣∣∣ d2 00 d3

∣∣∣∣ − 0∣∣∣∣ 0 00 d3

∣∣∣∣ + 0∣∣∣∣ 0 d2

0 0

∣∣∣∣ = d1d2d3

The matrix A has an inverse if its determinant d1d2d3 �= 0; thus, A is invertible if all of d1, d2, andd3 are non-zero. To find A−1, we could try to guess, or apply the usual procedure:⎡

⎣ d1 0 00 d2 00 0 d3

∣∣∣∣∣∣1 0 00 1 00 0 1

⎤⎦ (R1)

(R2)(R3)⎡

⎣ 1 0 00 1 00 0 1

∣∣∣∣∣∣1/d1 0 0

0 1/d2 00 0 1/d3

⎤⎦ (R4 ← R1/d1)

(R5 ← R2/d2)(R6 ← R3/d3)

Thus, ⎡⎣ d1 0 0

0 d2 00 0 d3

⎤⎦−1

=

⎡⎣ 1/d1 0 0

0 1/d2 00 0 1/d3

⎤⎦


25. Let

A =[

a b

c d

]Then (we use Theorem 8 to compute inverse matrices)

(A−1)t =(

1ad − bc

[d −b

−c a

])t

=1

ad − bc

[d −c

−b a

]and

(At)−1 =([

a c

b d

])−1

=1

ad − bc

[d −c

−b a

]Clearly, (A−1)t = (At)−1.

27. Suppose that

A =[

a b

c d

]is invertible. This means that (since AA−1 = A−1A = I) A−1 is also invertible. The calculation[

1 00 1

∣∣∣∣ 3 40 0

]implies that

A−1 =[

3 40 0

]Since the determinant of A−1 is zero, it means that A−1 is not invertible, which is a contradiction.

Now suppose that A is not invertible, i.e., ad − bc = 0. Then[a b

c d

∣∣∣∣ 1 00 1

][

a b

0 −bc/a + d

∣∣∣∣ 1 0−c/a 1

][

a b

0 ad − bc

∣∣∣∣ 1 0−c a

][

a b

0 0

∣∣∣∣ 1 0−c a

]shows that the matrix on the left is not the identity matrix; so we could not have arrived at[

1 00 1

∣∣∣∣ 3 40 0

]starting from a non-invertible matrix.

29. The determinant of the matrix is 4 − 6 = −2 �= 0; the system has only a trivial solution x = 0.

31. The determinant of the matrix is zero, which means that the system has a non-trivial solution.The equations −x + y = 0 and −2x + 2y = 0 are identical. Let x = t; then y = t, and thus there areinfinitely many solutions: x = t, y = t, where t ∈R.

33. The determinant of the matrix is zero, which means that the system has a non-trivial solution.The system consists of one equation, 3x+3y = 0. Let x = t; then y = −t, and thus there are infinitelymany solutions: x = t, y = −t, where t ∈R.


35. Since A is an invertible, there is a matrix B such that AB = BA = I. We claim that B2 is theinverse of A2. To prove it we verify

A2B2 = AABB = A(AB)B = AIB = AB = I

B2A2 = BBAA = B(BA)A = BIA = BA = I

(Note that, due to the associativity of matrix multiplication, we do not need to use parentheses.)


Section 10 Linear Transformations

1. We compute

A(9v1) = 9A(v1) = 9[

3−9

]=

[27

−81

]

A(−v2) = −A(v2) = −[

101

]=

[−10−1

]

A(2v1 − 3v2) = A(2v1) − A(3v2) = 2A(v1) − 3A(v2) = 2[

3−9

]− 3

[101

]=

[−24−21

]

3. Yes: A(−v) = −A(v) = − [ 3 −4 ] = [−3 4 ] .

5. We pick two vectors

v =[

11

]and w =

[01

]and compare A(v + w) with A(v) + A(w). The calculation

A(v + w) = A

[12

]=

[55

]

A(v) + A(w) = A

[11

]+ A

[01

]=

[36

]+

[23

]=

[59

]shows that A(v + w) �= A(v) + A(w) and so A is not linear.

Alternatively,

A(2v) = A

[22

]=

[68

]

2A(v) = 2A

[11

]= 2

[36

]=

[6

12

]Since A(2v) �= 2A(v), A is not linear.

Here is a slightly different reasoning:

A(0) = A

[00

]=

[04

]A linear transformation must map zero vector onto a zero vector (since A(0) = A · 0 = 0). Hence Ais not linear.

7. We compute

A(v) =[

3 −24 0

] [102

]=

[2640

]

A(w) =[

3 −24 0

] [−1−6

]=

[9−4

]

A(2v + 4w) = A(2v) + A(4w) = 2A(v) + 4A(w) = 2[

2640

]+ 4

[9−4

]=

[8864

]Alternatively,

2v + 4w = 2[

102

]+ 4

[−1−6

]=

[16

−20

]and thus

A(2v + 4w) =[

3 −24 0

] [16

−20

]=

[8864

]


9. Let

A =[

a b

c d

]From

A

[10

]=

[a b

c d

] [10

]=

[a

c

]=

[2

−7

]it follows that a = 2 and c = −7. Likewise, from

A

[01

]=

[a b

c d

] [01

]=

[b

d

]=

[−23

]we conclude that b = −2 and d = 3. Thus,

A =[

2 −2−7 3

]

11. Computing

A

[x

y

]=

[0 00 1

] [x

y

]=

[0y

]we see that A preserves the y coordinate of a vector and transforms its x-coordinate to 0. Thus, A isthe orthogonal projection onto the y-axis.

13. Computing

A

[x

y

]=

[−1 00 1

] [x

y

]=

[−x

y

]we see that A preserves the y coordinate of a vector and changes the sign of its x-coordinate. Thus,A is the reflection with respect to the y-axis.

15. Computing

A

[x

y

]=

[−2 00 1

] [x

y

]=

[−2x

y

]we see that A stretches the vector in the horizontal direction (i.e., stretches its x coordinate) by afactor of 2, and then reflects the stretched vector across the y-axis.

17. We replace θ by −θ, thus getting the matrix

A =[

cos(−θ) − sin(−θ)sin(−θ) cos(−θ)

]=

[cos θ sin θ

− sin θ cos θ

](Recall that sin(−θ) = − sin θ and cos(−θ) = cos θ.)

19. We compute

A

[x

y

]=

[1 11 1

] [x

y

]=

[x + y

x + y

]= (x + y)

[11

]The transformation A takes a vector v = [x, y] and maps it onto the vector w = Av = [x + y, x + y]which lies on the line of slope 1.

The length of w is |x + y|√

2. If x + y > 0, then w is in the first quadrant, and if x + y < 0, thenit is in the third quadrant.

If x + y = 0, then w is the zero vector.


21. Let v = [x, y]. A transforms v first to [x/4, y/4], and then to [−x/4,−y/4]. From A[x, y] =[−x/4,−y/4] we see that

A =[−1/4 0

0 −1/4

]To check:

Av =[−1/4 0

0 −1/4

] [x

y

]=

[−x/4−y/4

]

23. Let

A =[

a b

c d

]and compute the images of the unit vectors:

A

[10

]=

[0

−1

]and A

[01

]=

[−10

]Based on these two facts, we now recover the entries in A.

A

[10

]=

[a b

c d

] [10

]=

[a

c

]=

[0

−1

]it follows that a = 0 and c = −1. Likewise, from

A

[01

]=

[a b

c d

] [01

]=

[b

d

]=

[−10

]we conclude that b = −1 and d = 0. Thus,

A =[

0 −1−1 0

]If v = [x, y], then

v = x

[10

]+ y

[01

]and we obtain the general formula

A(v) = A

(x

[10

]+ y

[01

])= xA

[10

]+ yA

[01

]= x

[0

−1

]+ y

[−10

]=

[−y

−x

]

25. Using the formula of Example 10.3,

A =[

cos(5π/6) − sin(5π/6)sin(5π/6) cos(5π/6)

]=

[−√3/2 −1/21/2 −

√3/2

]

27. Define B(v) = A−1v (the matrix A−1 exists by assumption). From

A(B(v)) = A(A−1v) = AA−1v = Iv = v

B(A(v)) = B(Av) = A−1Av = Iv = v

we conclude that A(v) and B(v) are inverse transformations. So the matrix corresponding to theinverse transformation of A(v) is A−1.

29. Let v1 and v2 be vectors. Then

(A ◦ B)(v1 + v2) = A(B(v1 + v2))

the above is the definition of the composition; using the fact that B is linear:= A(B(v1) + B(v2))

now using the fact that A is linear:


= A(B(v1)) + A(B(v2))

by the definition of the composition:= (A ◦ B)(v1) + (A ◦ B)(v2)

As well (α is a real number)

(A ◦ B)(αv) = A(B(αv)) = A(αB(v)) = αA(B(v)) = α(A ◦ B)(v)

The two identities (A◦B)(v1 +v2) = (A◦B)(v1)+(A◦B)(v2) and (A◦B)(αv) = α(A◦B)(v) provethat the composition A ◦ B is a linear transformation.

31. We compute

A(5v) = 5A(v) = 5

⎡⎣ 1 0 −4

3 0 10 −2 1

⎤⎦

⎡⎣−1

11

⎤⎦ = 5

⎡⎣−5−2−1

⎤⎦ =

⎡⎣−25−10−5

⎤⎦

A(4w) = 4A(w) = 4

⎡⎣ 1 0 −4

3 0 10 −2 1

⎤⎦

⎡⎣ 0

02

⎤⎦ = 4

⎡⎣−8

22

⎤⎦ =

⎡⎣−32

88

⎤⎦

and, using the linearity of A,

A(5v + 4w) = A(5v) + A(4w) =

⎡⎣−25−10−5

⎤⎦ +

⎡⎣−32

88

⎤⎦ =

⎡⎣−57

−23

⎤⎦

33. We compute

Av =

⎡⎣ 1 0 0

0 1 00 0 −1

⎤⎦

⎡⎣ x

y

z

⎤⎦ =

⎡⎣ x

y

−z

⎤⎦

So A preserves the x and y coordinates of a vector, and changes the sign of its z coordinate. Thus, Ais the reflection across the xy-plane.

35. The reflection with respect to the xz-plane preserves the x and z coordinates of the vector, andchanges the sign of its y coordinate. Thus, the matrix of this transformation is

A =

⎡⎣ 1 0 0

0 −1 00 0 1

⎤⎦


Section 11 Eigenvalues and Eigenvectors

1. Any matrix of the form

A =[

3 b

c 7

]where one of (or both) b = 0 and c = 0 has eigenvalues 3 and 7.

Check: the characteristic polynomial of A is

det(A − λI) =∣∣∣∣ 3 − λ b

c 7 − λ

∣∣∣∣ = (3 − λ)(7 − λ) − bc = (3 − λ)(7 − λ)

The solutions of (3 − λ)(7 − λ) = 0 are λ = 3 and λ = 7.

3. Assume that

A =[

a b

c d

]has eigenvalues 1 and −3. Its characteristic polynomial is (λ − 1)(λ + 3) = 0, i.e., λ2 + 2λ − 3 = 0.On the other hand, the characteristic polynomial of A is λ2 − (a + d)λ + (ad − bc) = 0. Comparingthe two, we get a + d = −2 and ad− bc = −3. We have two equations with four variables (the entriesof A), so there are infinitely many solutions. For instance, take a = −1; then d = −1; the remainingequation ad − bc = −3 gives 1 − bc = −3 and bc = 4. Take b = 4 and c = 1; thus, the matrix

A =[−1 4

1 −1

]has eigenvalues 1 and −3.

To solve a + d = −2 and ad− bc = −3 in general, we use parameters. Let d = t; then a + d = −2implies that a = −2−t. Substituting into ad−bc = −3 we get (−2−t)t−bc = −3 and bc = −t2−2t+3.Let b = s (another parameter); then c = (−t2 − 2t + 3)/s. So, any matrix of the form

A =[ −2 − t s

(−t2 − 2t + 3)/s t

]where s �= 0 (and s and t are picked so that all entries are non-zero, as requested by the exercise) haseigenvalues 1 and −3.

5. Computing

A

[x

y

]=

[2 00 2

] [x

y

]=

[2x

2y

]= 2

[x

y

]we see that A stretches a vector by a factor of 2, and thus preserves all directions (so any non-zerovector is an eigenvector of A). The corresponding eigenvalue is 2.

7. A is a diagonal matrix, so its eigenvalues are 1 and −1. From

A

[x

y

]=

[1 00 −1

] [x

y

]=

[x

−y

]we conclude that A is a reflection across the x-axis. Which directions are preserved by this reflection?A vector sitting on the x-axis remains unchanged under A. Thus, [1, 0] is an eigenvector, and thecorresponding eigenvalue is 1. The vector [0, 1] is mapped to [0,−1]; this means that the directionof the y-axis is also preserved; so, the vector [0,−1] is an eigenvector of A and the correspondingeigenvalue is −1. To make it more obvious:

A

[01

]=

[0

−1

]= −1

[01

]


9. Since

Av1 =[

0 23 1

] [44

]=

[8

16

]is not parallel to v1, we conclude that v1 is not an eigenvector of A. From

Av2 =[

0 23 1

] [−22

]=

[4

−4

]= −2

[−22

]= −2v2

it follows that v2 is an eigenvector of A with eigenvalue −2. Since

Av3 =[

0 23 1

] [12

]=

[45

]�= λ

[12

]for any real number λ, v3 is not an eigenvector of A.

11. The matrix is diagonal, so the eigenvalues are λ1 = 7 and λ2 = −5.

To find an eigenvector corresponding to λ1 = 7, we need to solve the equation

Av = 7v[7 00 −5

] [x

y

]= 7

[x

y

]

for the coordinates x and y of the vector v. The corresponding linear system

7x = 7x

−5y = 7y

simplifies to

0x = 0y = 0

Thus, x can be any (non-zero) real number and y = 0. Let x = t; any vector of the form

v1 =[

t

0

]= t

[10

](with t �= 0) is an eigenvector corresponding to λ1 = 7. To find an eigenvector corresponding toλ2 = −5, we proceed as above:

Av = −5v[7 00 −5

] [x

y

]= −5

[x

y

]

7x = −5x

−5y = −5y

x = 00y = 0

This time, x = 0 and y can be any non-zero number. Let y = t; any vector of the form

v2 =[

0t

]= t

[01

](where t �= 0) is an eigenvector corresponding to λ2 = −5.

13. Call the given matrix A. The characteristic equation of A is

det(A − λI) =∣∣∣∣ 1 − λ 2

2 4 − λ

∣∣∣∣ = 0

i.e.,

(1 − λ)(4 − λ) − 4 = 0λ(λ − 5) = 0


Thus, the eigenvalues are λ1 = 0 and λ2 = 5. (Recall that, instead of recalculating the characteristicequation every time we need it, we can use the ready-made formula λ2 − (trA)λ + detA = 0.)


Av = 0v[1 22 4

] [x

y

]= 0

[x

y

]


x + 2y = 02x + 4y = 0

simplifies to

x + 2y = 0

Let y = t; then x = −2t and so any vector of the form

v1 =[−2t

t

]= t

[−21

](with t �= 0) is an eigenvector corresponding to λ1 = 0. To find an eigenvector corresponding to λ2 = 5,we solve

Av = 5v[1 22 4

] [x

y

]= 5

[x

y

]

x + 2y = 5x

2x + 4y = 5y

−4x + 2y = 02x − y = 0

The two equations are identical. Let x = t; then y = 2t, and any vector of the form

v2 =[

t

2t

]= t

[12

](where t �= 0) is an eigenvector corresponding to λ2 = 5.

15. Call the given matrix A. Using λ2 − (trA)λ + detA = 0 we find the characteristic equationλ2 − 12λ + 35 = 0. It follows that (λ − 5)(λ − 7) = 0, and the eigenvalues are λ1 = 5 and λ2 = 7.


Av = 5v[11 −212 1

] [x

y

]= 5

[x

y

]


11x − 2y = 5x

12x + y = 5y

simplifies to a single equation

6x − 2y = 012x − 4y = 0

Let x = t; then y = 3t, and any vector of the form

v1 =[

t

3t

]= t

[13

]


(with t �= 0) is an eigenvector corresponding to λ1 = 5. To find an eigenvector corresponding to λ2 = 7,we proceed as above:

Av = 7v[11 −212 1

] [x

y

]= 7

[x

y

]

11x − 2y = 7x

12x + y = 7y

4x − 2y = 012x − 6y = 0

Let x = t; then y = 2t, and any vector of the form

v2 =[

t

2t

]= t

[12


17. Let

A =[

1.5 3.53.5 1.5

]Using λ2 − (trA)λ + detA = 0 we find the characteristic equation λ2 − 3λ − 10 = 0. It follows that(λ + 2)(λ − 5) = 0, and the eigenvalues are λ1 = −2 and λ2 = 5.

To find an eigenvector corresponding to λ1 = −2, we solve the equation

Av = −2v[1.5 3.53.5 1.5

] [x

y

]= −2

[x

y

]


1.5x + 3.5y = −2x

3.5x + 1.5y = −2y

simplifies to

3.5x + 3.5y = 0

Let x = t; then y = −t and any vector of the form

v1 =[

t

−t

]= t

[1

−1

](with t �= 0) is an eigenvector corresponding to λ1 = −2. To find an eigenvector corresponding toλ2 = 5, we proceed in the same way as above:

Av = 5v[1.5 3.53.5 1.5

] [x

y

]= 5

[x

y

]

1.5x + 3.5y = 5x

3.5x + 1.5y = 5y

−3.5x + 3.5y = 03.5x − 3.5y = 0

Let x = t; then y = t, and any vector of the form

v2 =[

t

t

]= t

[11



19. From [2 00 −3

] [10

]=

[20

]= 2

[10

]we conclude that [1, 0] is an eigenvector with eigenvalue 2. Likewise,[

2 00 −3

] [01

]=

[0

−3

]= −3

[01

]shows that [0, 1] is an eigenvector with eigenvalue −3.

21. Yes. It is assumed that Av = 4v for some vector v �= 0. Then

(−A)v = −A · v = −(A · v) = −(4v) = −4v

23. The characteristic equation is λ2 − 9 = 0; the eigenvalues are λ1 = 3 and λ2 = −3.

To find an eigenvector corresponding to λ1 = 3, we need to solve the equation[ −9 6−12 9

] [x

y

]= 3

[x

y

]


−9x + 6y = 3x

−12x + 9y = 3y

simplifies to

−2x + y = 0

Let x = t; then y = 2t and any vector of the form

v1 =[

t

2t

]= t

[12

](with t �= 0) is an eigenvector corresponding to λ1 = 3. To find an eigenvector corresponding toλ2 = −3, we proceed as above: [ −9 6

−12 9

] [x

y

]= −3

[x

y

]

−9x + 6y = −3x

−12x + 9y = −3y

−6x + 6y = 0

Thus, −x + y = 0. Let x = t; then y = t, and any vector of the form

v2 =[

t

t

]= t

[11

](where t �= 0) is an eigenvector corresponding to λ2 = −3. Thus, the invariant directions are repre-sented by the line of slope 2 (eigenvalue 3) and the line of slope 1 (eigenvalue −3). See below.

y

x

v1=[1,2]

v2=[1,1]

λ1=3λ2=-3


25. It is assumed that Av = λv and Bv = μv for some real numbers λ and μ. The calculation

(A + B)v = Av + Bv = λv + μv = (λ + μ)v

proves that v is an eigenvector of A + B, and the corresponding eigenvalue is λ + μ.

27. It is assumed that Av = λv for a real number λ. The calculation

A2v = AAv = A(Av) = A(λv) = λA(v) = λ(λv) = λ2v

proves that v is an eigenvector of A2, and the corresponding eigenvalue is λ2.

29. We write the given system as v′(t) = Av(t), where v(t) = [ x(t) y(t) ] , and

A =[

1 22 4

]is the matrix of the system.

In Exercise 13 we showed that the eigenvalues of A are λ1 = 0 and λ2 = 5. The correspondingeigenvectors are: v1 = [−2 1 ] (for λ1 = 0) and v1 = [ 1 2 ] (for λ2 = 5). The solution of the givensystem is

v(t) = C1e0t

[−21

]+ C2e

5t

[12

]or, writing out the coordinates,

x(t) = −2C1 + C2e5t

y(t) = C1 + 2C2e5t

31. We write the given system as v′(t) = Av(t), where v(t) = [ x(t) y(t) ] , and

A =[

3 80 −1

]is the matrix of the system.

The matrix is upper-triangular, so its eigenvalues are the diagonal entries λ1 = 3 and λ2 = −1.To find an eigenvector corresponding to λ1 = 3, we need to solve the equation[

3 80 −1

] [x

y

]= 3

[x

y

]


3x + 8y = 3x

−y = 3y

simplifies to

y = 0

(and no conditions on x). We need one eigenvector, so we pick v1 = [ 1 0 ] . To find an eigenvectorcorresponding to λ2 = −1, we proceed as above:[

3 80 −1

] [x

y

]= −1

[x

y

]

3x + 8y = −x

−y = −y

4x + 8y = 0


and x + 2y = 0. Pick y = 1; then x = −2, and so an eigenvector for λ2 = −1 is v2 = [−2 1 ] .The solution of the given system is

v(t) = C1e3t

[10

]+ C2e

−t

[−21

]or, writing out the coordinates,

x(t) = C1e3t − 2C2e

−t

y(t) = C2e−t


Section 12 The Leslie Model: Age-Structured Population Dynamics

1. (a) The population is broken down into three age groups; we name them child, adult and senior.The entries in the first row are the birth parameters: an adult produces on average 2 offspring,whereas the average number of offspring per senior is 3. The remaining non-zero entries are thesurvival probabilities: a child survives to become adult with a 50% chance. An adult survives tobecome a senior with the probability of 0.3, i.e., with a 30% chance. The entry 0 in the first row andthe first column of L means that children do not produce offspring.(b) We compute:

P (1) = LP (0) =

⎡⎣ 0 2 3

0.5 0 00 0.3 0

⎤⎦

⎡⎣ 120

20030

⎤⎦ =

⎡⎣ 490

6060

⎤⎦

P (2) = LP (1) =

⎡⎣ 0 2 3

0.5 0 00 0.3 0

⎤⎦

⎡⎣ 490

6060

⎤⎦ =

⎡⎣ 300

24518

⎤⎦

3. All birth parameters are zero, which means that no offspring are produced. The population willgo extinct after the number of time intervals reaches the number of age groups. Indeed, for an initialpopulation P (0) = [ a b c ] , we find

P (1) = LP (0) =

⎡⎣ 0 0 0

0.5 0 00 0.5 0

⎤⎦

⎡⎣ a

b

c

⎤⎦ =

⎡⎣ 0

0.5a

0.5b

⎤⎦

P (2) = LP (1) =

⎡⎣ 0 0 0

0.5 0 00 0.5 0

⎤⎦

⎡⎣ 0

0.5a

0.5b

⎤⎦ =

⎡⎣ 0

00.25a

⎤⎦

P (3) = LP (2) =

⎡⎣ 0 0 0

0.5 0 00 0.5 0

⎤⎦

⎡⎣ 0

00.25a

⎤⎦ =

⎡⎣ 0

00

⎤⎦

5. The entry in the second row and the first column is supposed to represent a survival probability.However, a probability cannot be larger than 1 (i.e., larger than 100%).

7. (a) The polulation is broken down into four age groups; we name them child, adolescent, adultand senior. The entries in the first row are the birth parameters: an adult produces on average 3offspring, and no other age group produces offspring. The remaining non-zero entries are the survivalprobabilities: a child survives to become adolescent with a 20% chance. An adolescent survives tobecome an adult with a 60% chance, and an adult survives to become senior with a 40% chance.(b) With a 20% chance of survival, children have the highest mortality.(c) We compute

P (1) = LP (0) =

⎡⎢⎢⎢⎣

0 0 3 00.2 0 0 0

0 0.6 0 00 0 0.4 0

⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣

10020015040

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

45020

12060

⎤⎥⎥⎥⎦

P (2) = LP (1) =

⎡⎢⎢⎢⎣

0 0 3 00.2 0 0 0

0 0.6 0 00 0 0.4 0

⎤⎥⎥⎥⎦

⎡⎢⎢⎢⎣

45020

12060

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

360901248

⎤⎥⎥⎥⎦


9. (a) The characteristic equation is λ2 − 0.64 = 0 and the eigenvalues are λ1 = 0.8 and λ2 = −0.8.(b) The positive eigenvalue determines the behaviour of each age group in the long term. In thiscase, the relative growth rates within each age group approach 0.8, implying that the population size(within each age group, as well at total) will decrease in the long term.(c) The stable age distribution is extracted from the eigenvector corresponding to the positive eigen-value. To find an eigenvector corresponding to λ1 = 0.8, we need to solve the equation[

0 20.32 0

] [x

y

]= 0.8

[x

y

]

for the coordinates x and y of the eigenvector v. The corresponding linear system

2y = 0.8x

0.32x = 0.8y

simplifies to

5y = 2x

We need one eigenvector, so we pick v = [ 5 2 ] . Thus, the stable age distribution is given by thevector [ 5 2 ]; i.e., over time, the ratio of the first age group in the total population approaches 5/7,and the ratio of the second age group in the total population approaches 2/7.

11. (a) The characteristic equation is λ2 − 1.2λ − 0.13 = 0; thus, (λ − 1.3)(λ + 0.1) = 0, and theeigenvalues are λ1 = 1.3 and λ2 = −0.1.(b) The positive eigenvalue determines the behaviour of each age group in the long term. In thiscase, the relative growth rates within each age group approach 1.3, implying that the population size(within each age group, as well at total) will increase in the long term.(c) The stable age distribution is extracted from the eigenvector corresponding to the positive eigen-value. To find an eigenvector corresponding to λ1 = 1.3, we solve the equation[

1.2 0.50.26 0

] [x

y

]= 1.3

[x

y

]

for the coordinates x and y of the eigenvector v. The corresponding linear system

1.2x + 0.5y = 1.3x

0.26x = 1.3y

simplifies to

5y = x

We need one eigenvector, so we pick v = [ 5 1 ] . Thus, the stable age distribution is given by thevector [ 5 1 ] . Over time, the ratio of the first age group in the total population approaches 5/6, andthe ratio of the second age group in the total population approaches 1/6.

13. There are two age groups, call them child and adult. Only adults produce offspring, and therate is on average 1 offspring per adult. A child has a 50% chance of survival. Because only half ofthe children survive, there will be fewer adults to produce offspring in the next generation, so thepopulation will decrease. Indeed, starting with P (0) = [ 100 100 ] , we compute

P (1) = LP (0) =[

0 10.5 0

] [100100

]=

[10050

]=

[100

0.5 · 100

]

P (2) = LP (1) =[

0 10.5 0

] [10050

]=

[5050

]=

[0.5 · 1000.5 · 100

]

P (3) = LP (2) =[

0 10.5 0

] [5050

]=

[5025

]=

[0.5 · 100

0.52 · 100

]

P (4) = LP (3) =[

0 10.5 0

] [5025

]=

[2525

]=

[0.52 · 1000.52 · 100

]


P (5) = LP (4) =[

0 10.5 0

] [2525

]=

[25

12.5

]=

[0.52 · 1000.53 · 100

]

P (6) = LP (5) =[

0 10.5 0

] [25

12.5

]=

[12.512.5

]=

[0.53 · 1000.53 · 100

]· · ·

P (9) =[

0.54 · 1000.55 · 100

]

P (10) =[

0.55 · 1000.55 · 100

]≈

[3.13.1

]

Clearly, the population goes extinct.

15. (a) The matrix is diagonal, so its eigenvalues are λ1 = 2 and λ = 3. In the usual way (or byguessing) we find the corresponding eigenvectors v1 = [1, 0] and v2 = [0, 1].(b) No calculations are needed:

u =[

56

]= 5

[10

]+ 6

[01

]= 5v1 + 6v2

(c) Using formula (12.11), we get

L8u = 5(2)8v1 + 6(3)8v2 =[

5(2)8

6(3)8

]=

[1280

39366

]

17. (a)The characteristic equation is λ2 − 3λ− 4 = (λ− 4)(λ+1) = 0; the eigenvalues are λ1 = 4 andλ2 = −1.

To find an eigenvector corresponding to λ1 = 4, we need to solve the equation[1 23 2

] [x

y

]= 4

[x

y

]


x + 2y = 4x

3x + 2y = 4y

simplifies to

3x − 2y = 0

Let x = 2; then y = 3 and so v1 = [2, 3] is an eigenvector corresponding to λ1 = 4. To find aneigenvector corresponding to λ2 = −1, we proceed as above:[

1 23 2

] [x

y

]= −1

[x

y

]

x + 2y = −x

3x + 2y = −y

x + y = 0

Let x = 1; then y = −1 and so v2 = [1,−1] is an eigenvector corresponding to λ2 = −1.

(b) We need to find a and b so that

u =[

41

]= a

[23

]+ b

[1

−1

]The corresponding system is 2a + b = 4, 3a − b = 1. Adding the two equations, we get 5a = 5 anda = 1. Thus, b = 2, and

u = 1[

23

]+ 2

[1

−1

]= 1v1 + 2v2.


(c) Using formula (12.11), we get

L20u = 1(4)20v1 + 2(−1)20v2 = 420

[23

]+ 2

[1

−1

]=

[2 · 420 + 23 · 420 − 2

]

19. The matrix is diagonal, so its eigenvalues are λ1 = 7 and λ = −3. In the usual way (or byguessing) we find the corresponding eigenvectors v1 = [1, 0] and v2 = [0, 1]. Now

u =[

24

]= 2

[10

]+ 4

[01

]= 2v1 + 4v2

Using formula (12.11), we get

L13u = 5(7)13v1 + 4(−3)13v2 =[

5(7)13

−4(3)13

]

21. From [7535

]= α1

[61

]+ α2

[−34

]we get the system

6α1 − 3α2 = 75α1 + 4α2 = 35

Multiplying the second equation by −6 and adding to the first, we get −27α2 = −135 and α2 = 5.Using either equation, we get α1 = 15.

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