Linear Algebra

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Linear Algebra [Solutions]. = √. √. √. √. (. 8 + 2(2. √. 2). √ ..... (a) By the definition of the dot product, v · w = v1w1 + v2w2 and w · v = w1v1 + w2v2. Because.
Students’ Solutions Manual

Linear Algebra This manual contains solutions to odd-numbered exercises from the book Linear Algebra by Miroslav Lovri´c, published by Nelson Publishing. Keep in mind that the solutions provided represent one way of answering a question or solving an exercise. In many cases there are alternatives, so make sure that you don’t dismiss your solution just because 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, investigate different 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 you just read a solution you might fail to recognize the hard part(s); even worse, you might completely miss the point of the exercise. I accept full responsibility for errors in this text and will be grateful to anybody who brings them to my attention. Your comments and suggestions will be greatly appreciated.

Miroslav Lovri´c Department of Mathematics and Statistics McMaster University e-mail: [email protected]

September 2011

Section 1 [Solutions]

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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 + y 2  √ 2 2 The distance A = (1, 3) to the origin  from the point  is d(A, 0) = √(1) + (3) = 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, √ √ 6 ± 36 + 4 · 87 6 ± 2 9 + 87 a= = 2 2 √ √ √ i.e., a = 3 ± 96 = 3 ± 16 · 6 = 3 ± 4 6. 5. The distance is d=

7. From

 √ √ (4 − 1)2 + (5 − 2)2 + (6 − 3)2 = 9 · 3 = 3 3

 √ (6)2 + (1)2 + (−6)2 = 73  √ d((0, 0, 6), (9, 0, −4)) = (9)2 + (0)2 + (−10)2 = 181

d((3, −1, 2), (9, 0, −4)) = 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 the angle of π/4. All three points lie on a circle of radius 1 centred at the origin. See the figure below. y (1,π/2)

0

(1, 1) (1, π/4)

1

x

11. The Cartesian coordinates are (10 cos π, 10 sin π) = (−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, are √ 3π/4) √ (in polar coordinates). Their Cartesian √ coordinates √ A = (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   √ 2  √ 2  √ √ 2 2  d(A, B) = + 2 2− 2 2+ 2 2

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Linear Algebra [Solutions]

     √ √  √ √ √ 2 1 2 1  + + 8 − 2(2 2) + = 17 = 8 + 2(2 2) 2 2 2 2 Note: in Exercise 42 we derive the distance formula in polar coordinates (in which case no conversion to 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) = r12 + r22 − 2r1 r2 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 coordinates of (−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).

21. We compute r =

 √ 23. We compute r = (−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 polar coordinates 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 is a 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 polar coordinates are ( 2, 3π/4). 35. We get r = (2, 4π/3). 37. We get r = (2, 7π/6).





1 + 3 = 2, and tan θ =



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

√ 3 + 1 = 2, and tan θ = 1/ 3, i.e., θ = π/6 + π = 7π/6. The polar coordinates are

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 each set of parentheses, we get

2

2

2

2 a1 − b1 a2 − b2 a1 + b1 a2 + b2 2 + a2 − = + d (A, M ) = a1 − 2 2 2 2

2

2

2

2 a1 + b1 a2 + b2 a1 − b1 a2 − b2 d2 (M, B) = = + − b1 + − b2 2 2 2 2

Section 1 [Solutions]

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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 such numbers work: for instance, a = 4 and b = −2 satisfy a2 + b2 = 16 + 4 = 20 < 25; however, if a = −4 and 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 the most either population changes is by ±4 members. For instance, there could be 4 more rabbits and 2 fewer 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 foxes or 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, or rabbits, or both foxes and rabbits) are large.

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Linear Algebra [Solutions]

Section 2 Vectors 1. Let A = (10, 3) and C = (−4, −7). The representative directed line segment of v = [3, −1] which − − → starts at A is AB, where B = (10 + 3, 3 − 1) = (13, 2). The representative directed line segment which −−→ starts 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 line −−→ segment 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 = −4v, we get w = (−4v v) = − 4v v = 4v v = 4v2 = 4(3)2 = 36 Alternatively, simplify w first: w = −4v v = −12v; 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 vector √ the √ v v/v is a unit vector parallel to v. Thus, w = 10 v . Since v = 1 + 16 + 1 = 18 = 3 2, it √ [−1, 4, 1]. follows that w = 310 2 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

19. From d = αb + βc we get





 0 −1 −1 =α +β 4 1 0

and

√1 b 2

=

√1 [−1, 1]. 2

  0 −α − β = 4 α

Thus −α − β = 0 and α = 4; it follows that β = −4. 21. From a +√c = [3, −2] √= [2, −2] we compute a + c = √ + [−1, 0] a + c = 9 + 4 + 1 + 0 = 13 + 1.



4+4=

√ 8. On the other hand,

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 + 3b 11x = 4b − 8a 4b − 8a 1 x= = (4b − 8a) 11 11

Section 2 [Solutions]

27. We compute

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⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 0 −1 −8 0 4 −4 −2a + 3j − 4b = −2 ⎣ 0 ⎦ + 3 ⎣ 1 ⎦ − 4 ⎣ 6 ⎦ = ⎣ 0 ⎦ + ⎣ 3 ⎦ + ⎣ −24 ⎦ = ⎣ −21 ⎦ −2 0 0 4 0 0 4

29. Because b =



√ √ 1 + 36 + 0 = 37, the unit vector in the direction of b is given by √ ⎤ ⎡ ⎤ ⎡ −1 −1/ 37 √ 1 1 u= b = √ ⎣ 6 ⎦ = ⎣ 6/ 37 ⎦ b 37 0 0

31. From a = αc + βi + γk we get ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 4 2 1 0 ⎣ 0⎦ = α⎣ 7⎦ + β ⎣0⎦ + γ ⎣0⎦



⎤ ⎡ ⎤ 4 2α + β and ⎣ 0 ⎦ = ⎣ 7α ⎦ −2 −4 0 1 −2 −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 the second 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 / v12 + v22 , v2 / v12 + v22 ] = (1/ v12 + v22 )[v1 , v2 ], we compute  1 1  w =  2 [v , v ] = v12 + v22 = 1 1 2 v1 + v22 v12 + v22 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

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Linear Algebra [Solutions]

39. See the figure below.

z

( v+w ) +z y

v + (w + z)

v+w

v

w+z

y

w

w+z

v

z

z

z

w

w 0

x

0

x

Section 3 [Solutions]

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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 whose dot 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 a real number. 7. The dots in v · w and w · v represent the dot product, and thus both expressions in the parentheses are 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. ⎡ ⎤ ⎡ ⎤ 0 4 ⎣ ⎣ ⎦ 11. 8 · 0 ⎦ = (4)(0) + (8)(0) + (0)(3) = 0. 0 3 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 slope 2/5 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 plane x − 5y + 6z = 0. √ √ 19. From v · w = [ 1 2 0 ] · [ 0 3 1 ] = 6, v = 5, and w = 10, we get 6 v·w = √ √ ≈ 0.849 cos θ = v w 5 10 Using a calculator, θ ≈ 0.557 radians. √ √ 21. From v · w = (2i + 4k) · (−i − 2j + 5k) = 18, v = 20, and w = 30, we get v·w 18 cos θ = = √ √ ≈ 0.735 v w 20 30 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 in the triangle ABC are perpendicular. √ √ √ Alternatively, we calculate the lengths of the sides: v = 9, w = 20, and z = 29. Since z2 = v2 + w2 , the triangle ABC is a right-angled triangle (by the Pythagorean Theorem).

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Linear Algebra [Solutions]

25. (a) By the definition of the dot product, v · w = v1 w1 + v2 w2 and w · v = w1 v1 + w2 v2 . Because the 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 ) = v1 w1 + v1 z1 + v2 w2 + v2 z2 = (v1 w1 + v2 w2 ) + (v1 z1 + v2 z2 ) =v·w+v·z (b) Using the properties of real numbers, we get (tv) · w = [tv1 , tv2 ] · [w1 , w2 ] = (tv1 )w1 + (tv2 )w2 = t(v1 w1 ) + t(v2 w2 ) = t(v1 w1 + v2 w2 ) = t(v · w)

27. (a) It is assumed that v · w = w · v = 0. Using a2 = a · a, we compute v + w2 = (v + w) · (v + w) =v·v+v·w+w·v+w·w =v·v+w·w = v2 + w2 (b) Using the equality of the norm squared and the dot product as in (a), v + w2 = v2 + w2 (v + w) · (v + w) = v2 + w2 v · v + v · w + w · v + w · w = v2 + w2 v2 + 2v · w + w2 = v2 + w2 2v · w = 0 Thus, v · w = 0, i.e., v and w are orthogonal.

Section 4 [Solutions]

L1-9

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 t 0 1 = = +t y 4t + b b 4 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 not a real number. 5. The y-axis is characterized by x = 0 and z = 0; its parametric equations are ⎧x=0 ⎨ y=t ⎩ z=0 where t is a real number. 7. The xz-plane is characterized by y = 0; its parametric equations are ⎧x=s ⎨ y=0 ⎩ z=t where s, t ∈ R. 9. Let y = t (to avoid fractions). Then x = 5t − 9, and the vector equation is  



 5t − 9 −9 5 x = = +t y t 0 1 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 vector equation of the line is (we pick (4, 0) for a point on the line):

 

 x 4 −4 = +t y 0 −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 get x = 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]. A vector equation of the line is (we pick (0, 5, 0) for a point on the line): ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x 0 −1 ⎣ y⎦ = ⎣5⎦ + t⎣ 3⎦ z

0

2

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Linear Algebra [Solutions]

where t ∈ R. In parametric form,

⎧ x = −t ⎨ y = 5 + 3t ⎩ z = 2t

where t ∈ R. 15. The vector equation is

and the parametric equations are

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ −2 x 1 ⎣ y⎦ = ⎣2⎦ + t⎣ 0⎦ 4 z 7 ⎧ ⎪ ⎨ 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. To obtain parametric equations, we introduce a parameter: let x = t; then y = −x + 2 = −t + 2. Thus,  x=t y = −t + 2 where 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 3 2 = +t y −1 −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

and the parametric equations are

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x 3 0 0 ⎣ y⎦ = ⎣ 4⎦ + s⎣1⎦ + t⎣0⎦ z −1 4 2 ⎧ ⎪ ⎨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 vector from 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 hence are not parallel. The vector form of the equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x 1 2 0 ⎣ y ⎦ = ⎣ 3 ⎦ + s ⎣ 6 ⎦ + t ⎣ −3 ⎦ z 4 −10 −4

Section 4 [Solutions]

and the parametric equations are

L1-11

⎧ ⎪ ⎨ x = 1 + 2s y = 3 + 6s − 3t ⎪ ⎩ z = 4 − 10s − 4t

where s, t ∈ R. 25. From

⎡ ⎤ ⎡ ⎤ 3 x−4 ⎣0⎦ · ⎣y − 2⎦ = 0

1 z+5 we get 3(x − 4) + 1(z + 5) = 0, and 3x + z − 7 = 0. To obtain parametric form, we use one parameter for the missing variable y, since it can take on any value: y = s. When x = t, 3t + z − 7 = 0 and z = −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 to it. Thus, a vector equation of the plane passing through (2, 1, 3) parallel to the plane x + 4z − 1 = 0 is ⎤ ⎡ ⎤ ⎡ 1 x−2 ⎣0⎦ · ⎣y − 1⎦ = 0 4 z−3 We get 1(x − 2) + 4(z − 3) = 0, and x + 4z − 14 = 0. In parametric form, ⎧ x = 14 − 4t ⎨ ⎩

y=s z=t

where s, t ∈ R. 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 equation of the plane is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x 0 0 1 ⎣ y ⎦ = ⎣ 2 ⎦ + s ⎣ 4 ⎦ + t ⎣ −1 ⎦ z −6 2 2 and the parametric equations are ⎧x=t ⎨ ⎩ where s, t ∈ R.

y = 2 + 4s − t

z = −6 + 2s + 2t

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

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Linear Algebra [Solutions]

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 and t = 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 = 0 and y = 9t + 4. So, the parametric equations of the line are ⎧ x = 1 − 3t ⎨ ⎩

y = 4 + 9t z=t

where t ∈ R. 39. The normal vector to the plane, n = [1, −3, 1], is parallel to the line we are looking for (and thus can be taken as its direction vector). The vector equation of the line is ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 0 x 1 ⎣ y ⎦ = ⎣ 8 ⎦ + t ⎣ −3 ⎦ 2 z 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 [Solutions]

L1-13

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 equal slopes. Since they are not parallel, they intersect at a single point. Thus, the given system has a unique solution. To find the solution, compute y from the first equation (y = −x + 2) and substitute into the second equation: 2x − 2(−x + 2) = 5; it follows that 4x = 9 and x = 9/4. The unique solution is x = 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). To parametrize 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 therefore intersect 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 system is x = 1, y = 1. 11. Multiplying the second equation by 2 we obtain the first equation. Thus, the lines represented by the two equations overlap, i.e., are the same line. The system has infinitely many solutions, and the solution 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 that the two normal vectors are not parallel, we conclude that the planes 2x − y − z = 4 and 3x + z = −2 are 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 we get 2t − y − (−3t − 2) = 4 and y = 5t − 2. Thus, ⎧x=t ⎨ ⎩ where t ∈ R.

y = 5t − 2

z = −3t − 2

L1-14

Linear Algebra [Solutions]

15. First we compute x: (R1 ) (R2 )

Ax + By = P Cx + Dy = Q

(R3 ← DR1 ) (R4 ← (−B)R2 )

ADx + BDy = DP −BCx − BDy = −BQ (AD − BC)x = DP − BQ DP − BQ x= AD − BC

(R5 ← R3 + R4 ) (R6 ← R5 /(AD − BC))

In the same way, we compute y: Ax + By = P Cx + Dy = Q

(R1 ) (R2 ) (R3 ← CR1 )

ACx + BCy = CP −ACx − ADy = −AQ

(R4 ← (−A)R2 )

(−AD + BC)y = CP − AQ −(AQ − CP ) y= −(AD − BC) AQ − CP y= AD − BC

(R5 ← R3 + R4 ) (R6 ← R5 /(AD − BC))

17. The equations R1 and R4 below represent the system in upper-triangular form. 2x + 10y = 6 −x + y = 9

(R1 ) (R2 )

12y = 24 (R3 ← 2R2 + R1 ) 2x + 10y = 6 y=2

(R1 ) (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 7x − 3y = 2 y=0

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

(R1 ) (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 3x − y + 4z = 19

(R1 ) (R2 )

Section 5 [Solutions]

−2x − y + z = 0

L1-15

(R3 )

x + y + 2z = 7 −4y − 2z = −2 y + 5z = 14

(R1 ) (R4 ← −3R1 + R2 ) (R5 ← 2R1 + R3 )

x + y + 2z = 7 −4y − 2z = −2 18z = 54

(R1 ) (R4 ) (R6 ← R4 + 4R5 )

Thus, z = 3. From R4 we compute −4y = 4 and y = −1. The equation R1 reads x + (−1) + 2(3) = 7 and so x = 2. 25. We compute 7x − y − z = 0 y + 6z = 37 3x + 2y = 5

(R1 ) (R2 ) (R3 )

7x − y − z y + 6z 17 3 y+ z 7 7 17y + 3z

=0 = 37

(R1 ) (R2 )

=5

(R4 ← R3 − 3R1 /7)

= 35

(R5 ← 7R4 )

7x − y − z = 0 y + 6z = 37 −33y = −33

(R1 ) (R2 ) (R6 ← R2 − 2R5 )

(Note that this is upper-triangular form when the variables are ordered x, z, y.) From the last equation, 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 3x + z = 5 5x − 2y + 3z = 5 x−y+z =0 3y − 2z = 5 3y − 2z = 5

(R1 ) (R2 ) (R3 ) (R1 ) (R4 ← R2 − 3R1 ) (R5 ← R3 − 5R1 )

The equations R1 and R4 (or R5 ) represent the system in upper-triangular form. We need one parameter: 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 reads 2x − 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 Cx + Dy = Q C CB y + Dy = − P + Q A A (AD − BC)y = −CP + QA



(R1 ) (R2 ) (R3 ← R2 − CR1 /A) (R4 ← AR3 )

L1-16

Linear Algebra [Solutions]

By assumption, AD − BC = 0, and the upper-triangular form is: Ax + By = P 0y = −CP + QA

(R1 ) (R4 )

If −CP + QA = 0, the system has no solutions (that’s the answer to (b); note that −CP + QA = 0 can 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)): take y = 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 slopes and intersect at a single point (unique solution). If a = 1, the system y = x − 2 and y = x − 3 has no solutions (subtracting the second equation from the first, we get 0 = 1). There are no values of a for which the given system has infinitely many solutions. 35. First we compute x: Ax + By = 0 Cx + Dy = 0

(R1 ) (R2 )

ADx + BDy = 0 −BCx − BDy = 0

(R3 ← DR1 ) (R4 ← (−B)R2 )

(AD − BC)x = 0 0x = 0

(R5 ← R3 + R4 ) (R6 )

The equations R1 (or R2 ) and R6 represent the given system in upper triangular form. From R6 we conclude that x can be any real number: x = t, where t ∈ R. From R1 , y = −Ax/B = −At/B, if B = 0. If B = 0 then AD − BC = 0 implies that AD = 0 and D = 0 (since A = 0), and the system reads Ax = 0 and Cx = 0. In this case, the solution is x = 0 and y = t, t ∈ R.

Section 6 [Solutions]

L1-17

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, there should be a 0 in the second row below the leading 1 in the first row). To obtain the matrix in row echelon form:



 1 1 1 1 → ((−1)R1 + R2 → R2 ) 1 0 0 −1

5. The leading entry in the second row is not obtain the matrix in row echelon form: ⎡ ⎤ ⎡ 1 2 3 1 2 ⎣0 0 1⎦ → ⎣0 0 0 0 4 0 0

7. We need one step only:

to the left of the leading entry in the third row. To



 1 2 1 2 → 3 4 0 −2

⎤ 3 1⎦ 0

((−4)R2 + R3 → R3 )

((−3)R1 + R2 → R2 )

9. We need to work on the third row: ⎡ ⎤ ⎡ ⎤ 1 0 −3 1 0 −3 ⎣ 0 4 0⎦ → ⎣0 4 0⎦ −3 0 1 0 0 −8

(3R1 + R3 → R3 )

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

−2 1 1 −2 1 1 (4R1 + R2 → R2 ) → 0 4 4 8 0 0

13. We start by introducing 0 in the second row and the first column: ⎡ ⎤ ⎡ ⎤ 1 3 −2 1 3 −2 ⎣2 4 0 ⎦ → ⎣ 0 −2 4 ⎦ ((−2)R1 + R2 → R2 ) 0 4 1 0 4 1 ⎡ ⎤ 1 3 −2 → ⎣ 0 −2 4 ⎦ (2R2 + R3 → R3 ) 0 0 9 15. We solve the system of equations x − 3y + z = 2 and 2x + 3y − z = 5 by working with the augmented matrix: 

1 −3 1 2 (R1 ) 2 3 −1 5 (R2 ) 

1 −3 1 2 (R1 ) 0 9 −3 1 (R3 ← −2R1 + R2 )

L1-18

Linear Algebra [Solutions]

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

t 1 + x−3 +t=2 3 9 7 21 = x= 9 3 Thus, the set of points that belong to both planes is the line ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ x 7/3 7/3 0 ⎣ y ⎦ = ⎣ t/3 + 1/9 ⎦ = ⎣ 1/9 ⎦ + t ⎣ 1/3 ⎦ z t 0 1 where t ∈ R.

17. The matrix represents a system of three equations in three variables. The zeros in the last row imply that the system is consistent. To solve, we need to take one of the two variables appearing in the 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 we see that the system is consistent. There are four variables but only three equations, so the system has infinitely many solutions. (Note that the solution for the fourth variable (last row) is unique (equal to 5) and the solution for the third variable is unique (second row); thus, the parameter appears in the 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 we see 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 ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ⎤ ⎡ 4 1 3 8 ⎣ −4 ⎦ = a ⎣ 0 ⎦ + b ⎣ 2 ⎦ + c ⎣ 1 ⎦ 2 −1 1 8 Thus, we need to solve the system 3a + b + 4c = 8 2b + c = −4 a − b + 2c = 8 for a, b, and c. Using augmented ⎡ 3 ⎣0 1 ⎡ 0 ⎣0 1 ⎡ 1 ⎣0 0 ⎡ 1 ⎣0 0

matrix, ⎤ 1 4 8 (R1 ) ⎦ (R2 ) 2 1 −4 (R3 ) −1 2 8 ⎤ 4 −2 −16 (R4 ← −3R3 + R1 ) ⎦ 2 1 −4 (R2 ) −1 2 8 (R3 ) ⎤ −1 2 8 (R3 ) ⎦ 2 1 −4 (R2 ) 4 −2 −16 (R4 ) ⎤ −1 2 8 (R3 ) 2 1 −4 ⎦ (R2 ) 0 −4 −8 (R5 ← (−2)R2 + R4 )

Section 6 [Solutions]

L1-19

Thus, −4c = −8 and c = 2. from 2b + c = −4 we get b = −3, and then a − b + 2c = 8 it implies that a − (−3) + 2(2) = 8, i.e., a = 1. 25. To simplify working with the augmented matrix, we switch the first two equations: ⎤ ⎡ (R1 ) 1 1 2 7 ⎣ 3 −1 4 19 ⎦ (R2 ) −2 −1 1 0 (R3 ) ⎤ ⎡ (R1 ) 1 1 2 7 ⎣ 0 −4 −2 −2 ⎦ (R4 ← −3R1 + R2 ) 0 1 5 14 (R5 ← 2R1 + R3 ) ⎤ ⎡ (R1 ) 1 1 2 7 ⎦ ⎣0 1 5 14 (R5 ) 0 −4 −2 −2 (R4 ) ⎤ ⎡ (R1 ) 1 1 2 7 ⎣ 0 1 5 14 ⎦ (R5 ) (R6 ← 4R5 + R4 ) 0 0 18 54 Thus, 18z = 54 and z = 3. By back substitution, y + 5z = 14 and y = −1; from x + y + 2z = 7 we get x = 2. 27. We work with the augmented matrix: ⎤ ⎡ (R1 ) 1 1 −2 4 ⎣2 0 1 0 ⎦ (R2 ) 3 1 −1 4 (R3 ) ⎡ ⎤ 1 1 −2 4 (R1 ) ⎣ 0 −2 ⎦ 5 −8 (R4 ← −2R1 + R2 ) 0 −2 5 −8 (R5 ← −3R1 + R3 ) ⎤ ⎡ (R1 ) 1 1 −2 4 ⎦ ⎣ 0 −2 (R4 ) 5 −8 (R6 ← −R4 + R5 ) 0 0 0 0 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 = 4 and x = −t/2. 29. We work with the augmented matrix (to start, we switch the second and the third equation): ⎤ ⎡ (R1 ) 7 −1 −1 0 ⎣3 2 0 5 ⎦ (R2 ) 0 1 6 37 (R3 ) ⎤ ⎡ 7 −1 −1 0 (R1 ) ⎣ 0 17/7 3/7 5 ⎦ (R4 ← −3R1 /7 + R2 ) 0 1 6 37 (R3 ) ⎤ ⎡ 7 −1 −1 0 (R1 ) ⎣ 0 17 ⎦ 3 35 (R5 ← 7R4 ) 0 1 6 37 (R3 ) ⎤ ⎡ 7 −1 −1 0 (R1 ) ⎣0 ⎦ 1 6 37 (R3 ) 0 17 3 35 (R5 )

L1-20

Linear Algebra [Solutions]

⎤ 0 (R1 ) 7 −1 −1 ⎦ ⎣0 1 6 37 (R3 ) 0 0 −99 −594 (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: ⎤ ⎡ (R1 ) 1 −1 1 0 ⎦ ⎣3 0 1 5 (R2 ) 5 −2 3 5 (R3 ) ⎤ ⎡ 1 −1 1 0 (R1 ) ⎣0 ⎦ (R4 ← −3R1 + R2 ) 3 −2 5 (R5 ← −5R1 + R3 ) 0 3 −2 5 ⎤ ⎡ (R1 ) 1 1 −2 0 ⎣ 0 3 −2 5 ⎦ (R4 ) (R6 ← −R4 + R5 ) 0 0 0 0 The system has infinitely many solutions. Let z = t be a parameter. From 3y − 2z = 5 we get 3y − 2t = 5 and y = 2t/3 + 5/3. Finally, from x − y + z = 0 it follows that x − (2t/3 + 5/3) + t = 0 and x = −t/3 + 5/3. 33. We row reduce the augmented matrix: 

A B P (R1 ) C D Q (R2 ) 

P (R1 ) A B 0 −CB/A + D −CP/A + Q (R3 ← −CR1 /A + R2 ) 

P (R1 ) A B (R ← AR ) 0 AD − BC AQ − CP 4

3

Thus, (AD − BC)y = AQ − CP and y = (AQ − CP )/(AD − BC). Using Ax + By = P, we get AQ − CP Ax = P − B AD − BC P (AD − BC) − B(AQ − CP ) Ax = AD − BC P AD − ABQ A(DP − BQ) Ax = = AD − BC AD − BC DP − BQ x= AD − BC

35. We row reduce the augmented matrix: ⎤ ⎡ 3 1 5 (R1 ) ⎣4 ⎦ 3 5 (R2 ) 5 −1 11 (R3 ) ⎡ ⎤ 3 1 5 (R1 ) ⎣0 5/3 −5/3 ⎦ (R4 ← −4R1 /3 + R2 ) 0 −8/3 8/3 (R5 ← −5R1 /3 + R3 )

Section 6 [Solutions]



3 ⎣0 0 ⎡ 3 ⎣0 0 Thus y = −1. From 3x + y = 5 it 37. We need to solve the system

L1-21

⎤ (R1 ) 1 5 ⎦ 1 −1 (R6 ← R4 /(5/3) 1 −1 (R7 ← R5 /(8/3)) ⎤ (R1 ) 1 5 1 −1 ⎦ (R6 ) 0 0 (R8 ← R6 + R7 ) follows that x = 2.



⎤ ⎡ ⎤ 6−t 3s ⎣ −2 + t ⎦ = ⎣ 2 − 2s ⎦ 5−t 1 + 2s

i.e., 3s + t = 6 2s + t = 4 2s + t = 4 for s and t. By subtracting the second equation from the first equation we get s = 2. From either equation it follows that t = 0. Thus, the two lines intersect at a single point (6, −2, 5) (we obtained this point by substituting either s = 2 or t = 0 into the corresponding equation of the line).

L1-22

Linear Algebra [Solutions]

Section 8 Matrices 1. The product is a 1 × 1 matrix (i.e., a matrix with a single entry), which can be identified with a real number. 3. For example, if



   0 2 0 2 0 2 0 0 , then A2 = = 0 0 0 0 0 0 0 0 There are many other matrices with the same property, such as:



  0 a 0 0 a a , , and , 0 0 a 0 −a −a where a is a non-zero real number. A=

5. Recall that a matrix D is called symmetric if that (A + At )t = A + At . Let ⎡ a11 ⎣ A = a21 a31 Then ⎤ ⎡ ⎡ a11 a21 a11 a12 a13 t ⎦ ⎣ ⎣ A + A = a21 a22 a23 + a12 a22 a31 a32 a33 a13 a23

Dt = D; replacing D by A + At , we have to show a12 a22 a32

⎤ a13 a23 ⎦ a33

⎤ ⎡ 2a11 a31 ⎦ ⎣ a32 = a21 + a12 a33 a31 + a13

a12 + a21 2a22 a32 + a23

⎤ a13 + a31 a23 + a32 ⎦ 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



      1 2 4 0 1 0 7 14 −8 0 1 0 0 14 7A − 2B + I2 = 7 −2 + = + + = 0 4 0 −2 0 1 0 28 0 4 0 1 0 33

9. The matrix operations involved are defined. We compute ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ t

1 2 3 6 1 1 −1 1 0 0 = ⎣9 0⎦ − 4⎣0 2⎦ = ⎣ 9 3E − 4F t = 3 ⎣ 3 0 ⎦ − 4 1 2 3 2 5 6 15 0 3 6

⎤ 2 −8 ⎦ 3

11. A is a 2 × 2 matrix, and E is a 3 × 2 matrix. The product AE is not defined. (Note: the product EA is defined, and is a 3 × 2 matrix.) 13. E is a 3 × 2 matrix, and so the power E 2 = E · E is not defined. Thus, D − E 2 is not defined. 15. The product EF is defined, and is a 3 × 3 matrix. Thus, D − EF is defined. We compute: ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎤ ⎡ ⎤ ⎡ 0 3 2 3 4 6 −3 −1 −4 0 3 2 

1 2 1 0 0 = ⎣ 0 0 6 ⎦ − ⎣ 3 0 0 ⎦ = ⎣ −3 0 6⎦ D − EF = ⎣ 0 0 6 ⎦ − ⎣ 3 0 ⎦ · 1 2 3 2 5 0 0 0 7 10 15 −7 −10 −15 0 0 0

Section 8 [Solutions]

17. Since D is a 3 × 3 matrix, any power ⎡ 0 D2 = D · D = ⎣ 0 0 we get ⎡ 0 3 2 ⎣ D =D ·D = 0 0

of 3 0 0

L1-23

D is defined (and is a 3 × 3 matrix). From ⎤ ⎡ ⎤ ⎡ ⎤ 2 0 3 2 0 0 18 6⎦ · ⎣0 0 6⎦ = ⎣0 0 0⎦ 0 0 0 0 0 0 0

⎤ ⎡ ⎤ ⎡ ⎤ 0 18 0 3 2 0 0 0 0 0⎦ · ⎣0 0 6⎦ = ⎣0 0 0⎦ 0 0 0 0 0 0 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 is defined, and is a 2 × 2 matrix. From



 5 25 30 t · [5 6] = GG = 6 30 36 we obtain

   4 0 25 30 100 120 t t BGG = B(GG ) = · = 0 −2 30 36 −60 −72

21. Let

a11 A= a21 Then

AS =

and

a11 a21

a12 a22



  a12 s 0 sa11 · = a22 sa21 0 s

sa12 sa22



   sa11 sa12 s 0 a11 a12 = SA = · a21 a22 sa21 sa22 0 s Thus, AS = SA. To multiply a matrix by a scalar matrix (whose non-zero entry is s), we multiply each entry in the matrix by s.

23. Let



a11 ⎣ A = a21 a31 We compute:



a12 a22 a32

⎤ a13 a23 ⎦ a33

⎤ ⎡ ⎤ ⎡ ⎤ a11 a12 a13 d1 d1 a11 d2 a12 d3 a13 0 0 AD = ⎣ a21 a22 a23 ⎦ · ⎣ 0 d2 0 ⎦ = ⎣ d1 a21 d2 a22 d3 a23 ⎦ a31 a32 a33 d1 a31 d2 a32 d3 a33 0 0 d3 To compute AD, we multiply the first column of A by d1 , the second column of A by d2 , and the third column of A by d3 . In reversed order: ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ a11 a12 a13 d1 a11 d1 a12 d1 a13 d1 0 0 0 ⎦ · ⎣ a21 a22 a23 ⎦ = ⎣ d2 a21 d2 a22 d2 a23 ⎦ DA = ⎣ 0 d2 a31 a32 a33 d3 a31 d3 a32 d3 a33 0 0 d3 To compute DA, we multiply the first row of A by d1 , the second row of A by d2 , and the third row of A by d3 . Clearly, AD = DA.

L1-24

Linear Algebra [Solutions]

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

  cos θ − sin θ cos θ − sin θ A2 = · sin θ cos θ sin θ cos θ  

2 2 cos 2θ − sin 2θ cos θ − sin θ − sin θ cos θ − sin θ cos θ = = sin 2θ cos 2θ sin θ cos θ + sin θ cos θ − sin2 θ + cos2 θ

27. From 4(X + D) = BC we get 4X + 4D = BC and 1 X = (BC − 4D) 4  

−2 0 8 1 0 1 = −4 4 0 −2 5 1 2  

−16 −2 0 0 1 = − = 4 −10 −2 8 0

 −4 −1/2 = −9/2 −1/2



0 0

1 −16 4 −18

−2 −2



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



  −1 2 8 1 −41/4 −3/4 1 1 −41 −3 1 −5 = = X = (A − 5C) = 4 4 4 −25 −1 0 4 5 1 −25/4 −1/4

31. Let

A=

Then

a11 a21

a12 a22



and B =

b11 b21

b12 b22



   b11 b12 αa11 αa12 b11 a11 a12 = α a21 a22 b21 b22 αa21 αa22 b21

 (αa11 )b11 + (αa12 )b21 (αa11 )b12 + (αa12 )b22 = (αa21 )b11 + (αa22 )b21 (αa21 )b12 + (αa22 )b22

(αA)B =

and

α(AB) = α

a11 a21

a12 a22



b11 b21

b12 b22

b12 b22





 a11 b11 + a12 b21 a11 b12 + a12 b22 =α a21 b11 + a22 b21 a21 b12 + a22 b22

 α(a11 b11 + a12 b21 ) α(a11 b12 + a12 b22 ) = α(a21 b11 + a22 b21 ) α(a21 b12 + a22 b22 ) 

α(a11 b11 ) + α(a12 b21 ) α(a11 b12 ) + α(a12 b22 ) = α(a21 b11 ) + α(a22 b21 ) α(a21 b12 ) + α(a22 b22 ) 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,

1 −2 = (1)(4) − (−3)(−2) = −2 det(A) = −3 4

Section 8 [Solutions]

35. Using the definition, 1 2 3 3 4 − 2 2 4 + 3 2 3 det(A) = 2 3 4 = 1 3 5 3 7 5 7 3 5 7 = 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=

and

ca11 det(cA) = ca21

a11 a21

 a12 ; a22

ca11 ca21

ca12 ca22



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

39. We compute

AB =

and

−1 3 1 4

2 3 BA = 1 1 Thus, AB = BA. However,

and

then cA =



1 det(AB) = 6 1 det(BA) = 0



  2 3 1 0 = 1 1 6 7

  −1 3 1 18 = 1 4 0 7 0 =1·7−6·0=7 7

18 = 1 · 7 − 0 · 18 = 7 7

L1-25

L1-26

Linear Algebra [Solutions]

Section 9 Matrices and Linear Systems 1. Yes. Think of ABC = CAB = I as (AB)C = C(AB) = I; i.e., the matrix AB multiplied by 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 the identity matrix. Thus, A−1 = A2 . 5. The determinant of the given matrix is a 2 − a = a − a(2 − a) = a[1 − (2 − a)] = a(−1 + a) a 1 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 1 2 3 = −2 = 0 the matrix is invertible. Using the formula given in Theorem 8,

  0 1 −1 3 −1 −3/2 1 = = −2 −2 2 3 0 1

1/2 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 0.2 0.4 −0.2 = (0.1)(−0.2) − (0.4)(0.2) = −0.1 = 0 the matrix is invertible. Using the formula given in Theorem 8,

 

−1 −0.2 −0.2 2 2 0.1 0.2 1 = = −0.1 −0.4 0.1 4 −1 0.4 −0.2

13. The determinant of the given matrix is 1 −2 3 0 1 0 = 1 1 0 − (−2) 0 0 + 3 0 1 = 0 1 0 0 0 0 1 0 1 0 We conclude that the matrix is non-invertible (i.e., singular). 15. We use Algorithm 2: ⎡ ⎤ 1 −2 3 1 0 0 (R1 ) ⎣4 ⎦ (R2 ) 1 0 0 1 0 (R3 ) 0 1 1 0 0 1 ⎡ ⎤ 1 −2 3 1 0 0 (R1 ) ⎣0 9 −12 −4 1 0 ⎦ (R4 ← −4R1 + R2 ) 0 1 1 0 0 1 (R3 ) ⎡ ⎤ 1 0 5 1 0 2 (R5 ← 2R3 + R1 ) ⎣ 0 0 −21 −4 1 −9 ⎦ (R6 ← −9R3 + R4 ) 0 1 1 0 0 (R3 ) 1

Section 9 [Solutions]



1 ⎣0 0 ⎡ 1 ⎣0 0 Thus,

L1-27

⎤ 1 0 2 (R5 ) 0 5 ⎦ 0 0 1 1 1 (R3 ) 0 1 4/21 −1/21 9/21 (R7 ← R6 /(−21)) ⎤ 5/21 −3/21 (R8 ← −5R7 + R5 ) 0 0 1/21 1/21 12/21 ⎦ (R9 ← −R7 + R3 ) 1 0 −4/21 9/21 0 1 4/21 −1/21 (R7 ) ⎡

1 ⎣4 0

⎤ ⎡ ⎤ −2 3 −1 1/21 5/21 −3/21 1 0 ⎦ = ⎣ −4/21 1/21 12/21 ⎦ 1 1 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 two lines have different slopes, they intersect at a single point; therefore, the given system has a unique solution. Write the system as Ax = b, where





 2 −6 x 13 A= , x= and b = 3 5 y 7 The determinant of A is detA = 10 − (−18) = 28, and thus (Theorem 8)

 5 6 1 A−1 = 28 −3 2 The solution is

 

 5 6 13 107 1 1 · = x = A−1 b = 28 −3 2 28 −25 7 i.e., x = 107/28 and y = −25/28. 19. The matrix of the system is



⎤ 1 3 −1 A = ⎣ −1 0 1⎦ 2 2 1

Since the determinant 1 3 −1 = 1 0 1 − 3 −1 1 + (−1) −1 0 = 1(−2) − 3(−3) − 1(−2) = 9 −1 0 1 2 2 2 1 2 1 2 2 1 is non-zero, the corresponding system has a unique solution; it is given by x = A−1 b where ⎡ ⎤ ⎡ ⎤ x 2 ⎣ ⎦ ⎣ x= y and b = 3 ⎦ z 1 Next, we find A−1 :

⎤ (R1 ) 1 3 −1 1 0 0 ⎦ ⎣ −1 0 (R2 ) 1 0 1 0 (R3 ) 2 2 1 0 0 1 ⎡ ⎤ 1 3 −1 1 0 0 (R1 ) ⎣0 ⎦ (R4 ← R1 + R2 ) 3 0 1 1 0 (R5 ← −2R1 + R3 ) 0 −4 3 −2 0 1 ⎡ ⎤ 1 3 −1 1 0 0 (R1 ) ⎣0 ⎦ 1 0 1/3 1/3 0 (R6 ← R4 /3) 0 −4 3 −2 (R5 ) 0 1 ⎡

L1-28

Linear Algebra [Solutions]



1 ⎣0 0 ⎡ 1 ⎣0 0 ⎡ 1 ⎣0 0 We are done. The inverse is

0 −1 0 −1 1 0 1/3 1/3 0 3 −2/3 4/3 0 −1 0 −1 1 0 1/3 1/3 0 1 −2/9 4/9 0 0 −2/9 −5/9 1/3 1 0 1/3 −2/9 4/9 0 1

A−1

⎤ 0 (R7 ← −3R6 + R1 ) ⎦ 0 (R6 ) 1 (R8 ← 4R6 + R5 ) ⎤ 0 (R7 ) 0 ⎦ (R6 ) 3/9 (R9 ← R8 /3) ⎤ 3/9 (R10 ← R9 + R7 ) 0 ⎦ (R6 ) 3/9 (R9 )

⎡ ⎤ ⎤ −2 −5 3 1 3 −1 −1 1 3 0⎦ = ⎣ −1 0 1⎦ = ⎣ 3 9 −2 4 3 2 2 1 ⎡

and

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ x −2 −5 3 2 −16 1 1 ⎣ y ⎦ = A−1 b = ⎣ 3 3 0 ⎦ ⎣ 3 ⎦ = ⎣ 15 ⎦ 9 9 z −2 4 3 1 11 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 claim that A−1 = (1/α)I. To prove:

1 1 −1 I = α I · I = 1I = I A · A = (αI) α α

1 1 −1 I (αI) = αI · I = 1I = I A ·A= α α We are done. 23. A 3 × 3 diagonal matrix is of the form



⎤ d1 0 0 A = ⎣ 0 d2 0⎦ 0 0 d3 where d1 , d2 , and d3 are real numbers. The determinant of A is d1 0 0 0 d2 0 = d1 d2 0 − 0 0 0 + 0 0 d2 = d1 d2 d3 0 d 0 0 0 d 3 3 0 0 d 3 The matrix A has an inverse if its determinant d1 d2 d3 = 0; thus, A is invertible if all of d1 , d2 , and d3 are non-zero. To find A−1 , we could try to guess, or apply the usual procedure: ⎤ ⎡ (R1 ) 0 0 1 0 0 d1 ⎣ 0 d2 0 0 1 0 ⎦ (R2 ) 0 0 d3 0 0 1 (R3 ) ⎡ ⎤ 1 0 0 1/d1 (R4 ← R1 /d1 ) 0 0 ⎣0 1 0 ⎦ 0 0 1/d (R 2 5 ← R2 /d2 ) 0 0 1 0 0 1/d3 (R6 ← R3 /d3 ) Thus, ⎡ ⎤−1 ⎡ ⎤ d1 1/d1 0 0 0 0 ⎣ 0 d2 0⎦ = ⎣ 0⎦ 0 1/d2 0

0

d3

0

0 1/d3

Section 9 [Solutions]

25. Let

L1-29

 a b A= c d Then (we use Theorem 8 to compute inverse matrices)

 t

 d −b d −c 1 1 (A−1 )t = = ad − bc −c ad − bc −b a a and  −1

 a c d −c 1 = (At )−1 = ad − bc −b b d a

Clearly, (A−1 )t = (At )−1 . 27. Suppose that

a A= c

b d



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

 1 0 3 4 0 1 0 0 implies that

 3 4 −1 A = 0 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 1 0 c d 0 1 

1 0 a b 0 −bc/a + d −c/a 1 

a b 1 0 0 ad − bc −c a 

a b 1 0 0 0 −c a shows that the matrix on the left is not the identity matrix; so we could not have arrived at

 1 0 3 4 0 1 0 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 are infinitely 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 infinitely many solutions: x = t, y = −t, where t ∈ R.

L1-30

Linear Algebra [Solutions]

35. Since A is an invertible, there is a matrix B such that AB = BA = I. We claim that B 2 is the inverse of A2 . To prove it we verify A2 B 2 = AABB = A(AB)B = AIB = AB = I B 2 A2 = 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 [Solutions]

L1-31

Section 10 Linear Transformations 1. We compute

  3 27 = A(9v1 ) = 9A(v1 ) = 9 −9 −81

  10 −10 A(−v2 ) = −A(v2 ) = − = 1 −1

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



  3 10 −24 −3 = −9 1 −21

3. Yes: A(−v) = −A(v) = − [ 3 −4 ] = [ −3 4 ] . 5. We pick two vectors

 1 v= 1

and

 0 w= 1

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

  1 5 A(v + w) = A = 2 5

   

 0 3 2 5 1 = + = A(v) + A(w) = A +A 1 6 3 9 1 shows that A(v + w) = A(v) + A(w) and so A is not linear. Alternatively,

  2 6 A(2v) = A = 2 8



  1 3 6 2A(v) = 2A =2 = 1 6 12 Since A(2v) = 2A(v), A is not linear. Here is a slightly different reasoning:

  0 0 A(0) = A = 0 4 A linear transformation must map zero vector onto a zero vector (since A(0) = A · 0 = 0). Hence A is not linear. 7. We compute



  10 26 = 2 40

   3 −2 −1 9 A(w) = = 4 0 −6 −4 A(v) =

3 −2 4 0



  26 9 88 A(2v + 4w) = A(2v) + A(4w) = 2A(v) + 4A(w) = 2 +4 = 40 −4 64 Alternatively,  



−1 16 10 = 2v + 4w = 2 +4 −6 −20 2 and thus

   3 −2 16 88 A(2v + 4w) = = 4 0 −20 64

L1-32

Linear Algebra [Solutions]

9. Let

A=

a c

b d



From

     1 a b 1 a 2 = = = 0 c d 0 c −7 it follows that a = 2 and c = −7. Likewise, from

     0 a b 0 b −2 A = = = 1 c d 1 d 3 we conclude that b = −2 and d = 3. Thus,

 2 −2 A= −7 3 A

11. Computing

    0 x 0 0 x A = = y y 0 1 y we see that A preserves the y coordinate of a vector and transforms its x-coordinate to 0. Thus, A is the orthogonal projection onto the y-axis. 13. Computing

    x −1 0 x −x A = = y 0 1 y 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

    x −2 0 x −2x A = = y 0 1 y y we see that A stretches the vector in the horizontal direction (i.e., stretches its x coordinate) by a factor of 2, and then reflects the stretched vector across the y-axis. 17. We replace θ by −θ, thus getting the matrix 

cos θ cos(−θ) − sin(−θ) = A= cos(−θ) − sin θ sin(−θ) (Recall that sin(−θ) = − sin θ and cos(−θ) = cos θ.) 19. We compute

sin θ cos θ



   

 x 1 1 x x+y 1 A = = = (x + y) y 1 1 y x+y 1 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, then it is in the third quadrant. If x + y = 0, then w is the zero vector.

Section 10 [Solutions]

L1-33

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

 −1/4 0 A= 0 −1/4 To check:

   −1/4 0 x −x/4 Av = = 0 −1/4 y −y/4

23. Let

 a b A= c d and compute the images of the unit vectors: 

 

 0 −1 1 0 = A = and A 1 0 0 −1 Based on these two facts, we now recover the entries in A.

     1 a b 1 a 0 A = = = 0 c d 0 c −1 it follows that a = 0 and c = −1. Likewise, from

     0 a b 0 b −1 A = = = 1 c d 1 d 0 we conclude that b = −1 and d = 0. Thus,

 0 −1 A= −1 0 If v = [x, y], then



 0 1 +y v=x 1 0 and we obtain the general formula 









  1 0 1 0 0 −1 −y A(v) = A x +y = xA + yA =x +y = 0 1 0 1 −1 0 −x

25. Using the formula of Example 10.3,

 √  cos(5π/6) − sin(5π/6) −1/2 − 3/2 √ A= = sin(5π/6) cos(5π/6) 1/2 − 3/2 27. Define B(v) = A−1 v (the matrix A−1 exists by assumption). From A(B(v)) = A(A−1 v) = AA−1 v = Iv = v B(A(v)) = B(Av) = A−1 Av = Iv = v we conclude that A(v) and B(v) are inverse transformations. So the matrix corresponding to the inverse 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:

L1-34

Linear Algebra [Solutions]

= 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) prove that the composition A ◦ B is a linear transformation. 31. We compute



⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 1 0 −4 −1 −5 −25 A(5v) = 5A(v) = 5 ⎣ 3 0 1 ⎦ ⎣ 1 ⎦ = 5 ⎣ −2 ⎦ = ⎣ −10 ⎦ 0 −2 1 1 −1 −5 ⎤ ⎡ ⎡ ⎤⎡ ⎤ ⎡ ⎤ −8 1 0 −4 0 −32 A(4w) = 4A(w) = 4 ⎣ 3 0 1⎦⎣0⎦ = 4⎣ 2⎦ = ⎣ 8⎦ 0 −2 1 2 2 8

and, using the linearity of A,

⎤ ⎡ ⎤ ⎡ ⎤ −25 −32 −57 A(5v + 4w) = A(5v) + A(4w) = ⎣ −10 ⎦ + ⎣ 8 ⎦ = ⎣ −2 ⎦ −5 8 3

33. We compute





1 ⎣ Av = 0 0 So A preserves the x and y coordinates of a is the reflection across the xy-plane.

⎤ ⎤⎡ ⎤ ⎡ x x 0 0 1 0⎦⎣y⎦ = ⎣ y⎦ 0 −1 z −z vector, and changes the sign of its z coordinate. Thus, A

35. The reflection with respect to the xz-plane preserves the x and z coordinates of the vector, and changes the sign of its y coordinate. Thus, the matrix of this transformation is ⎡ ⎤ 1 0 0 A = ⎣ 0 −1 0 ⎦ 0

0 1

Section 11 [Solutions]

L1-35

Section 11 Eigenvalues and Eigenvectors 1. Any matrix of the form

 3 b A= c 7 where one of (or both) b = 0 and c = 0 has eigenvalues 3 and 7. Check: the characteristic polynomial of A is 3 − λ b = (3 − λ)(7 − λ) − bc = (3 − λ)(7 − λ) det(A − λI) = c 7 − λ The solutions of (3 − λ)(7 − λ) = 0 are λ = 3 and λ = 7. 3. Assume that

a A= c

b 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. Comparing the two, we get a + d = −2 and ad − bc = −3. We have two equations with four variables (the entries of A), so there are infinitely many solutions. For instance, take a = −1; then d = −1; the remaining equation ad − bc = −3 gives 1 − bc = −3 and bc = 4. Take b = 4 and c = 1; thus, the matrix

 −1 4 A= 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 = −2 implies 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 

−2 − t s A= (−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) has eigenvalues 1 and −3. 5. Computing

   

 x 2 0 x 2x x A = = =2 y 0 2 y 2y y we see that A stretches a vector by a factor of 2, and thus preserves all directions (so any non-zero vector is an eigenvector of A). The corresponding eigenvalue is 2.

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

    x 1 0 x x A = = y 0 −1 y −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 the corresponding eigenvalue is 1. The vector [0, 1] is mapped to [0, −1]; this means that the direction of the y-axis is also preserved; so, the vector [0, −1] is an eigenvector of A and the corresponding eigenvalue is −1. To make it more obvious:

 

 0 0 0 A = = −1 1 −1 1

L1-36

Linear Algebra [Solutions]

9. Since

   0 2 4 8 = Av1 = 3 1 4 16 is not parallel to v1 , we conclude that v1 is not an eigenvector of A. From

  

 0 2 −2 4 −2 = = −2 = −2v2 Av2 = 3 1 2 −4 2 it follows that v2 is an eigenvector of A with eigenvalue −2. Since

  

 0 2 1 4 1 = = λ Av3 = 3 1 2 5 2 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 0 x x =7 0 −5 y y

for the coordinates x and y of the vector v. The corresponding linear system 7x = 7x −5y = 7y simplifies to 0x = 0 y=0 Thus, x can be any (non-zero) real number and y = 0. Let x = t; any vector of the form



 t 1 v1 = =t 0 0 (with t = 0) is an eigenvector corresponding to λ1 = 7. To find an eigenvector corresponding to λ2 = −5, we proceed as above:

Av = −5v  

 7 0 x x = −5 0 −5 y y 7x = −5x −5y = −5y x=0 0y = 0

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



 0 0 =t v2 = t 1 (where t = 0) is an eigenvector corresponding to λ2 = −5. 13. Call the given matrix A. The characteristic equation of A is 1 − λ 2 det(A − λI) = =0 2 4 − λ i.e., (1 − λ)(4 − λ) − 4 = 0 λ(λ − 5) = 0

Section 11 [Solutions]

L1-37

Thus, the eigenvalues are λ1 = 0 and λ2 = 5. (Recall that, instead of recalculating the characteristic equation every time we need it, we can use the ready-made formula λ2 − (trA)λ + detA = 0.) To find an eigenvector corresponding to λ1 = 0, we need to solve the equation

1 2

Av = 0v  

 2 x x =0 4 y y

for the coordinates x and y of the vector v. The corresponding linear system x + 2y = 0 2x + 4y = 0 simplifies to x + 2y = 0 Let y = t; then x = −2t and so any vector of the form



 −2t −2 v1 = =t t 1 (with t = 0) is an eigenvector corresponding to λ1 = 0. To find an eigenvector corresponding to λ2 = 5, we solve

1 2

Av = 5v  

 2 x x =5 4 y y x + 2y = 5x 2x + 4y = 5y −4x + 2y = 0 2x − y = 0

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



 t 1 v2 = =t 2t 2 (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 equation λ2 − 12λ + 35 = 0. It follows that (λ − 5)(λ − 7) = 0, and the eigenvalues are λ1 = 5 and λ2 = 7. To find an eigenvector corresponding to λ1 = 5, we need to solve the equation

Av = 5v  

 11 −2 x x =5 12 1 y y

for the coordinates x and y of the vector v. The corresponding linear system 11x − 2y = 5x 12x + y = 5y simplifies to a single equation 6x − 2y = 0 12x − 4y = 0 Let x = t; then y = 3t, and any vector of the form



 t 1 =t v1 = 3t 3

L1-38

Linear Algebra [Solutions]

(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 −2 x x =7 12 1 y y 11x − 2y = 7x 12x + y = 7y 4x − 2y = 0 12x − 6y = 0

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



 t 1 v2 = =t 2t 2 (where t = 0) is an eigenvector corresponding to λ2 = 7. 17. Let

1.5 3.5 A= 3.5 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 3.5 x x = −2 3.5 1.5 y y

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



 t 1 =t v1 = −t −1 (with t = 0) is an eigenvector corresponding to λ1 = −2. To find an eigenvector corresponding to λ2 = 5, we proceed in the same way as above:

1.5 3.5

Av = 5v  

 3.5 x x =5 1.5 y y 1.5x + 3.5y = 5x 3.5x + 1.5y = 5y

−3.5x + 3.5y = 0 3.5x − 3.5y = 0 Let x = t; then y = t, and any vector of the form



 t 1 v2 = =t t 1 (where t = 0) is an eigenvector corresponding to λ2 = 5.

Section 11 [Solutions]

19. From

L1-39

  

 2 0 1 2 1 = =2 0 −3 0 0 0 we conclude that [1, 0] is an eigenvector with eigenvalue 2. Likewise,

  

 2 0 0 0 0 = = −3 0 −3 1 −3 1 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 x x =3 −12 9 y y for the coordinates x and y of the vector v. The corresponding linear system −9x + 6y = 3x −12x + 9y = 3y simplifies to −2x + y = 0 Let x = t; then y = 2t and any vector of the form



 t 1 =t v1 = 2t 2 (with t = 0) is an eigenvector corresponding to λ1 = 3. To find an eigenvector corresponding to λ2 = −3, we proceed as above:

 

 −9 6 x x = −3 −12 9 y 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



 t 1 v2 = =t t 1 (where t = 0) is an eigenvector corresponding to λ2 = −3. Thus, the invariant directions are represented by the line of slope 2 (eigenvalue 3) and the line of slope 1 (eigenvalue −3). See below. y λ 1 =3

λ 2= - 3

v 1 =[1,2] v 2 =[1,1] x

L1-40

Linear Algebra [Solutions]

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 A2 v = AAv = A(Av) = A(λv) = λA(v) = λ(λv) = λ2 v 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

 1 2 A= 2 4 is the matrix of the system. In Exercise 13 we showed that the eigenvalues of A are λ1 = 0 and λ2 = 5. The corresponding eigenvectors are: v1 = [ −2 1 ] (for λ1 = 0) and v1 = [ 1 2 ] (for λ2 = 5). The solution of the given system is



 −2 1 + C2 e5t v(t) = C1 e0t 1 2 or, writing out the coordinates, x(t) = −2C1 + C2 e5t y(t) = C1 + 2C2 e5t

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

 3 8 A= 0 −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 8 x x =3 0 −1 y y for the coordinates x and y of the vector v. The corresponding linear system 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 eigenvector corresponding to λ2 = −1, we proceed as above:

 

 3 8 x x = −1 0 −1 y y 3x + 8y = −x −y = −y 4x + 8y = 0

Section 11 [Solutions]

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



 3t 1 −t −2 v(t) = C1 e + C2 e 0 1 or, writing out the coordinates, x(t) = C1 e3t − 2C2 e−t y(t) = C2 e−t

L1-41

L1-42

Linear Algebra [Solutions]

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 the survival probabilities: a child survives to become adult with a 50% chance. An adult survives to become a senior with the probability of 0.3, i.e., with a 30% chance. The entry 0 in the first row and the first column of L means that children do not produce offspring. (b) We compute: ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 2 3 120 490 P (1) = LP (0) = ⎣ 0.5 0 0 ⎦ ⎣ 200 ⎦ = ⎣ 60 ⎦ 30 60 0 0.3 0 ⎤ ⎡ ⎤ ⎤⎡ 490 300 0 2 3 P (2) = LP (1) = ⎣ 0.5 0 0 ⎦ ⎣ 60 ⎦ = ⎣ 245 ⎦ 0 0.3 0 60 18 ⎡

3. All birth parameters are zero, which means that go extinct after the number of time intervals reaches population P (0) = [ a b c ] , we find ⎡ 0 0 P (1) = LP (0) = ⎣ 0.5 0 0 0.5 ⎡ 0 0 ⎣ P (2) = LP (1) = 0.5 0 0 0.5 ⎡ 0 0 P (3) = LP (2) = ⎣ 0.5 0 0 0.5

no offspring are produced. The population will the number of age groups. Indeed, for an initial ⎤ ⎤⎡ ⎤ ⎡ 0 0 a 0 ⎦ ⎣ b ⎦ = ⎣ 0.5a ⎦ 0.5b c 0 ⎤⎡ ⎤ ⎡ ⎤ 0 0 0 0 ⎦ ⎣ 0.5a ⎦ = ⎣ 0⎦ 0 0.5b 0.25a ⎤ ⎡ ⎤ ⎤⎡ 0 0 0 0⎦⎣ 0⎦ = ⎣0⎦ 0

0.25a

0

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, adult and senior. The entries in the first row are the birth parameters: an adult produces on average 3 offspring, and no other age group produces offspring. The remaining non-zero entries are the survival probabilities: a child survives to become adolescent with a 20% chance. An adolescent survives to become 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 ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 0 3 0 100 450 ⎢ 0.2 ⎢ ⎥ ⎢ ⎥ 0 0 0⎥ ⎢ ⎥ ⎢ 200 ⎥ ⎢ 20 ⎥ P (1) = LP (0) = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 0 0.6 0 0 ⎦ ⎣ 150 ⎦ ⎣ 120 ⎦ 0 0 0.4 0 40 60 ⎡ ⎤⎡ ⎤ ⎡ ⎤ 0 0 3 0 450 360 ⎢ 0.2 ⎢ ⎥ ⎢ ⎥ 0 0 0⎥ ⎢ ⎥ ⎢ 20 ⎥ ⎢ 90 ⎥ P (2) = LP (1) = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 0 0.6 0 0 ⎦ ⎣ 120 ⎦ ⎣ 12 ⎦ 0 0 0.4 0 60 48

Section 12 [Solutions]

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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 this case, 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 eigenvalue. To find an eigenvector corresponding to λ1 = 0.8, we need to solve the equation

 

 0 2 x x = 0.8 0.32 0 y 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 the vector [ 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 the eigenvalues are λ1 = 1.3 and λ2 = −0.1. (b) The positive eigenvalue determines the behaviour of each age group in the long term. In this case, 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 eigenvalue. To find an eigenvector corresponding to λ1 = 1.3, we solve the equation

 

 1.2 0.5 x x = 1.3 0.26 0 y 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 the vector [ 5 1 ] . Over time, the ratio of the first age group in the total population approaches 5/6, and the 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 the rate is on average 1 offspring per adult. A child has a 50% chance of survival. Because only half of the children survive, there will be fewer adults to produce offspring in the next generation, so the population will decrease. Indeed, starting with P (0) = [ 100 100 ] , we compute    

100 0 1 100 100 = P (1) = LP (0) = = 0.5 · 100 0.5 0 100 50

    0 1 100 50 0.5 · 100 P (2) = LP (1) = = = 0.5 0 50 50 0.5 · 100

    0 1 50 50 0.5 · 100 P (3) = LP (2) = = = 0.5 0 50 25 0.52 · 100 

   2 0 1 50 25 0.5 · 100 P (4) = LP (3) = = = 0.52 · 100 0.5 0 25 25

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Linear Algebra [Solutions]

   2  0 1 25 25 0.5 · 100 = = 0.5 0 25 12.5 0.53 · 100

    3 0 1 25 12.5 0.5 · 100 P (6) = LP (5) = = = 0.5 0 0.53 · 100 12.5 12.5 ···  0.54 · 100 P (9) = 0.55 · 100

5   0.5 · 100 3.1 P (10) = ≈ 3.1 0.55 · 100 P (5) = LP (4) =

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 by guessing) we find the corresponding eigenvectors v1 = [1, 0] and v2 = [0, 1]. (b) No calculations are needed:





 5 1 0 u= =5 +6 = 5v1 + 6v2 6 0 1 (c) Using formula (12.11), we get

5(2)8 L u = 5(2) v1 + 6(3) v2 = 6(3)8 8

8

8



1280 = 39366



17. (a)The characteristic equation is λ2 − 3λ − 4 = (λ − 4)(λ + 1) = 0; the eigenvalues are λ1 = 4 and λ2 = −1. To find an eigenvector corresponding to λ1 = 4, we need to solve the equation

 

 1 2 x x =4 3 2 y y for the coordinates x and y of the vector v. The corresponding linear system 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 an eigenvector corresponding to λ2 = −1, we proceed as above:

 

 1 2 x x = −1 3 2 y 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





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



 2 1 u=1 +2 = 1v1 + 2v2 . 3 −1

Section 12 [Solutions]

(c) Using formula (12.11), we get 20

20

20

20

L u = 1(4) v1 + 2(−1) v2 = 4

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 2 1 2 · 420 + 2 +2 = 3 · 420 − 2 3 −1

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





 2 1 0 u= =2 +4 = 2v1 + 4v2 4 0 1 Using formula (12.11), we get

 5(7)13 L13 u = 5(7)13 v1 + 4(−3)13 v2 = −4(3)13

21. From





  6 −3 75 + α2 = α1 1 4 35

we get the system 6α1 − 3α2 = 75 α1 + 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.