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CHAPTER I XI -2t
1.1 Answers to more theoretical questions are omitted. X2 t
1. (x. y) =(-1, 1) 3. No solutions^ 7.^ I^ X3^ =^0
5. (x, y) = (0,0) 7. No solutions 9. (x, y, z) = (t. ~ - 2t, t), where t is arbitrary 11. (x, y) = (4, 1) 13. No solutions 15. (x, y, z) = (0,0,0) 17. (x, y) = (-Sa + 2b, 3a - b) 9. 19.a.Ifk= b. If k = 7, there are infinitely many solutions. c. If k = 7, the solutions are (x, y, z) = (1 - t, 2t - 3, t).
- 1l,13,and
X4 0
X5 0
Xl I I t - s 2r x2 I I r
x3 -t^ +s^ +^ I
X4 t-2s+ X5 s X
X
;:^1 ~^ r31~:^ -2t^ 1
- a. Products are competing. [ b. PI = 26, P2 = 46 25. a = 400, b = 300
- No solutions
- a. (x, y) = (t, 2t); (^) 15 m~ m b. (x, y) = (t, -3t); c. (x,y)=(O,O)^ _x,
- f(t)=1-5t+3t 2_ 33. f(t)=2t 2 -3t+4^ _X
- f(t)_ = at^2 + (l - 4a)t + 3a. for arbitrary a^ 17.^ I X _X
- f(t)_ = 2e3t^ - e 2t X
- -20 - 2x - 4v + x 2 + i = 0, the circle centered at (I, 2) with radius 5
- If a - 2b + c = 0
- a. The intercepts of the line x + y = 1 are (1.0)^ 19.^ [1^ aod^ [ and (0. I). The intercepts of the line x + & y = tare (t, 0) and (0,2). The lines intersect if (^) 21. b=O,c= l,d=O,withabeingarbitrary t oj- 2. t 2t - 2 23.^ 4 types b. x = ----, )I = - t-2' t-2 27. Yes; perform the operations backwards.
- There are many correct answers. Example: (^) 29. No; you cannot make the last column zero by ele
x -5z = -41 mentary row operations. I^ y^ -3z^ =-2 31. a = 2, b = c = d = I
- Twenty $1 bills, eight $5 bills, and four $10 bills. (^) 33. f(t) = I - 5t + 4t (^2) + 3t 3 - 2t 4
1.2 Answers to more theoretical questions are omitted. (^) 35. f(t) = -5 + 13t - IOt (^2) + 3t 3
IOt+13] -8t - 8 I. [~] [ (^) t 37. _:~. -t]6t^ where t ..IS arbitrary.
X] [4 - 2s - 3t]
r
3. y = s [Z t XI] [500] 39. X2 = 300 [X3 400
- a. Neither the manufacturing nor the energy sector makes demands on agriculture.
- [} r~~ b. XI :0:0 18.67, X2 :0:0 22.60, X3 :0:0 3. 471
43. In\ = ~m2^ II.^ Undefined
- a"'" 12.17, b "'" -1.15. c "'" 0.18. The longest day (^) 13. 15. 70 is about 13.3 hours. [~~]
- a. If k is neither I nor 2 b. If k = I c. If k = 2 17.^ Undefined^ 19.
- a. Xl = 3X3 - 2X4, x2 = 2X3 - X4, for arbitrary X3^ m and x^4 b. Yes, x) = I,X2 = 5,.q = 9,X4 = 13. I
- C = 25 P~1 (^123) r~ 0
r
53. xv = 0, the union of the two coordinate axes 25. The system A.r = c has infinitely many solutions or 55. a(xy - y) + b(y2 - y) = 0, where a#-O or b #- (^0) none. 57. a(x 2 - x) + b(i - y) = 0, where a#-O or b #- 0 0 0
- 25 - lOx - lOy + x 2 + i = 0, the circle of radius (^) 27. rIOo^ .1^0 o (^0) I 29.^ [~^0 5 centered at (5,5) (^0) o 0 0 ~]^ J
- No solutions
- Statistics: $86; Set Theory: $92; Psychology: $55 2
- Beginning: 120 liberal. ]40 conservative. 31.^3 End: 140 liberal, 120 conservative. [~^0 -il^ fO'mmpk
- Cow: 34121 liang; sheep: 20/21 liang 33.^ Ax^ =^ x
- Swallow: 24/19 liang; sparrow: 32/19 liang^ 35.^ Aej^ is the^ ith^ column^ of^ A.
- A: 265; B: 191; C: ]48; D: 129; E: 76 37. x ~ [2: 2t lwh~e t i' "'bitr",
- Gaussian elimination shows that Pigeons = -250 + ~(Swans) + 20(Peacocks) and I 0 0 Sarasas = 350 - .!J(Swans) - 21 (Peacocks). (^) 39. 0 I 0 [ (^) :] One solution (the one given by Mahavira) is 15 pi 001 geons, 28 sarasabirds, 45 swans, and 12 peacocks (^) 41. One solution 43. No solutions (spending 9,20,35, and 36 panas, respectively). 47. a. x = 0 is a solution.
- 53 sheep, 42 goats, and 5 hogs (^) b. By part (a) and Theorem 1.3.
- (^) Full Half Empty c. A(xj +X2) = Ax\ +AX2 = 0+0 = 0 d. A (k."i) = k(Ax) = kG = 0 1st Son p 10 - 2p p 2nd Son q 10 - 2q q 49.^ a.^ Infinitely many solutions^ or^ none 3rd Son lO-p-q 2p+2q -10 lO-p-q b.^ One^ solution^ or^ none c. No solutions Here, p and q are integers between 0 and 5 such that d. Infinitely many solutions
p + q 2: 5.
1.3 Answers to more theoretical questions are omitted.^ 51.^ If^ In^ =^ rand^ s^ =^ p I. a. No solutions b. One solution 53.^ Yes c. Infinitely many solutions
- Rank is 1. (^) 55 Yo<; m~ -1 [~] +2 [~]
- a. x l~] + y l~] = lIn b. x = 3, y = 2 57.^ [I~]^ [~]+[~]
- One solution (^) 59. c = 9, d = 11
- For c = 2 and for c = 3
- Line through the endpoint of v in the direction of in
474 ANSWERS TO ODD·NUMBERED EXERCISES
23. [~ ~ I]
- [~ -~] ; you shear back. 27. A is a reflection; B is a scaling; C is a projection; D is a rotation; E is a shear.
31 k n~ ~]
- Use a parallelogram. 35. Yes
37. a. b. °
c. -2:s tr A :S 2 d. 2
- a. Projection with scaling b. Shear with scaling c. Reflection with scaling
- refQ.r = -refpx
- cose. e^ Sine] e^ ,a clockwise rotatiOn through the.^. [- sm cos angle e
- A-I = A. The inverse of a reflection is the reflec tion itself.
- Write T(.r) = [~ :] x. Express f(t) in terms of a, b, c,d.
49. c = °(or c = JT: "2)'^ VI^4 = [I]°^ ,and V2^4 = [0]I
51. c =^ JT: 4'^ 4 VI^ = v0.I^ [I] 1 ,and- V2^ = v0.I [-I]I
- The image is an ellipse with semimajor axes ±5el and semiminor axes ±2e2.
- The curve C is the image of the unit circle under the
transformation with matrix [ W I W2].
2.3 Answers to more theoretical questions are omitted.
- Undefined [~ ~]
1
- 3
[~ ~] P-6 -2 J
- [10] 13. [h] [~ ~]
- I~ I ~ 1
~
- All matrices of the form [~ ~]
- All matrices of the form [_~ ~]
- All matrices of the form [~ (^) a ~ b]
- All matrices of the form (^) [~ab
~b ~a]
- All matrices of the form c
[~
~]
29. a. The matrices DaD fJ and D ° fJ D a
both represent the counterclockwise rotation through the angle ex + 13. b. DaDfJ = DfJDa COS(ex + 13) - sin(ex + /3)]
- [^ sin(ex + 13) cos(ex + 13)
- (ith row of AB) = {ith row of A)B
33. An = h for even n and An = -12 for odd n. A
represents the rotation through rr.
35. An = h for even n and A" = A for odd n. A repre
sents the reflection about the line spanned by [~].
37. A" = [ 1 0]. A represents a vertical shear. -n 1
39. A" represents the rotation through nrr/4 in the
clockwise direction, so that A^8 = h. Now A 1001 = (A 8 ) 125 A = A.
41. An = h for even n and A" = A for odd n. A repre sents the reflection about a line.
- A = - h, for example, or any matrix representing a reflection about a line
- The matrix representing the rotation through 2rr/3, cOS(2rr /3) - sin(2rr /3)] for example, A = [ sin(2rr /3) cos(2rr /3) =
-)3]
-I
- Any projection or reflection matrix with nonzero
entries will do, for example, A = [~:~ _~:~].
49. AF = [~ _~], FA = [-~ ~l We compose a
rotation with a reflection to obtain a reflection.
51. F.I =.IF = [-: =~].NotethatFrepresents
the rotation through rr /2 while .I represents the rotation through rr / 4 combined with a scaling by ../2. The products F.I and .I F both represent the rotation through 3rr /4 combined with a scaling by../2.
[
53. CD = 0 I -1] 0 and DC = [ 0-1 O'I] We com
pose two reflections to obtain a rotation.
- X = [^ -s2s^ -2t]t ,where sand t are arbitrary
constants
- X = [-~ (^) - ~]
- No such matrix X exists.
- X = ['^ -:s +,^ -2+1]I^ ~^ 2t^ ,^ where^ sand^ t^ are arbitrary
constants
- No such matrix X exists.
- X = [~ ~ ] , where t is an arbitrary constant
67 a. [ I 1 I] A = [I I 1]
69a. Of the surfers who are on Page 1 initially, 25% will be on Page 3 after following two consecutive links.
69b. The ijth entry of A 2 is °if there is no path of length 2 from vertex j to vertex i in the graph of the mini-Web, meaning that a surfer cannot get from Page j to Page i by following two consecutive links.
- A straightforward computation shows that the only nonzero entry of A^4 is the first component of the third column. Thus there is no path of length 4 from Page 3 to Page I in the graph of the mini-Web, while there is a path of length 4 linking any other two pages. We cannot get from Page 3 to Page 1 by fol lowing four consecutive links.
73. lim (Amx) _ ( lim Am) X
m~oo m~oo
[ i,~" X~"" ] X -^ (XI^ + ... +^ x^ ll^ )^ Xequ '-v-"
components of X
X (^) equ
75. The ijth entry of Am+ 1 = Am A -is the dot product
of the ith row tV or Am with the jth column i! of A.
Since all components of tV and some components of
i! are positive (with all components of i! being non negative), this dot product is positive, as claimed. 0.5003 OA985 0.5000]
- We find A 10 ~ 0.0996 0.1026 0.0999. This [ (^) OA002 0.3989 OAOO shows that A is regular and it suggests then xequ =
0.5]0.1. To be sure, we verify that A [0'5]0.1 = [ (^) 0.4 0.
0.5] 0..
[ OA
- An extreme example is the identity matrix In. In this
case, Inx = X for all distribution vectors.
81. Ami! = 5'lltj 83. By Definition 2.3.10, there exists a positive integer m such that Am is a positive transition matrix. Note that Am X = X. Thus, the ith component of x is the
dot product of the ith row tV of Am with.r. Since
all components of tV and some components of x are positive (with all components of x being nonnega tive), this dot product is positive, as claimed.
- There is one and only one such X.
2.4 Answers to more theoretical questions are omitted.
[
8 -3]
- -5 (^2 3) [-: ~]
- Not invertible 7. Not invertible
[
- Not invertible I^ °^ -1]
- ° I ° ° ° 1
-2 ° 1 13 r-f
o
I
°
~
15. 9 -1^ -
[ (^1 2) -3 -~
- Not invertible
- Xl = 3YI - 2.5Y2 + 0.5Y X2 = -3YI +4Y2 - Y X3 = YI - 1.5Y2 + 0.5Y
- Not invertible 23. Invertible
- Invertible 27. Not invertible
- For all k except k = land k = 2
- It's never invertible.
33. If a^2 + b^2 = I
35. a. Invertible if a, d, f are all nonzero
b. Invertible if all diagonal entries are nonzero c. Yes; use Theorem 2A.5. d. Invertible if all diagonal entries are nonzero I
- (cA)-1 = -A-I C
39. M is invertible; if mij = k (where i =P j), then the
ijth entry of M- 1 is -k; all other entries are the same.
- The transformations in parts a, c, and d are invert ible, while the projection in part b is not.
- Yes; x = B- 1 (A -I y)
- ker A = (O)
- n~
I 0 0
o '
0 I 0
- (^) [:],[~] 17. AlloflR?
- The line spanned by [_~]
- All oflR?
- kernel is (O), image is all of lR?^2.
- Same as Exercise 23 27. fix) = x 3 - x
¢ [Sin ¢ cos fI]
29. f [fI] = sin¢sinfl cos¢ (compare with spherical coordinates)
[
-,., -3]
31.A= ~ ~
33 T m~ x + 2y +3<
- ker T is the plane with normal vector v; imT =lR?.
- im A = Span(el, e2); ker A = span(el);
im(A 2 ) = Span(el); ker(A 2 ) = Span(el, e);
A 3 = 0 so ker(A 3 ) = lR?3 and im(A 3 ) = {O}
- a. ker B is contained in ker(AB), but they need not be equal. b. im(AB) is contained in im A, but they need not be equal.
- a. im A is the line spanned by [~], and ker A is the
perpendicular line, spanned by [ -~].
b. A 2 = A; if v is in im A, then A v = v. c. Orthogonal projection onto the line spanned by
[~]
- Suppose A is an n x m matrix of rank r. Let B be the matrix you get when you omit the first r rows and the first m columns of ITef [A! Ill]' (What can you do when r = II?)
- There are m - r nonleading variables, which can be chosen freely. The general vector in the kernel can be written as a linear combination of m - r vectors, with the nonleading variables as coefficients.
47. im T = L2 and kerT = LI
51. ker(AB) = (O) I 0 I 1 1 I 0 I
- a.
I
II
I
I
o I 0 0 001 0 000 1
b. ker H span(vl, V2, V3, V4), by part a,
and im M = span(v, V2, V3, V4), by Theo
rem 3.1.3. Thus ker H = im(A1). H(Mx) = O. since Mx is in im M = ker H.
3.2 Answers to more theoretical questions are omitted.
- Not a subspace 3. W is a subspace.
- Yes 9. Dependent II. Independent 13. Dependent
- Dependent 17. Independent
- r~1 ,," r~1 M'~""""L 21. V2 = v I, or, v I - V2 = 0, so that [_ ~] is in the kernel. 23. VI^ ~^ = O~^ , so that [I] 0. .IS In^ the kerne.I
- (^) '" "J, 0', '" '- "' ~ ii, '0 th" [_~] i' io the
kernel.
- (^) [ij,[i] 29. [~], m
r~H!1 33 r~H~H!~
35. Suppose^ there is a nontrivial relation^ CI^ V I^ +^ ...^ +
ci Vi + ... + c (^) m Vm = 0, with Ci i=- O. We can solve
this equation for Vi and thus express Vi as a linear
combination of the other vectors in the list.
Conversely, if Vi is a linear combination of the
other vectors, Vi = '" , then we can subtract Vi
from both sides of the equation to obtain a nontri
vial relation (the coefficient of Vi will be -1).
- The vectors T(v]), .... T(vm ) are not necessarily independent.
39. The vectors V], ... , Vm , V are linearly independent.
- The columns of B are linearly independent, while the columns of A are dependent.
- The vectors are linearly independent.
- Yes 47. f~ !~ 000
o I -
-I]
51. a. Consider a relation C]VI +",+cpVp+d]llil +
... + dqwq = 6. Then CIV] + ... + cpvp = -dl W l - ... - d (^) q wq is 6, because this vector is both in V and in W. The claim follows. b. From part a we know that the vectors
VI, .... Vp,WI, ... ,Wq are linearly indepen
dent. Consider a vector x in V + W. By the definition of V + W, we can write x = v + W for a v in V and a W in W. The v is a linear
combination of the Vi, and W is a linear com
bination of the ilij. This shows that the vectors v] , ... , v p, WI, ... , Wq span V + W.
I o o o
- 0 I o o o o I o o o o I
- For j = 1,3,6, and 7, corresponding to the columns that do not contain leading I's
3.3 (^) Answers to more theoretical questions are omitted.
1. V2 = 3v]; basis of image: [~];
basis of kernel: [-n
- No redundant vectors; basis of image: [~], [~]; basis of kernel: 0
- V3=3vl;basisofimage: [~], [-~];
b"" of k'md nJ
- v2=2vj;basisofimage: [~], [~];
b"" ofk'md nJ
9. V2 = 2v I; basis of image:
[:],m
b"" o'k'md nj
II. V3 = VI; basis of image:
[~HIl
b"i'ork,,'" [-~J
13. V2 = 2Vh V3 = 3v I; basis of image: [ I] ;
b"i,ofk'm,j nl nJ
- (^) V3 = 2v I + 2V2, V4 = 0;
basis of image:
ft1' fJ
b"i, of mod f=fl f~
- (^) VI = 6, V3 = 2V2, V5 = 3V2 + 4V4;
basis of image:
o
basis of kernel: o
I
19. V3 = 5VI +4V2, V4 = 3Vl + 2V 2;
basis of image:
f~l fn f! -5 -
basis of kernel: I 0 0 I 0 0
21 rr,r A ~ [~
I 4 3]^ ; 0 0
480 ANSWERS TO ODD·NUMBERED EXERCISES
Each such cubic is the union of the y-axis, the 3.4 Answers to more theoretical questions are omitted. x-axis, and any line through the point (1, 1).
- Plugging the points (1, 1), (2, I), (I. 2), and (3, 2) into the solutions of Problem 44, we find that Cs = C7 = Cs = C9 = 0, so that the solutions are of the form CIOY(Y - l)(y - 2) = 0, where CIO -=I- 0. Division by CIO produces the unique so lution y(y - 1)(y - 2) = 0, the union of the three horizontal lines y = 0, y = I, and y = 2.
- Plugging the points (1, 1), (2, I), (1, 2), (2, 2), and (3,3) into the solutions of Problem 44, we find that Cs = Cs = C9 = °and C7 + CIO = 0, so that the solutions are of the form C7X(X - l)(x - 2) C7Y(y - I)(y - 2) = 0, where C7 -=I- O. Division by C7 produces the unique solution x (x - I) (x - 2) = y(y - 1)(y - 2). the union of the diagonal y = x with an ellipse.
- See Exercises 41 and 56. Since the kernel of the 8 x IO matrix A is at least two-dimensional, and because everyone-dimensional subspace of ker A defines a unique cubic (compare with Exercise 40), there will be infinitely many such cubics.
- There may be no such cubic [as in Exercise 49], ex actly one [take the 9 points in Exercise 47 and add ( -1, -I)]. or infinitely many [as in Exercise 51].
- A basis of V is also a basis of W, by Theorem 3.3.4c. 65. dim(V+W) = dim V+dim W, by Exercise 3.2.51.
- The first p columns of ITef A contain leading l' s be
cause the Vi are linearly independent. Now apply
Theorem 3.3.5.
- [0 I 0 2 (I], [0 0 3 0] , [0 (^0 0 0) 1]
- a. A and E have the same row space, since ele mentary row operations leave the row space un changed. b. rank A = dim(rowspace(A»), by part a and Ex ercise 72.
- Suppose rank A = n. The submatrix of A consisting of the n pivot columns of A is invertible, since the pivot columns are linearly independent. Conversely, if A has an invertible II x n subma trix, then the columns of that submatrix span lR", so im A = lR" and rank A = n.
- Let m be the smallest number such that Am = O. By Exercise 78, there are m linearly independent vectors in lR"; therefore, m :s II, and AI1 = Am^ AI1^ - m = O.
- a.3,4,or5 b. 0,I,or
- a. rank(AB).:s rank A b. rank(AB):S rank B
I. (^) [x]9.1 = m
- (^) [x] 9.1 = [~]
- (^) [x]9. 1 = (^) [-~] 7. (^) [x] 9.1 = (^) [~]
- x isn't in V. 11. (^) [xh; = [~j~]
- (^) [i]" ~ [-1] 15. [xb = H]
- (^) [xl" ~ [ :] 19. B= (^) [~ . -I -~] 21. B = (^) [~ ~] 23. B= (^) [~ -~] 25. B = (^) [-~ (^) -~] 27. B = 0 [~
° ~] °
29. B = (^1) 31. B = 0 [~
~] (^) U
0 0 ~]
33. B = I 35. B = 1 [~
~]
(^0) H^ ~] °
45. If V is any vector in the plane that is not parallel
to x, then V, ~(x - 2v) is a basis with the desired
pmp«". Fo' "=ple. "~ mgi", th' b"",
m·~ [=;j
47. A = [~ ~] (^) 49. [=:] 53. x = [~~] 55. ~ [-: ~]
- Consider a basis with two vectors parallel to the (^) b.. [I plane, and one vector perpendicular. 27.^ A^ aSls^ IS^0 0]o '^ [0 0 ~^ ]. so that the dimension
- Yes is^ 2.
- [-~], [~],forexamPle 29.^ A basis is^ [-~^ ~]^ , so that the dimen ~J. [-~ sion is 2.
- Yes 67. B = [~ b~ ~ :d] (^) 31. A baSIS^ oo IS [1I - ~ ] , so that the dimension ~J. [~ is 2. [
- lfS= 2 I 2]I ,thenS -I AS= [3°^ -I'0]
- Only the zero matrix has this property, so that the basis is 0, and the dimension is 0.
- a. If Bx = S-1 ASx = 0, then A(S.~) = O. b. Since we have the p linearly independent vec (^1) ° 0] [0 ° 0] tors Sv 1, SV2, ... , Sv p in ker A, we know that 35.^ A basis^ is^ [° ° ° , °^1 ° , dim(ker B) = p :s dim(ker A), by Theo (^) ° ° ° ° ° ° rem 3.3.4a. Reversing the roles of A and B, we find that dim(ker A) :s dim(ker B). Thus, 0 °^ 0] [° ° ° .and the dimension is^ 3. nullity A = dim(ker A) = dim(ker B) = 001 nullity B. (^) n n(n + 1) [ 0.36 0.48 0.8] 37.^ 3,5,or9^ 39.^ Lk^ = 2
- 0.48 0.64 -0.6 k= -0.8 0.6 (^0) 41. 0,3,6,or9 (^) 43. 2 45. dim(V) = 3
- [~ -~] 47.^ Yes^ and yes^ 49.^ Yes 51. f(x) = ae^ 3x^ + _be4x
- bij_ = a (^) n+l-i,ll+l-j
- By Theorem 3.4.7, we seek a basis VI, V2 such that 4.2^ Answers to more theoretical questions are omitted. AVI = VI and AV2 = -V2. Solving the linea~ sys 1.^ Nonlinear tems Ax = x and Ax = -.~ [or (A - h)"x = °and (^) 3. Linear, not an isomorphism (A + h)x = OJ, we find VI = [n and V2 = [~], 5. Nonlinear 7. Isomorphism for example. 9.^ Isomorphism^ 11.^ Isomorphism
- Linear, not an isomorphism
CHAPTER 4
- Isomorphism 4.1 Answers to more theoretical questions are omitted.
- Linear, not an isomorphism I. Not a subspace
- Subspace with basis I - t, 2 - t 2 19.^ Isomorphism^ 21.^ Isomorphism
- Linear, not an isomorphism
- Subspace with basis t 7. Subspace
- Linear, not an isomorphism
- Not a subspace II. Not a subspace
- Isomorphism
- Not a subspace 15. Subspace
- Linear, not an isomorphism
- Matrices with one entry equal to I and all other en (^) 31. Linear, not an isomorphism tries equal to 0. The dimension is mn.
- Linear, not an isomorphism
- A basis is [~J. [~], [~], [~] ,so that the dimension 35.^ Linear, not an isomorphism is 4. 37.^ Linear, not an isomorphism
- Linear, not an isomorphism
- A baSIS.^ IS.^ [I°^ 0]0'^ [0°^ 0]I ' so that t eh^ d'ImenSlOn^. 41. Nonlinear 43. Isomorphism is 2. 45.^ Linear. not an isomorphism
- Linear, not an isomorphism
- A basis is [~ ~], [~ ~], [~ ~], so that the (^) 49. Linear, not an isomorphism
dimension is 3. (^) 51. ker T consists of all matrices of the form [~ ~ ] ,
- A basis is I - t, I - t 2 , so that the dimension is 2. (^) so that the nullity is 2.
p2 = [a
65. a. +^ bc^ b(a^ +^ (~)]^ and
e(a + d) be + d~ U [p2]^ =^ [be^ -^ ad]^
,,"I a +d
- By Pythagoras, Ilxll = J 49 + 9 + 4 + 1 + 1 = 8. b. B (^) = 0 I be-ad]a + d. T" IS^ an Isomorph' Ism I^ 'fB [ (^) 31. P ::; Ilxll 2. Equality holds if (and only if) x is a
is invertible. linear combination^ of^ the vectors^ Ui.
c. In this case, im T is spanned by P and ker T is (^) 33. The vector whose n components are all1/n
[ spanned by (a + d)h - p = d^ -b]. -c a (^) 35. __ 1 [~] Jl4 (^3) 0 0 o 0 -2^ o 2] 0 37.^ R(x)^ =^ 2(1/1^.^ X)lll^ +^ 2(U2^.^ X)U2^ -.r
- a. b. f(t) = !t sin(t) [ o^0 o^0 39.^ No;^ if^ u is a unit vector in^ L,^ then^ x.^ ProjLx^ = o 0 o (^0) x. (u. x)u = (u. x)2 2: O. 41. arccos(20/2l):::::; 0.31 radians [ 73. b.A= -1^1 -12] c.B= [0 1 -1] 0 43. 9l'2 5-^ 45. 25-4Tl'2^ - 4Tl'3^1
d. S = [~-n e. AS = SA 1'. No 5.2 Answers to more theoretical questions are omitted. CHAPTER 5
5.1 Answers to more theoretical questions are omitted. (^) 2/3] (^) 4/5] [ 3/5]
- 1/3 3.^ 0,^0 [ (^) -2/3 [3/5 -4/
- JI76 3. J
[
- arccos (~) :::::; 0.12 (radians) (^) 5. 2/3]2/3, /10 I [-1]-
1/3 v 18 4
- obtuse 9. acute
I I. arccos (.:n) ---+ ~ (as n ---+ CXJ) (^) 2/3] [-2/3] [ 1/3]
- 2/3, 1/3, -2/ [
- 2 arccos(0.8) :::::; 74° 1/3^ 2/3^ 2/
[
-2] [-3] [-4] 1/2]1/2 [-1/1017/
- ~. r' ~ 9.^ [ 1/2'^ -7/ 1/2 1/
4/5]o [-3/15]2/ 17 [-n [-~ [
- a. Orthogonal projection onto L ~ b. Reflection about L ~ c. Reflection about L 1/2]1/2^ [^ -1/2^ 1/2]^ [^ 1/2]1/ I 13.^ 1/2'^ -1/2^ '^ -1/ [
- For example: b = d = e = g = 0, a (^) 1/2 1/2 -1/ 2 J3 J
e= -2 ,f=--2 2/3]
15. 1/3 [3]
- a. IIkull = yI'-(k-'v-).-(-kv-') = Vk2(v. v) =^ [-2/ v1Zi-./v· v = Iklllvil 4/5 3/5] [5 5]
- 0 0 'i b. Byparta,IIII~llvll= 1I~lIllvll=1. [^ 3/5 -4/5 0 3.
A]
12]
-I~
10] 15
- [~j~], [ ~j~] 37. - 2 I r:I -:]I 2/3 -2/ I I
4. Q is diagonal with % = I if % > 0 and qii = -
if aii < O. You can get R from A by multiplying the ith row of A with -1 whenever aii is negative.
43. Write the QR factorization of A in partitioned form
as A = [AI A2] = [QJ Q2] [Rb ~~].
Then A 1 = Q J R J is the QR factorization of A,.
- Yes
5.3 Answers to more theoretical questions are omitted.
. Not orthogonal 3. Orthogonal
- Not orthogonal 7. Orthogonal
- Orthogonal 11. Orthogonal
- Symmetric
- Not necessarily symmetric
- Symmetric 19. Symmetric
2. Symmetric
- Not necessarily symmetric
- Symmetric
- (Av)· w = (Av)T W = v T^ AT u) = v· (AT w) _ _ L(v). L(w) 29. L(L(V),~(~») = arccos IIL(v)IIIIL(w)11 = v· w __ arccos _ _ = L (v, w) [The equation
Ilvllllwll
L(v). L(w) = v. w is shown in Exercise 28.]
- Yes, since AAT^ = In.
- The first column is a unit vector; we can write it
as VI = [c~s e] for some e. The second col sme umn is a .unit vector orthogonal to vJ; there are
two choices: [- sin~] and [ sin ~]. Solution: cos {7 - cos ( COS e - sin e] d [cos e sin e] t b' [ sin e cos e an sin e _ cos e ' or ar 1 trarye.
- For example, T(.r) = ~ [-~ I (^3 2 )
- No, by Theorem 5.3. 39. (ijth entry of A) = Ui U j 1 4. All entries of A are -. n
- A represents the reflection about the line spanned by u (compare with Example 2), and B represents the reflection about the plane with normal vector u.
- dim (ker A) = m - rank A (by Theorem 3.3.7) and dim(ker(A T^ ») = n - rank(A T^ ) = n - rank A (by Theorem 5.3.9c). Therefore, the dimensions of the two kernels are equal if (and only if) m = n, that is, if A is a square matrix. 47. ATA=(QR)TQR=RTQTQR=RTR
- By Exercise 5.2.45, we can write AT = QL, where
Q is orthogonal and L is lower triangular. Then
A = (QL/ = L T^ QT does the job. 5. a.lm=QfQJ=STQIQ2S=STS,sothatSis orthogonal. b. R2Ri1 is both orthogonal (by part a) and up per triangular, with positive diagonal entries. By Exercise 50a, we have R2Ri' = (^1) m , so that
R2 = RJ and QI = Q2, as claimed.
53 H~ ~l U~ ~H~ -~ ~l dimension 3 n(n + 1)
2
- Yes, and yes
486 ANSWERS TO ODD·NUMBERED EXERCISES
5.5 Answers to more theoretical questions are omitted. 45. They are the same.
- a. If S is invertible (^) 49. The kernel is the plane span(v, w) and the image b. If S is orthogonal is R.
- Yes 7. For positive k (^) 51. Let au be the first diagonal entry that does not be
- True 11. The angle is 8 long to the pattern. The pattern must contain an en try in the ith row to the right of au as well as an
- The two norms are equal, by Theorem 5.5.6. (^) entry in the ith column below au.
- If b = c and b^2 < d 17. Ifker T = [O} (^) 53. Only one pattern has a nonzero product, and that product is I. Since there are n^2 inversions in that
- The matrices A = l~ ~ J such that b = c, a > 0, (^) pattern, we have det A = (_l)n^2 = _(1)n.
and b^2 < ad. (^) 55. Yes, since the determinants of all principal subma
Yes,^ (v,^ It')^ =^ 2(v^.^ w) trices are nonzero. See Exercise 2.4.93.
1, 2t - 1 57.^ Only one pattern has a nonzero product, and that product is 1. Thus, del A = I or del A = -1.
(^) VI + ~ + ~ + ... = ~ 59. a. Yes b. No c. No
(^1) 0]
(^) ao = viz' Ck = 0 tor all k 61. (^) F,;I.<lo be ,"'emati ", ..,; "" F [~ () 1 = I but o 0 2 if k is odd bk = (^) { Okn F [~ ~ ~] = O. if k is even 1 () () 1 n^2
L --
k odd k^2 - 8 6.2^ Answers to more theoretical questions are omitted.
33. b. IIfl1 2 = (f, f) = i~ w(t)dt = 1, so that
IIIII = 1 7.^1 9.^24 11.^ -
If
- a. IItl132 = ~ ill r^2 dt = and IItl134 = (^) 19. -1 21. 1 _23. (1)n(I1-I)/2. This is 1 if either nor (n - I) is di )~jl ~t2dt=V~'~=~ visible by 4, and -1 otherwise. V n -J Jr (^8 2) 25. 16 27. a 2 + b 2
b. For f(t) = ~we have Ilfll32 = VI and 29.^ det(P))^ =^ 1 and^ det(Pn)^ =^ det(Pn_I),^ by expan sion down the first column, so det( Pn ) = 1 for all n.
IIfl134 = If (^) 31. a. det [ 1 1 J = a I - ao
ao al
CHAPTER (^6) b. Use Laplace expansion down the last column to
6.1 Answers to more theoretical questions are omitted. see that f (t) is a polynomial of degree "S: n. The
- 0 3. -2 5. 110 coefficient^ k^ of^ t^ n^ is^ IT^ (ai^ -^ a)).^ Now n-I~i»
7. 0 9. -36 11. k =F 3/2 det A = f(a n ) = k(a n - ao)(an - al) ... (an
13. Lie () 15. (^) k =F 1/2^ an-I)^ =^ IT^ (ai^ -^ a)^ ).^ as claimed. n,?:-i>j
- [f k is neither I nor -1^ n
- IT ai' IT (ai -a)) (use linearity in the columns and
- [f k is neither 0 nor I (^) i=1 i»
- If k is neither 1 nor -2 Exercise 31)
- If A is 1 or 4 25. If A is 2 or 8 (^) 35. A A^1 [ ~21]^ [aa21]^ and^ [~21]^ [
hb2] are solutions. The
- If A is 2, 3, or 4 29. If A is 3 or 8 equation is of the form pXl + qX2 + b = 0; that is,
- 24 33. 99 35. (^18) it defines a line.
- 55 39. 120 41. 24 37.^ ±l
- det(-A) = (-l)ndetA 2 39. det(A T^ A) = (detA) > 0
- detA = det(A T^ ) = det(-A) = (-l)ndetA b.^ V(VI,^ V2,^ V3,^ VI^ x^ V2^ x^ V3) -det A, so det A = 0 =ldet[vl xv^2 XV^3 VI V2 V3]! (^2) = Ilvl x V2 x v311 2 , by definition of the cross
- AT^ A^ =^ [IIVI1^ v·^ w]^ product
v·iL, Ilw11 2 '
c. V(VI, V2. V3) = Ilvl x V2 x v311, by parts (a) so det(A T^ A) = IIvI1 2 11wll 2 - (v·w)2 ~ 0, by the (^) and (b) Cauchy-Schwarz inequality. Equality holds only if v and ware parallel. 19.^ det^ [VI^ V2^ V3]^ =^ VI^.^ (V2^ x^ V3)^ is positive^ if (and only if) VI and V2 x V3 enclose an acute angle.
- Expand down the first column: j(x) = (^) 21. a. Reverses b. Preserves c. Reverses -xdet(A41) + constant, so j'(x) = -det(A41) = -24. det [~ -~] (^7) 47. T is linear in the rows and columns. (^) 23. XI = 17' 1 det [^ -6^5 -~] 49. A = 2 ;^ ],^ [0<^ ",mp[e.^ Start^ with^ a (^1 14) det [_~ triangular
[: matrix with determinant 13, such as ~]^6 X2 =
[~
det [_~ (^) -~] I (^) :]. 'od add the r.", row to the ,~ood 0 13 and to the third to make all entries nonzero. .[]
- adj A ~ [ ~ -1^ o ;
- det A = _(I)n -2 0 1
- a. Note that det(A) det(A -I) = 1, and both factors (^) A-I = ---adj^1 A = -adj A are integers. det A b. Use the formula for the inverse of a 2 x 2 matrix (Theorem 2.4.9b). -1^0 I] o 1 0
- No [^2 0 -
- Take the determinant of both sides of^ a^ -b
- X = -2--2 > 0; Y = -2--2 < 0; X decreases In 0] [A B ] [ A B]^ a^ +b^ a^ +b [ (^) -C A C D - 0 AD - CB ' as b increases. and divide by det A. 29.^ dXI^ =^ -D-^
I (^) R2(1 - RIHI - a)2de dYI = D- I^ (1-a)R2(RI(1-a) +a)de2 > 0
- a. dn = dn-I + dn -2, a Fibonacci sequence (^) dp = D- I (^) RI R2de2 > 0 b. dl = I, d2 = 2, d3 = 3, d4 = 5, .... dlO = 89 c. Invertible for all positive integers n (^) [ ·6 (^0) [24 12 0
- -~ (^5) -~] 33. ~ (^0 )
6.3 Answers to more theoretical questions are omitted. -5 0 0 ~j
I. 50 3. 13 7. 110 35.^ det(adj^ A)^ =^ (detA)/-
- Idet A I = 12, the expansion factor of T on the par 37.^ adj(A-^ I^ )^ =^ (adj^ A)-I^ =^ (detA)-lA allelogram defined by v 1 and V2 (^) 39. Yes. Use Exercises 38: If AS = S^ B,^ then
- ,j2O (^) (adj S)(adj A) = (adj B)(adj S).
- We need to show that if VI, .. , , vm are linearly dependent, then (a) V(VI .... ,V 43.^ A(adj^ A)^ =^ (adj^ A)A^ =^ (detA)I^ n^ =^0 m )^ =^0 and (b) det(A T^ A) = O.
- [~ -~] a. One of the Vi is redundant. so that vf = 0 and V (v 1, ... , Vm ) = 0, by Definition 6.3.5. b. ker A i- (OJ and ker A ~ ker(A T A), so (^) CHAPTER 7 that ker(A T^ A) i- (OJ. Thus, AT A fails to be (^) 7.1 Answers to more theoretical questions are omitted. invertible. (^) I. Yes; the eigenvalue is A. 3.
- a. V(VI. V2. V3, VI x V2 x V3) (^) 3. Yes; the eigenvalue is A. + 2. = V(VI,V2,v3)11vI x V2 x v311 because VI x V2 x v 3 is orthogonal to v I , V2, and V3 5.^ Yes
- Eigenbasis (^) [-~],[n,s 1] B [-~ I^ '
[~ ~]
- To form an eigenbasis for this matrix of rank I, con catenate a basis of the kernel with a nonzero vector
i"theim'ge,!o'mmp!e, n], [-~], [H
-2 -3 I] The matrices S = 1 0 2 and B [ (^) o I 3
(^0) o 00 0] 0 diagonalize A. [ (^) o 0 14
- S = (^) [^34 -4]3' B = [1 0 0] 0
- Matrix A represents a reflection, with eigenvalues I and -1. Solve the equation Ax = x to find an eigenvector with eigenvalue 1. For example, S
[_i ~] and B = [~ _ ~ ].
59. Pick two linearly independent vectors VI, V2 on the
plane and one perpendicular vector V3, for exam
ple, S = [vi] ~2 v13] = [i ~ -~] and
I I I 0 I 2
1 0 0] B = 0 I 0. [ 000
61. Pick two linearly independent solutions VI, V2 of
the equation Ax = x, (^) let V3 = VI X V2, and
-2 -3 I] m,keS ~ [~1 H] 1 0 2 and [ (^) o I 3
1 0 0] B = 0 I O. [o 0 -
- Matrix A represents the orthogonal projection onto
[
span 2 I]. We can let S [-2 1 -3 0 1] 2 and 3 0 I 3
B=^00 00 0] 0.
[ 001
- The subspaces spanned by an eigenvector 67. eCt) = 300(1.1)1 - 200(0.9) ret) = 900(1.1)1 - 100(0.9)
- a. e(t) = 100(1.5)1, ret) = 200(1.5)
b. e(t) = 100(0.75)1, ret) = 100(0.75)
ANSWERS TO ODD·NUMBERED EXERCISES 489
c. e(t) = 300(0.75)1 + 200(1.5)1, ret) = 300(0.75)1 + 400(1.5)
aCt + J)] 1 [0 1 1] [aCt)]
71. bet + 1) = 2 1 0 1 bCt). The three [ (^) e(t + 1) I 1 0 e(t) '-.,.-" A given vectors are eigenvectors of A, with eigenval ues I^ , - 2:'I^ - 2:'^ I^ respectIVe^. Iy.
a. aCt) = 3 +3 (-!r, bet) = 3 -2 (-!y,
e(t) = 3 - ( - n 1 b. Benjamin will have the most.
7.2 Answers to more theoretical questions are omitted.
- 1,3 3. 1,3 5. None
- 1, 1, I 9. I, 2, 2 II. -1 13. 1
- Eigenvalues A 1,2 = 1 ±./k. Two distinct real eigen values if k is positive; none, if k is negative. 17. A represents a reflection followed by a scaling, with a scaling factor of Ja^2 + b^2_._ The eigenvalues are ±Ja^2 +b^2_._
- True [the discriminant (tr A)2 - 4detA is positive]
- Write IA(A) = (AI - A)'" (An - A) to show that
the coefficient of (_A)n-] is A1+ ... + An. But that
coefficient is tr A. by Theorem 7.2.5.
23. A and B have the same characteristic polynomial and the same eigenvalues, with the same algebraic multiplicities.
- (^) A [~] = [~] and A [~] = (a -b) [~lNote
that la - bl < I. Phase portrait when a > b:
line spanned by [~J
line spanned by ~ ~J
- a. (^) x(t) ~^ = 3"1[1] 2 + 2 3" (1)1 4 [1]-1 for Xu_ = e, ~
x(t)^ _^ = 1 3" [I] 2 - 3"^1 (1)1 4 [I]-I for Xu~^ = e2 _
490 ANSWERS TO ODD·NUMBERED EXERCISES
- Eigenbasis: e I, e2, e" with eigenvalues 1. 2, 3 0 0
~]
- (^) s ~ [~ I (^) -I]o and B = (^) [' 0 I 0 1 0 0
II. Eigenbasis:^
[:],Hl
[ _~], with ,lg,,,,,I,,,
3,0, I 0 b. Al = [AIel Al ih] approaches ~ 3 [;_ 13. (^) S ~ [~ -3^ i] '''d R ~ [~ I 0 I 0 J] part a.
c. A^ I^ -+--^1 [b b +c c ~]^ 15.^ Elg,,,,,clo~ m, [i], willi ,Ig'"vm", 0 1;
- Ae = e, so that e is an eigenvector with associated eigenvalue 1. no eigenbasis
- A and AT have the same eigenvalues, by Exer 17.^ Eigenbasis:^ C2,^ e4,^ el,^ e3^ -^ e2,^ with eigenvalues^ I, cise 22. Since the row sums of AT are I, we can 1,0, use the results of Exercises 29 and 30: I is an eigenvalue of A; if A is an eigenvalue of A, then (^) 19. Eg,,,,,,,to" [ j] ,[~] with ,Igm,'ue, 0, 1. -1 < A ~ 1. e need not be an eigenvector of A; . [0.9 0.9]^ Matrix^ A^ fails to^ be^ diagonalizable. consIder A = 0.1 0..
- We want A [~] = [;] and A m=2 [~] [:];
33. a. fA (A) = _A^3 + CA^2 + bA + a
that is, A [~ ~] = [~ : ]. The unique solution b. M = [~ ~ ~l IT -5 17 is A = [~ =~J. -1 0 o 0 23.^ The^ only eigenvalue^ of^ A^ is^ l,^ with^ EI^ =^ span(el).
- (^) There is no eigenbasis. A represents a horizontal o 0 o I -~]^ shear.
- We can write fA (A) = (A - AO)2 (^) g (A). By the prod^ 25.^ The^ geometric multiplicity is always^ I. uct rule, f~ (A) = 2(A - AO)g(A) + (A - Ao)2l (A), (^) 27. fA(A) = A^2 - 5A + 6 = (A - 2)(A - 3), so that the
so that f~ (AO) = O. eigenvalues are 2. 3.
- It's a straightforward computation. (^) 29. Both multiplicities are tl - r.
- tr(S-I(AS)) = tr((AS)S-t) = trA (^3) I. They are the same.
- No. since tr(AB - BA) = 0 and tr(ln) = 11 33. If B = S-I AS, then B - M n = S-1 (A - Aln)S.
- For k = 3 35. No (consider the eigenvalues)
- If we write M = [v w], then it is required that 37. a. AD· II) = (AiJ)T w = iJTATw = iJTAw = Ai! = 2i! and Aw = 3w. Thus, a nonzero M with iJ· Au) the given property exists if 2 or 3 is an eigenvalue (^) b. Suppose Ai! = AV and AU'} = Jlw. Then of A. (^) Av· w = i-,(v· w) and v· AUI = (L(V' UI). By part
- If 2, 3, or 4 is an eigenvalue of A a,^ A(V'^ U})^ =^ {i(v,^ UI),^ so that^ (),.^ -/l)(V'^ w)^ =^ O. Since A i {i, it follows that v. w = 0, as
7.3 Answers to more theoretical questions are omitted. claimed.
- a. E, = V and Eo = V-L, so that the geometric
- Eigenbasis: [~J, [~], with eigenvalues 7, 9 (^) multiplicity of ] is 111 and that of 0 is 11 -111. The algebraic multiplicities are the same. See Exer
- S = [_~ qand B = [~ ~] cise 31. b. £) = VandE_J = V-L,sothatthemuItiplicity
- No real eigenvalues. (^) of lis 111 and that of -1 is 11 - m.
