Elementary Linear Algebra

Transcription

1 Instructor s Manual with Sample Tests for Elementary Linear Algebra Fourth Edition Stephen Andrilli and David Hecker

2 Dedication To all the instructors who have used the various editions of our book over the years

3 Table of Contents Preface...iii Answers to Exercises. Chapter Answer Key. Chapter Answer Key... Chapter Answer Key... Chapter Answer Key...8 Chapter Answer Key...9 Chapter Answer Key. Chapter Answer Key.9 Chapter 8 Answer Key. Chapter 9 Answer Key.8 Appendix B Answer Key.0 Appendix C Answer Key.0 Chapter Tests...0 Chapter Chapter Tests 0 Chapter Chapter Tests Chapter Chapter Tests Chapter Chapter Tests Chapter Chapter Tests Chapter Chapter Tests 0 Chapter Chapter Tests Answers to Chapter Tests.9 Chapter Answers for Chapter Tests...9 Chapter Answers for Chapter Tests... Chapter Answers for Chapter Tests... Chapter Answers for Chapter Tests... Chapter Answers for Chapter Tests... Chapter Answers for Chapter Tests... Chapter Answers for Chapter Tests... ii

4 Preface This Instructor s Manual with Sample Tests is designed to accompany Elementary Linear Algebra, th edition, by Stephen Andrilli and David Hecker. This manual contains answers for all the computational exercises in the textbook and detailed solutions for virtually all of the problems that ask students for proofs. The exceptions are typically those exercises that ask students to verify a particular computation, or that ask for a proof for which detailed hints have already been supplied in the textbook. A few proofs that are extremely trivial in nature have also been omitted. This manual also contains sample Chapter Tests for the material in Chapters through, as well as answer keys for these tests. Additional information regarding the textbook, this manual, the Student Solutions Manual, and linear algebra in general can be found at the web site for the textbook, where you found this manual. Thank you for using our textbook. Stephen Andrilli David Hecker August 009 iii

5 Andrilli/Hecker - Answers to Exercises Section. Answers to Exercises Chapter Section. () (a) [9; ], distance = p 9 (b) [ ; ; ], distance = p 0 () (a) (; ; ) (see Figure ) (b) (0; ; ) (see Figure ) (c) [ ; ; ; ; ], distance = p (c) (; ; 0) (see Figure ) (d) (; 0; 0) (see Figure ) () (a) (; ) (b) ( ; ; ) (c) ( ; ; ; ; ) () (a) ; ; 8 (b) ; 0; ; h () (a) p 0 ; p 0 ; p 0 i; shorter, since length of original vector is > h (b) p ; p ; 0; p i; shorter, since length of original vector is > (c) [0:; h (d) 0:8]; neither, since given vector is a unit vector p ; p ; p ; p ; = p < p i; longer, since length of original vector () (a) Parallel (b) Parallel (c) Not parallel (d) Not parallel () (a) [ ; ; ] (b) [; 0; ] (c) [ ; ; 8] (d) [ ; ; ] (e) [; 0; ] (f) [ ; ; ] (8) (a) x + y = [; ], x y = [ ; 9], y x = [; 9] (see Figure ) (b) x + y = [; ], x y = [; ], y x = [ ; ] (see Figure ) (c) x+y = [; 8; ], x y = [; ; ], y x = [ ; ; ] (see Figure ) (d) x+y = [ ; ; ], x y = [; 0; ], y x = [ ; 0; ] (see Figure 8) (9) With A = (; ; ), B = (; ; ), and C = (0; ; 8), the length of side AB = length of side AC = p 9. The triangle is isosceles, but not equilateral, since the length of side BC is p 0.

6 Andrilli/Hecker - Answers to Exercises Section.

7 Andrilli/Hecker - Answers to Exercises Section.

8 Andrilli/Hecker - Answers to Exercises Section. (0) (a) [0; 0] (b) [ p ; ] (c) [0; 0] = 0 () See Figures. and.8 in Section. of the textbook. () See Figure 9. Both represent the same diagonal vector by the associative law of addition for vectors. Figure 9: x + (y + z) () [0: 0: p ; 0: p ] [ 0:8; 0:] () (a) If x = [a ; a ] is a unit vector, then cos = a kxk = a = a. Similarly, cos = a. (b) Analogous to part (a). () Net velocity = [ p ; + p ], resultant speed.8 km/hr h i () Net velocity = ; speed.8 km/hr p ; 8 p () [ 8 p ; p ] (8) Acceleration = 0 [ ; (9) Acceleration = ; 0; ; 9 (0) 80 p [ ; ; ] [ 9:; :; 8:] () a = [ mg + p ; mg + p ]; b = [ mg + p ; mg p + p ] ] [0:0; 0:; 0:0] () Let a = [a ; : : : ; a n ]. Then, kak = a + +a n is a sum of squares, which must be nonnegative. The square root of a nonnegative real number is a nonnegative real number. The sum can only be zero if every a i = 0.

9 Andrilli/Hecker - Answers to Exercises Section. () Let x = [x ; x ; : : : ; x n ]. Then kcxk = p (cx ) + + (cx n ) = p c (x + + x n) = jcj p x + + x n = jcj kxk: () In each part, suppose that x = [x ; : : : ; x n ], y = [y ; : : : ; y n ], and z = [z ; : : : ; z n ]. (a) x + (y + z) = [x ; : : : ; x n ] + [(y + z ); : : : ; (y n + z n )] = [(x + (y + z )); : : : ; (x n + (y n + z n ))] = [((x + y ) + z ); : : : ; ((x n + y n ) + z n )] = [(x + y ); : : : ; (x n + y n )] + [z ; : : : ; z n ] = (x + y) + z (b) x + ( x) = [x ; : : : ; x n ] + [ x ; : : : ; x n ] = [(x + ( x )); : : : ; (x n + ( x n ))] = [0; : : : ; 0]. Also, ( x) + x = x + ( x) (by part () of Theorem.) = 0, by the above. (c) c(x+y) = c([(x +y ); : : : ; (x n +y n )]) = [c(x +y ); : : : ; c(x n +y n )] = [(cx +cy ); : : : ; (cx n +cy n )] = [cx ; : : : ; cx n ]+[cy ; : : : ; cy n ] = cx+cy (d) (cd)x = [((cd)x ); : : : ; ((cd)x n )] = [(c(dx )); : : : ; (c(dx n ))] = c[(dx ); : : : ; (dx n )] = c(d[x ; : : : ; x n ]) = c(dx) () If c = 0, done. Otherwise, ( c )(cx) = c (0) ) ( c c)x = 0 (by part () of Theorem.) ) x = 0: Thus either c = 0 or x = 0. () Let x = [x ; x ; : : : ; x n ]. Then [0; 0; : : : ; 0] = c x c x = [(c c )x ; : : : ; (c c )x n ]. Since c c = 0, we have x = x = = x n = 0. () (a) F (b) T (c) T (d) F (e) T (f) F (g) F Section. () (a) arccos( p ), or about :, or. radians (b) arccos( p p ), or about :, or. radians (c) arccos(0), which is 90, or radians (d) arccos( ), which is 80, or radians (since x = y) () The vector from A to A is [; ; ], and the vector from A to A is [; ; ]. These vectors are orthogonal. () (a) [a; b] [ b; a] = a( b) + ba = 0: Similarly, [a; b] [b; a] = 0: (b) A vector in the direction of the line ax + by + c = 0 is [b; a vector in the direction of bx ay + d = 0 is [a; b]. a], while () (a) joules (b) 00p 9, or 8: joules (c) joules () Note that yz is a scalar, so x(yz) is not de ned.

10 Andrilli/Hecker - Answers to Exercises Section. () In all parts, let x = [x ; x ; : : : ; x n ] ; y = [y ; y ; : : : ; y n ] ; and z = [z ; z ; : : : ; z n ] : Part (): x y = [x ; x ; : : : ; x n ] [y ; y ; : : : ; y n ] = x y + + x n y n = y x + + y n x n = [y ; y ; : : : ; y n ] [x ; x ; : : : ; x n ] = y x Part (): x x = [x ; x ; : : : ; x n ] [x ; x ; : : : ; x n ] = x x + + x n x n = x + +x n. Now x + +x n is a sum of squares, each of which must be nonnegative. Hence, the sum is also nonnegative, and so its square root is de ned. Thus, 0 x x = x + + x n = p x + + x n = kxk. Part (): Suppose x x = 0. From part (), 0 = x x = x + + x n x i, for each i, since the sum of the remaining squares (without x i ), which is nonnegative. Hence, 0 x i for each i. But x i 0, because it is a square. Hence each x i = 0. Therefore, x = 0. Next, suppose x = 0. Then x x = [0; : : : ; 0] [0; : : : ; 0] = (0)(0) + + (0)(0) = 0. Part (): c(x y) = c ([x ; x ; : : : ; x n ] [y ; y ; : : : ; y n ]) = c (x y + + x n y n ) = cx y + + cx n y n = [cx ; cx ; : : : ; cx n ] [y ; y ; : : : ; y n ] = (cx) y. Next, c(x y) = c(y x) (by part ()) = (cy) x (by the above) = x (cy), by part (): Part (): (x + y) z = ([x ; x ; : : : ; x n ] + [y ; y ; : : : ; y n ]) [z ; z ; : : : ; z n ] = [x + y ; x + y ; : : : ; x n + y n ] [z ; z ; : : : ; z n ] = (x + y )z + (x + y )z + + (x n + y n )z n = (x z + x z + + x n z n ) + (y z + y z + + y n z n ): Also, (x z) + (y z) = ([x ; x ; : : : ; x n ] [z ; z ; : : : ; z n ]) + ([y ; y ; : : : ; y n ] [z ; z ; : : : ; z n ]) = (x z + x z + + x n z n ) + (y z + y z + + y n z n ). Hence, (x + y) z = (x z) + (y z). () No; consider x = [; 0], y = [0; ], and z = [; ]. (8) A method similar to the rst part of the proof of Theorem. in the textbook yields: ka bk 0 ) (a a) (b a) (a b) + (b b) 0 ) (a b) + 0 ) a b. (9) Note that (x + y)(x y) = (xx) + (yx) (xy) (yy) = kxk kyk. (0) Note that kx + yk = kxk + (xy) + kyk, while kx yk = kxk (xy) + kyk. () (a) This follows immediately from the solution to Exercise 0 above. (b) Note that kx + y + zk = k(x + y) + zk = kx + yk + ((x + y)z) + kzk = kxk + (xy) + kyk + (xz) + (yz) + kzk. (c) This follows immediately from the solution to Exercise 0 above.

11 Andrilli/Hecker - Answers to Exercises Section. () Note that x(c y + c z) = c (xy) + c (xz). () cos = a p a +b +c, cos = () (a) Length = p s b p a +b +c, and cos = (b) angle = arccos( p ) :, or 0:9 radians () (a) [ ; 0 ; ] (c) [ (b) [ 90 ; 0; ; ] ; ; 0] [:9; :8; 0] c p a +b +c () (a) 0 (zero vector). The dropped perpendicular travels along b to the common initial point of a and b. (b) The vector b. The terminal point of b lies on the line through a, so the dropped perpendicular has length zero. () ai, bj, ck (8) (a) Parallel: [ 0 9 ; 0 9 ; 0 9 ], orthogonal: [ 9 9 ; 88 9 ; 9 ] (b) Parallel: [ ; ], orthogonal: [ ; ; ] (c) Parallel: [ 0 9 ; 0 9 ; 0 9 ], orthogonal: [ 9 ; 8 9 ; 9 ] (9) From the lower triangle in the gure, we have (proj r x) + (proj r x x) = re ection of x (see Figure 0). Figure 0: Re ection of a vector x through a line. (0) For the case kxk kyk: j kxk kyk j = kyk kxk = kx+y+( x)k kxk kx+yk+k xk kxk (by the Triangle Inequality) = kx+yk+kxk kxk = kx + yk. The case kxk kyk is done similarly, with the roles of x and y reversed. () (a) Let c = (xy)=(kxk ), and w = y proj x y. (In this way, cx = proj x y.)

12 Andrilli/Hecker - Answers to Exercises Section. (b) Suppose cx + w = dx + v. Then (d c)x = w v, and (d c)x(w v) = (w v)(w v) = kw vk. But (d c)x(w v) = (d c)(xw) (d c)(xv) = 0, since v and w are orthogonal to x. Hence kw vk = 0 =) w v = 0 =) w = v. Then, cx = dx =) c = d, from Exercise in Section., since x is nonzero. () If is the angle between x and y, and is the angle between proj x y and proj y x, then cos = xy yx kxk kyk (xy) = jxyj jyxj kxk kxk kyk kyk proj x y proj y x kproj x ykkproj y xk = (xy) kxk kyk (xy) kxkkyk xy kxk x yx kyk y xy kxk x yx = kyk y = (xy) kxkkyk = cos. Hence =. () (a) T (b) T (c) F (d) F (e) T (f) F Section. () (a) We have kx + yk kxk + kyk = kxk + kyk kxk + kyk = (kxk + kyk). (b) Let m = maxfjcj; jdjg. Then kcx dyk m(kxk + kyk). () (a) Note that j = ((j )) +. Let k = (j ). (b) Consider the number. () Note that since x = 0 and y = 0, proj x y = 0 i (xy)=(kxk ) = 0 i xy = 0 i yx = 0 i (yx)=(kyk ) = 0 i proj y x = 0. () If y = cx (for c > 0), then kx+yk = kx+cxk = (+c)kxk = kxk+ckxk = kxk + kcxk = kxk + kyk. On the other hand, if kx + yk = kxk + kyk, then kx + yk = (kxk + kyk). Now kx+yk = (x+y)(x+y) = kxk +(xy)+kyk, while (kxk+kyk) = kxk + kxkkyk + kyk. Hence xy = kxkkyk. By Result, y = cx, for some c > 0. () (a) Consider x = [; 0; 0] and y = [; ; 0]. (b) If x = y, then x y = kxk. (c) Yes () (a) Suppose y = 0. We must show that x is not orthogonal to y. Now kx + yk = kxk, so kxk + (xy) + kyk = kxk. Hence kyk = (xy). Since y = 0, we have kyk = 0, and so xy = 0. (b) Suppose x is not a unit vector. We must show that xy =. Now proj x y = x =) ((xy)=(kxk ))x = x =) (xy)=(kxk ) = =) xy = kxk. But then kxk = =) kxk = =) xy =. () Use Exercise (a) in Section.. 8

13 Andrilli/Hecker - Answers to Exercises Section. (8) (a) Contrapositive: If x = 0, then x is not a unit vector. Converse: If x is nonzero, then x is a unit vector. Inverse: If x is not a unit vector, then x = 0. (b) (Let x and y be nonzero vectors.) Contrapositive: If y = proj x y, then x is not parallel to y. Converse: If y = proj x y, then x k y. Inverse: If x is not parallel to y, then y = proj x y. (c) (Let x, y be nonzero vectors.) Contrapositive: If proj y x = 0, then proj x y = 0. Converse: If proj y x = 0, then proj x y = 0. Inverse: If proj x y = 0, then proj y x = 0. (9) (a) Converse: Let x and y be nonzero vectors in R n. If kx + yk > kyk, then xy 0. (b) Let x = [; ] and y = [0; ]. (0) (a) Converse: Let x, y, and z be vectors in R n. If y = z, then xy = xz. The converse is obviously true, but the original statement is false in general, with counterexample x = [; ], y = [; ], and z = [ ; ]. (b) Converse: Let x and y be vectors in R n. If kx + yk kyk, then x y = 0. The original statement is true, but the converse is false in general. Proof of the original statement follows from kx + yk = (x + y) (x + y) = kxk + (x y) + kyk = kxk + kyk kyk. Counterexample to converse: let x = [; 0], y = [; ]. (c) Converse: Let x, y be vectors in R n, with n >. If x = 0 or y = 0, then xy = 0. The converse is obviously true, but the original statement is false in general, with counterexample x = [; ] and y = [; ]. () Suppose x? y and n is odd. Then x y = 0. Now x y = P n i= x iy i. But each product x i y i equals either or. If exactly k of these products equal, then x y = k (n k) = n + k. Hence n + k = 0, and so n = k, contradicting n odd. () Suppose that the vectors [; a], [; b], and [z ; z ] as constructed in the hint are all mutually orthogonal. Then z + az = 0 = z + bz. Hence (a b)z = 0, so either a = b or z = 0. But if a = b, then [; a] and [; b] are not orthogonal. Alternatively, if z = 0, then z + az = 0 =) z = 0, so [z ; z ] = [0; 0], a contradiction. () Base Step (n = ): x = x. Inductive Step: Assume x +x + +x n +x n = x n +x n + +x +x, for some n : Prove: x + x + + x n + x n + x n+ = x n+ + x n + x n + + x + x : But x + x + + x n + x n + x n+ = (x + x + + x n + x n ) + x n+ = x n+ + (x + x + + x n + x n ) = x n+ + (x n + x n + + x + x ) (by the assumption) = x n+ + x n + x n + + x + x : 9

14 Andrilli/Hecker - Answers to Exercises Section. () Base Step (m = ): kx k kx k. Inductive Step: Assume kx + + x m k kx k + + kx m k, for some m. Prove kx + +x m +x m+ k kx k+ +kx m k+kx m+ k. But, by the Triangle Inequality, k(x + +x m )+x m+ k kx + +x m k+kx m+ k kx k + + kx m k + kx m+ k. () Base Step (k = ): kx k = kx k. Inductive Step: Assume kx + + x k k = kx k + + kx k k. Prove kx + +x k +x k+ k = kx k + +kx k k +kx k+ k. Use the fact that k(x + +x k )+x k+ k = kx + +x k k +((x + +x k )x k+ )+kx k+ k = kx + +x k k +kx k+ k, since x k+ is orthogonal to all of x ; : : : ; x k. () Base Step (k = ): We must show (a x )y ja j kyk. But (a x )y j(a x )yj ka x k kyk (by the Cauchy-Schwarz Inequality) = ja j kx k kyk = ja j kyk, since x is a unit vector. Inductive Step: Assume (a x + + a k x k )y (ja j + + ja k j)kyk, for some k. Prove (a x + + a k x k + a k+ x k+ )y (ja j + + ja k j + ja k+ j)kyk. Use the fact that ((a x + + a k x k ) +a k+ x k+ )y = (a x + + a k x k )y + (a k+ x k+ )y and the Base Step. () Base Step (n = ): We must show that k[x ]k jx j. This is obvious since k[x ]k = p x = jx j. q Pk Inductive Step: Assume i= x i P q k i= jx Pk+ ij and prove i= x i P k+ i= jx ij. Let y = [x ; : : : ; x k ; x k+ ], z = [x ; : : : ; x k ; 0], and w = [0; : : : ; 0; x k+ ]. Note q that y = z+w. So, by the Triangle Inequality, kyk Pk+ q kzk+kwk. Thus, i= x i Pk q i= x i + x k+ P q k i= jx ij+ x k+ (by the inductive hypothesis) = P k i= jx ij + jx k+ j = P k+ i= jx ij. (8) Step cannot be reversed, because y could equal (x + ). Step cannot be reversed, because y could equal x + x + c. Step cannot be reversed, because in general y does not have to equal x +. Step cannot be reversed, since dy dx could equal x + c. All other steps remain true when reversed. (9) (a) For every unit vector x in R, x [; ; ] = 0. (b) x = 0 and xy 0, for some vectors x and y in R n. (c) x = 0 or kx + yk = kyk, for all vectors x and y in R n. (d) There is some vector x R n for which xx 0. (e) There is an x R such that for every nonzero y R, x y = 0. (f) For every x R, there is some y R such that xy = 0. 0

15 Andrilli/Hecker - Answers to Exercises Section. (0) (a) Contrapositive: If x = 0 and kx yk kyk, then x y = 0. Converse: If x = 0 or kx yk > kyk, then x y = 0. Inverse: If x y = 0, then x = 0 and kx yk kyk. (b) Contrapositive: If kx yk kyk, then either x = 0 or xy = 0. Converse: If kx yk > kyk, then x = 0 and xy = 0. Inverse: If x = 0 or xy = 0, then kx yk kyk. () Suppose x = 0. We must prove xy = 0 for some vector y R n. Let y = x. () The contrapositive is: If u and v are (nonzero) vectors that are not in opposite directions, then there is a vector x such that u x > 0 and v x > 0. If u and v are in the same direction, we can let x = u. So, we can assume that u and v are not parallel. Consider a vector x that bisects the angle between u and v. Geometrically, it is clear that such a vector makes an acute angle with both u and v, which nishes the proof. (Algebraically, a formula for such a vector would be x = u kuk + v kvk. Then, u x = u u kuk + v kvk = uu kuk + uv kvk kuk the Cauchy-Schwarz inequality) = kuk similar proof shows that v x > 0.) () Let x = [; ], y = [; ]. uv > kuk kvk kukkvk kvk (by kuk = 0. Hence u x > 0. A () Let y = [; ; ]. Then since xy 0, Result implies that kx + yk > kyk =. () (a) F (b) T (c) T (d) F (e) F (f) F (g) F (h) T (i) F Section. () (a) 9 0 (b) Impossible 8 (c) (d) (e) Impossible (f) (g) (h) Impossible (i) 8 8 (j) (k) (l) Impossible (m) Impossible (n) We used > rather than in the Cauchy-Schwarz inequality because equality occurs in the Cauchy-Schwarz inequality only when the vectors are parallel (see Result ).

16 Andrilli/Hecker - Answers to Exercises Section. () Square: B; C; E; F; G; H; J; K; L; M; N; P; Q Diagonal: B; G; N Upper triangular: B; G; L; N Lower triangular: B; G; M; N; Q Symmetric: B; F; G; J; N; P Skew-symmetric: H (but not E; C; K) Transposes: A T = on () (a) (b) (c) (d) , B T = B, C T = , and so () If A T = B T, then A T T = B T T, implying A = B, by part () of Theorem.. () (a) If A is an m n matrix, then A T is n m. If A = A T, then m = n. (b) If A is a diagonal matrix and if i = j, then a ij = 0 = a ji. (c) Follows directly from part (b), since I n is diagonal. (d) The matrix must be a square zero matrix; that is, O n, for some n. () (a) If i = j, then a ij + b ij = = 0. (b) Use the fact that a ij + b ij = a ji + b ji. () Use induction on n. Base Step (n = ): Obvious. Inductive Step: Assume A ; : : : ; A k+ and B = P k i= A i are upper triangular. Prove that D = P k+ i= A i is upper triangular. Let C = A k+. Then D = B + C. Hence, d ij = b ij + c ij = = 0 if i > j (by the inductive hypothesis). Hence D is upper triangular.

17 Andrilli/Hecker - Answers to Exercises Section. (8) (a) Let B = A T. Then b ij = a ji = a ij = b ji. Let D = ca. Then d ij = ca ij = ca ji = d ji. (b) Let B = A T. Then b ij = a ji = a ij = b ji. Let D = ca. Then d ij = ca ij = c( a ji ) = ca ji = d ji. (9) I n is de ned as the matrix whose (i; j) entry is ij. (0) Part (): Let B = A + ( A) (= ( A) + A, by part ()). Then b ij = a ij + ( a ij ) = 0. Part (): Let D = c(a+b) and let E = ca+cb. Then, d ij = c(a ij +b ij ) = ca ij + cb ij = e ij. Part (): Let B = (cd)a and let E = c(da). Then b ij = (cd)a ij = c(da ij ) = e ij. () Part (): Let B = A T and C = (A T ) T. Then c ij = b ji = a ij. Part (): Let B = c(a T ), D = ca and F = (ca) T. Then f ij = d ji = ca ji = b ij. () Assume c = 0. We must show A = O mn. But for all i; j, i m, j n, ca ij = 0 with c = 0. Hence, all a ij = 0. () (a) ( (A + AT )) T = (A + AT ) T = (AT + (A T ) T ) = (AT + A) = (A + AT ) (b) ( (A AT )) T = (A AT ) T = (AT (A T ) T ) = (AT A) = (A AT ) (c) (A + AT ) + (A AT ) = (A) = A (d) S V = S V (e) Follows immediately from part (d). (f) Follows immediately from parts (d) and (e). (g) Parts (a) through (c) show A can be decomposed as the sum of a symmetric matrix and a skew-symmetric matrix, while parts (d) through (f) show that the decomposition for A is unique. () (a) Trace (B) =, trace (C) = 0, trace (E) =, trace (F) =, trace (G) = 8, trace (H) = 0, trace (J) =, trace (K) =, trace (L) =, trace (M) = 0, trace (N) =, trace (P) = 0, trace (Q) = (b) Part (i): Let D = A + B. Then trace(d) = P n P i= d ii = n i= a ii + P n i= b ii = trace(a) + trace(b). Part (ii): Let B = ca. Then trace(b) = P n i= b ii = P n i= ca ii = c P n i= a ii = c(trace(a)). Part (iii): Let B = A T. Then trace(b) = P n i= b ii = P n i= a ii (since b ii = a ii for all i) = trace(a). (c) Not necessarily: consider the matrices L and N in Exercise. (Note: If n =, the statement is true.)

18 Andrilli/Hecker - Answers to Exercises Section. () (a) F (b) T (c) F (d) T (e) T Section. () (a) Impossible (e) [ 8] 8 (k) (b) 9 (f) 8 8 (c) Impossible (g) Impossible (l) (h) [; 8] 0 (d) (i) Impossible 9 (j) Impossible (m) [98] 0 (o) Impossible (n) () (a) No (b) Yes () (a) [; ; 8] (b) (c) No (d) Yes (c) [] (d) [; 8; ; ] (e) No () (a) Valid, by Theorem., part () (b) Invalid (c) Valid, by Theorem., part () (d) Valid, by Theorem., part () (e) Valid, by Theorem. (f) Invalid (g) Valid, by Theorem., part () (h) Valid, by Theorem., part () (i) Invalid (j) Valid, by Theorem., part (), and Theorem. () Outlet Outlet Outlet Outlet Salary $00 $8000 $000 $8000 $000 $000 $0000 $00 Fringe Bene ts () June July August Tickets Food Souvenirs $00 $0900 $900 $0000 $00 $800 $9800 $900 $9000

19 Andrilli/Hecker - Answers to Exercises Section. () Nitrogen Phosphate Potash Field Field Field :00 0: 0: 0:90 0: 0: (in tons) 0:9 0: 0:8 (8) Rocket Rocket Rocket Rocket Chip Chip Chip Chip (9) (a) One example: 0 (b) One example: (0) (a) Third row, fourth column entry of AB (b) Fourth row, rst column entry of AB (c) Third row, second column entry of BA (d) Second row, fth column entry of BA (c) Consider () (a) P n k= a kb k (b) P m k= b ka k () (a) [ ; ; ] (b) 8 () (a) [ ; ; 0; 9] () (a) Let B = Ai. Then B is m and b k = P m j= a kji j = (a k )() + (a k )(0) + (a k )(0) = a k. (b) Ae i = ith column of A. (c) By part (b), each column of A is easily seen to be the zero vector by letting x equal each of e ; : : : ; e n in turn. (b) () Proof of Part (): The (i; j) entry of A(B + C) = (ith row of A)(jth column of (B + C)) = (ith row of A)(jth column of B + jth column of C) = (ith row of A)(jth column of B) + (ith row of A)(jth column of C) = (i; j) entry of AB + (i; j) entry of AC = (i; j) entry of (AB + AC). The proof of Part () is similar. For the rst equation in part (), the

20 Andrilli/Hecker - Answers to Exercises Section. (i; j) entry of c(ab) = c ((ith row of A)(jth column of B)) = (c(ith row of A))(jth column of B) = (ith row of ca)(jth column of B) = (i; j) entry of (ca)b. The proof of the second equation is similar. () Let B = AO np. Clearly B is an m p matrix and b ij = P n k= a ik0 = 0. () Proof that AI n = A: Let B = AI n. Then B is clearly an m n matrix and b jl = P n k= a jki kl = a jl, since i kl = 0 unless k = l, in which case i kk =. The proof that I m A = A is similar. (8) (a) We need to show c ij = 0 if i = j. Now, if i = j, both factors in each term of the formula for c ij in the formal de nition of matrix multiplication are zero, except possibly for the terms a ii b ij and a ij b jj. But since the factors b ij and a ij also equal zero, all terms in the formula for c ij equal zero. (b) Assume i > j. Consider the term a ik b kj in the formula for c ij. If i > k, then a ik = 0. If i k, then j < k, so b kj = 0. Hence all terms in the formula for c ij equal zero. (c) Let L and L be lower triangular matrices. U = L T and U = L T are then upper triangular, and L L = U T U T = (U U ) T (by Theorem.). But by part (b), U U is upper triangular. So L L is lower triangular. (9) Base Step: Clearly, (ca) = c A. Inductive Step: Assume (ca) n = c n A n, and prove (ca) n+ = c n+ A n+. Now, (ca) n+ = (ca) n (ca) = (c n A n )(ca) (by the inductive hypothesis) = c n ca n A (by part () of Theorem.) = c n+ A n+. (0) Proof of Part (): Base Step: A s+0 = A s = A s I = A s A 0. Inductive Step: Assume A s+t = A s A t for some t 0. We must prove A s+(t+) = A s A t+. But A s+(t+) = A (s+t)+ = A s+t A = (A s A t )A (by the inductive hypothesis) = A s (A t A) = A s A t+. Proof of Part (): (We only need prove (A s ) t = A st.) Base Step: (A s ) 0 = I = A 0 = A s0. Inductive Step: Assume (A s ) t = A st for some integer t 0. We must prove (A s ) t+ = A s(t+). But (A s ) t+ = (A s ) t A s = A st A s (by the inductive hypothesis) = A st+s (by part ()) = A s(t+). () (a) If A is an m n matrix, and B is a p r matrix, then the fact that AB exists means n = p, and the fact that BA exists means m = r. Then AB is an m m matrix, while BA is an n n matrix. If AB = BA, then m = n. (b) Note that by the Distributive Law, (A+B) = A +AB+BA+B. () (AB)C = A(BC) = A(CB) = (AC)B = (CA)B = C(AB).

21 Andrilli/Hecker - Answers to Exercises Section. () Use the fact that A T B T = (BA) T, while B T A T = (AB) T. () (AA T ) T = (A T ) T A T (by Theorem.) = AA T. (Similarly for A T A.) () (a) If A; B are both skew-symmetric, then (AB) T = B T A T = ( B)( A) = ( )( )BA = BA. The symmetric case is similar. (b) Use the fact that (AB) T = B T A T = BA, since A; B are symmetric. () (a) The (i; i) entry of AA T = (ith row of A)(ith column of A T ) = (ith row of A)(ith row of A) = sum of the squares of the entries in the ith row of A. Hence, trace(aa T ) is the sum of the squares of the entries from all rows of A. (c) Trace(AB) = P n i= (P n k= a ikb ki ) = P n P n P i= k= b kia ik = n P n k= i= b kia ik. Reversing the roles of the dummy variables k and i gives P n P n i= k= b ika ki, which is equal to trace(ba). 0 () (a) Consider any matrix of the form. x 0 (c) (I n A) = I n I n A AI n + A = I n A A + A = I n A. (d) 0 0 (e) A = (A)A = (AB)A = A(BA) = AB = A. (8) (a) (ith row of A) (jth column of B) = (i; j)th entry of O mp = 0. (b) Consider A = and B = 0. 0 (c) Let C = (9) See Exercise 0(c) (0) Throughout this solution, we use the fact that Ae i = (ith column of A). (See Exercise (b).) Similarly, if e i is thought of as a row matrix, e i A = (ith row of A). (a) (jth column of A ij) = A(jth column of ij) = Ae i = (ith column of A). And, if k = j, (kth column of A ij) = A(kth column of ij) = A0 = 0. (b) (ith row of ija) = (ith row of ij)a = e j A = (jth row of A). And, if k = i, (kth row of ija) = (kth row of ij)a = 0A = 0.

22 Andrilli/Hecker - Answers to Exercises Chapter Review (c) Suppose A is an n n matrix that commutes with all other n n matrices. Suppose i = j. Then a ij = ith entry of (jth column of A) = ith entry of (ith column of A ji) (by reversing the roles of i and j in part (a)) = (i; i)th entry of A ji = (i; i)th entry of jia (since A commutes with ji) = ith entry of (ith row of jia) = 0 (since all rows other than the jth row of jia have all zero entries, by part (b)). Thus, A is diagonal. Next, a ii = ith entry of (ith column of A) = ith entry of (jth column of A ij) (by part (a)) = (i; j)th entry of A ij = (i; j)th entry of ija (since A commutes with ij) = jth entry of (ith row of ija) = jth entry of (jth row of A) (by part (b)) = a jj. Thus, all the main diagonal terms of A are equal, and so A = ci n for some c R. () (a) T (b) T (c) T (d) F (e) F (f) F Chapter Review Exercises () Yes. Vectors corresponding to adjacent sides are orthogonal. Vectors corresponding to opposite sides are parallel, with one pair having slope and the other pair having slope. h i () u = p 9 ; p 9 ; p 9 [0:8; 0:9; 0:]; slightly longer. () Net velocity = p ; p [0:9; :9]; speed :9 mi/hr. () a = [ 0; 9; 0] m/sec () jx yj = kxk kyk 90:9 () () proj a b = ; 8 ; 9 ; = [:; :; 0:; :8]; b proj a b = ; ; ; = [ 0:; :8; :; :8]; a (b proj a b) = 0. (8) 8 joules (9) We must prove that (x + y) (x y) = 0 ) kxk = kyk : But, (x + y) (x y) = 0 ) x x x y + y x y y = 0 ) kxk kyk = 0 ) kxk = kyk = kxk = kyk (0) First, x = 0, or else proj x y is not de ned. Also, y = 0, since that would imply proj x y = y. Now, assume xky. Then, there is a scalar c = 0 such xy kxk xcx that y = cx. Hence, proj x y = x = x = = y, contradicting the assumption that y = proj x y. kxk ckxk kxk x = cx 8

23 Andrilli/Hecker - Answers to Exercises () (a) A C T = () S = 9 0 ; AB = 0 BA is not de ned; AC = ; 0 CA = 0 8 ; A is not de ned; B = (b) Third row of BC = [ 8]. ; V = Chapter Review () (a) ((A B) T ) T = (A B). Also, ((A B) T ) = ( )(A T B T ) = ( )(( A) ( B)) (de nition of skew-symmetric) = (A B). Hence, ((A B) T ) T = ((A B) T ), so (A B) T is skew-symmetric. (b) Let C = A + B. Now, c ij = a ij + b ij. But for i < j, a ij = b ij = 0. Hence, for i < j, c ij = 0. Thus, C is lower triangular. ; () Company I Company II Company III Price Shipping Cost $800 $00 $000 $900. $000 $00 () Take the transpose of both sides of A T B T = B T A T to get BA = AB. Then, (AB) = (AB)(AB) = A(BA)B = A(AB)B = A B. () Negation: For every square matrix A, A = A. Counterexample: A = [ ]. () If A = O, then some row of A, say the ith row, is nonzero. Apply Result in Section. with x = (ith row of A). (8) Base Step (k = ): Suppose A and B are upper triangular n n matrices, and let C = AB. Then a ij = b ij = 0, for i > j. Hence, for i > j, c ij = P n m= a imb mj = P i m= 0b mj +a ii b ij + P n m=i+ a im 0 = a ii (0) = 0. Thus, C is upper triangular. Inductive Step: Let A ; : : : ; A k+ be upper triangular matrices. Then, the product C = A A k is upper triangular by the Induction Hypothesis, 9

24 Andrilli/Hecker - Answers to Exercises Chapter Review and so the product A A k+ = CA k+ is upper triangular by the Base Step. (9) (a) Let A and B be n n matrices having the properties given in the exercise. Let C = AB. Then we know that a ij = 0 for all i < j, b ij = 0 for all i > j, c ij = 0 for all i = j, and that a ii = 0 and b ii = 0 for all i. We need to prove that a ij = 0 for all i > j. We use induction on j. Base Step (j = ): Let i > j =. Then 0 = c i = P n k= a ikb k = a i b + P n k= a ikb k = a i b + P n k= a ik 0 = a i b. Since b = 0, this implies that a i = 0, completing the Base Step. Inductive Step: Assume for j = ; ; : : : ; m that a ij = 0 for all i > j: That is, assume we have already proved that, for some m, that the rst m columns of A have all zeros below the main diagonal. To complete the inductive step, we need to prove that the mth column of A has all zeros below the main diagonal. That is, we must prove that a im = 0 for all i > m. Let i > m. Then, 0 = c im = P n k= a ikb km = P m k= a ikb km +a im b mm + P n k=m+ a ikb km = P m k= 0 b km + a im b mm + P n k=m+ a ik 0 = a im b mm. But, since b mm = 0, we must have a im = 0, completing the proof. (b) Apply part (a) to B T and A T to prove that B T is diagonal. Hence B is also diagonal. (0) (a) F (b) T (c) F (d) F (e) F (f) T (g) F (h) F (i) F (j) T (k) T (l) T (m) T (n) F (o) F (p) F (q) F (r) T 0

25 Andrilli/Hecker - Answers to Exercises Section. Chapter Section. () (a) f( ; ; )g (b) f(; ; )g (c) fg (d) f(c + ; c ; c; ) j c Rg; (; ; 0; ); (8; ; ; ); (9; ; ; ) (e) f(b d ; b; d + ; d; ) j b; d Rg; ( ; 0; ; 0; ); ( ; ; ; 0; ); ( ; 0; ; ; ) (f) f(b + c ; b; c; 8) j b; c Rg; ( ; 0; 0; 8); ( ; ; 0; 8); (; 0; ; 8) (g) f(; ; )g (h) fg () (a) f(c + e + ; c + e + ; c; e + ; e) j c; e Rg (b) f(b d + 0e ; b; d 8e + ; d; e; ) j b; d; e Rg (c) f( 0c+9d f 8; c d+f +; c; d; f +; f) j c; d; f Rg (d) fg () nickels, dimes, quarters () y = x x + () y = x + x x + () x + y x 8y = 0, or (x ) + (y ) = () In each part, R(AB) = (R(A))B, which equals the given matrix. (a) 0 0 (b) 0 8 (8) (a) To save space, we write hci i for the ith row of a matrix C. For the Type (I) operation R : hii chii: Now, hr(ab)i i = chabi i = chai i B (by the hint in the text) = hr(a)i i B = hr(a)bi i. But, if k = i, hr(ab)i k = habi k = hai k B (by the hint in the text) = hr(a)i k B = hr(a)bi k. For the Type (II) operation R : hii chji + hii: Now, hr(ab)i i = chabi j + habi i = chai j B + hai i B (by the hint in the text) = (chai j + hai i )B = hr(a)i i B = hr(a)bi i. But, if k = i, hr(ab)i k = habi k = hai k B (by the hint in the text) = hr(a)i k B = hr(a)bi k.

26 Andrilli/Hecker - Answers to Exercises Section. For the Type (III) operation R : hii! hji: Now, hr(ab)i i = habi j = hai j B (by the hint in the text) = hr(a)i i B = hr(a)bi i. Similarly, hr(ab)i j = habi i = hai i B (by the hint in the text) = hr(a)i j B = hr(a)bi j. And, if k = i and k = j, hr(ab)i k = habi k = hai k B (by the hint in the text) = hr(a)i k B = hr(a)bi k. (b) Use induction on n, the number of row operations used. Base Step: The case n = is part () of Theorem., and is proven in part (a) of this exercise. Inductive Step: Assume that R n ( (R (R (AB))) ) = R n ( (R (R (A))) )B and prove that R n+ (R n ( (R (R (AB))) )) = R n+ (R n ( (R (R (A))) ))B: Now, R n+ (R n ( (R (R (AB))) )) = R n+ (R n ( (R (R (A))) )B) (by the inductive hypothesis) (by part (a)). = R n+ (R n ( (R (R (A))) ))B (9) Multiplying a row by zero changes all of its entries to zero, essentially erasing all of the information in the row. (0) Follow the hint in the textbook. First, A(X + c(x X )) = AX + ca(x X ) = B + cax cax = B + cb cb = B. Then we want to show that X + c(x X ) = X + d(x X ) implies that c = d to prove that each real number c produces a di erent solution. But X + c(x X ) = X + d(x X ) ) c(x X ) = d(x X ) ) (c d)(x X ) = 0 ) (c d) = 0 or (X X ) = 0. However, X = X, and so c = d. () (a) T (b) F (c) F (d) F (e) T (f) T Section. () Matrices in (a), (b), (c), (d), and (f) are not in reduced row echelon form. Matrix in (a) fails condition of the de nition. Matrix in (b) fails condition of the de nition. Matrix in (c) fails condition of the de nition. Matrix in (d) fails conditions,, and of the de nition. Matrix in (f) fails condition of the de nition.

27 Andrilli/Hecker - Answers to Exercises Section. () (a) (b) I (c) () (a) (e) (g) ; ( ; ; ) ; f(; ; )g (d) (e) (f) I ; f(b d ; b; d + ; d; ) j b; d Rg () (a) Solution set = f(c d; d; c; d) j c; d Rg; one particular solution = ( ; ; ; ) (b) Solution set = f(e; d e; d; d; e) j d; e Rg; one particular solution = (; ; ; ; ) (c) Solution set = f( b + d f; b; d + f; d; f; f) j b; d; f Rg; one particular solution = ( ; ; 0; ; ; ) () (a) f(c; c; c) j c Rg = fc(; ; ) j c Rg (b) f(0; 0; 0)g (c) f(0; 0; 0; 0)g (d) f(d; d; d; d) j d Rg = fd(; ; ; ) j d Rg () (a) a =, b =, c =, d = (b) a =, b =, c =, d = 8 (c) a =, b =, c =, d =, e = (d) a =, b =, c =, d =, e =, f = () (a) A =, B =, C = (b) A =, B =, C =, D = 0 (8) Solution for system AX = B : (; ; ); solution for system AX = B : ( ; 98; 9 ). (9) Solution for system AX = B : ( ; ; ; 9); solution for system AX = B : ( 0; ; 8; 0 ).

28 Andrilli/Hecker - Answers to Exercises Section. (0) (a) R : (III): hi! hi R : (I): hi hi R : (II): hi hi R : (II): hi hi + hi ; (b) AB = ; 0 R (R (R (R (AB)))) = R (R (R (R (A))))B = 0 9 () (a) A(X + X ) = AX + AX = O + O = O; A(cX ) = c(ax ) = co = O. (b) Any nonhomogeneous system with two equations and two unknowns that has a unique solution will serve as a counterexample. For instance, consider x + y = x y = : This system has a unique solution: (; 0). Let (s ; s ) and (t ; t ) both equal (; 0). Then the sum of solutions is not a solution in this case. Also, if c =, then the scalar multiple of a solution by c is not a solution. (c) A(X + X ) = AX + AX = B + O = B. (d) Let X be the unique solution to AX = B. Suppose AX = O has a nontrivial solution X. Then, by (c), X + X is a solution to AX = B, and X = X + X since X = O. This contradicts the uniqueness of X. a b () If a = 0, then pivoting in the rst column of 0 c d yields 0 " # b a 0. There is a nontrivial solution if and only if the 0 d c b 0 a (; ) entry of this matrix is zero, which occurs if and only if ad bc = 0. a b Similarly, if c = 0, swapping the rst and second rows of 0 c d 0 " # d c 0 and then pivoting in the rst column yields. There 0 b a d 0 c is a nontrivial solution if and only if the (; ) entry of this matrix is zero, which occurs if and only if ad bc = 0. Finally, if both a and c equal 0, then ad bc = 0 and (; 0) is a nontrivial solution. () (a) The contrapositive is: If AX = O has only the trivial solution, then A X = O has only the trivial solution. Let X be a solution to A X = O. Then O = A X = A(AX ). Thus AX is a solution to AX = O. Hence AX = O by the premise. Thus, X = O, using the premise again. :

29 Andrilli/Hecker - Answers to Exercises Section. (b) The contrapositive is: Let t be a positive integer. If AX = O has only the trivial solution, then A t X = O has only the trivial solution. Proceed by induction. The statement is clearly true when t =, completing the Base Step. Inductive Step: Assume that if AX = O has only the trivial solution, then A t X = O has only the trivial solution. We must prove that if AX = O has only the trivial solution, then A t+ X = O has only the trivial solution. Let X be a solution to A t+ X = O. Then A(A t X ) = O. Thus A t X = O, since it is a solution to AX = O. But then X = O by the inductive hypothesis. () (a) T (b) T (c) F (d) T (e) F (f) F Section. () (a) A row operation of type (I) converts A to B: hi hi. (b) A row operation of type (III) converts A to B: hi! hi. (c) A row operation of type (II) converts A to B: hi hi + hi. () (a) B = I. The sequence of row operations converting A to B is: (I): hi hi (II): hi hi + hi (II): hi hi + hi (III): hi! hi (II): hi hi + hi (b) The sequence of row operations converting B to A is: (II): hi hi + hi (III): hi! hi (II): hi hi + hi (II): hi hi + hi (I): hi hi () (a) The common reduced row echelon form is I. (b) The sequence of row operations is: (II): hi hi + hi (I): hi hi (II): hi 9 hi + hi (II): hi hi + hi 9 (II): hi hi + hi

30 Andrilli/Hecker - Answers to Exercises Section. (II): hi hi + hi (I): hi hi (II): hi (II): hi (I): hi hi + hi hi + hi hi () A reduces to ; while B reduces to () (a) (b) (c) (d) (e) (f) () Corollary. cannot be used in either part (a) or part (b) because there are more equations than variables. (a) Rank =. Theorem. predicts that there is only the trivial solution. Solution set = f(0; 0; 0)g (b) Rank =. Theorem. predicts that nontrivial solutions exist. Solution set = f(c; c; c) j c Rg () In the following answers, the asterisk represents any real entry: (a) Smallest rank = Largest rank = : (b) Smallest rank = Largest rank = (c) Smallest rank = Largest rank =

31 Andrilli/Hecker - Answers to Exercises Section. (d) Smallest rank = Largest rank = (8) (a) x = a + a (c) Not possible (b) x = a a + a (d) x = a a + a (e) The answer is not unique; one possible answer is x = a +a +0a. (f) Not possible (g) x = a a a (h) Not possible (9) (a) Yes: (row ) (row ) (row ) (b) Not in row space (c) Not in row space (d) Yes: (row ) + (row ) (e) Yes, but the linear combination of the rows is not unique; one possible expression for the given vector is (row ) + (row ) + 0(row ). (0) (a) [; ; 0] = q + q + q (b) q = r r r ; q = r + r r ; q = r r + r (c) [; ; 0] = r r + r () (a) (i) B = ; (ii) [; 0; ; ] = 8 [0; ; ; 8] + [; ; 9; 8] + 0[; ; ; ]; [0; ; ; ] = [0; ; ; 8] + 0[; ; 9; 8] + 0[; ; ; ]; (iii) [0; ; ; 8] = 0[; 0; ; ] + [0; ; ; ]; [; ; 9; 8] = [; 0; ; ] + [0; ; ; ]; [; ; ; ] = [; 0; ; ]+ [0; ; ; ] 0 (b) (i) B = ; (ii) [; ; ; 0; ] = [; ; ; ; ] [ ; ; ; ; ]+0[; ; 9; ; ]+0[; ; ; ; ]; [0; 0; 0; ; ] = [; ; ; ; ] [ ; ; ; ; ] + 0[; ; 9; ; ] + 0[; ; ; ; ]; (iii) [; ; ; ; ] = [; ; ; 0; ] [0; 0; 0; ; ]; [ ; ; ; ; ] = [; ; ; 0; ] + [0; 0; 0; ; ]; [; ; 9; ; ] = [; ; ; 0; ] + [0; 0; 0; ; ]; [; ; ; ; ] = [; ; ; 0; ] [0; 0; 0; ; ]

32 Andrilli/Hecker - Answers to Exercises Section. () Suppose that all main diagonal entries of A are nonzero. Then, for each i, perform the row operation hii (=a ii )hii on the matrix A. This will convert A into I n. We prove the converse by contrapositive. Suppose some diagonal entry a ii equals 0. Then the ith column of A has all zero entries. No step in the row reduction process will alter this column of zeroes, and so the unique reduced row echelon form for the matrix must contain at least one column of zeroes, and so cannot equal I n. () (a) Suppose we are performing row operations on an m n matrix A. Throughout this part, we will write hbi i for the ith row of a matrix B. For the Type (I) operation R : hii chii: Now R is hii c hii. Clearly, R and R change only the ith row of A. We want to show that R R leaves hai i unchanged. But hr (R(A))i i = c hr(a)i i = c (chai i) = hai i. For the Type (II) operation R : hii chji+hii: Now R is hii chji + hii. Again, R and R change only the ith row of A, and we need to show that R R leaves hai i unchanged. But hr (R(A))i i = chr(a)i j +hr(a)i i = chai j +hr(a)i i = chai j +chai j +hai i = hai i. For the Type (III) operation R : hii! hji: Now, R = R. Also, R changes only the ith and jth rows of A, and these get swapped. Obviously, a second application of R swaps them back to where they were, proving that R is indeed its own inverse. (b) An approach similar to that used for Type (II) operations in the abridged proof of Theorem. in the text works just as easily for Type (I) and Type (III) operations. However, here is a di erent approach: Suppose R is a row operation, and let X satisfy AX = B. Multiplying both sides of this matrix equation by the matrix R(I) yields R(I)AX = R(I)B, implying R(IA)X = R(IB), by Theorem.. Thus, R(A)X = R(B), showing that X is a solution to the new linear system obtained from AX = B after the row operation R is performed. () The zero vector is a solution to AX = O, but it is not a solution for AX = B. () Consider the systems x + y = x + y = 0 and x y = x y = : The reduced row echelon matrices for these inconsistent systems are, respectively, and : Thus, the original augmented matrices are not row equivalent, since their reduced row echelon forms are di erent. 8

33 Andrilli/Hecker - Answers to Exercises Section. () (a) Plug each of the points in turn into the equation for the conic. This will give a homogeneous system of equations in the variables a, b, c, d, e, and f. This system has a nontrivial solution, by Corollary.. (b) Yes. In this case, there will be even fewer equations, so Corollary. again applies. () The fact that the system is homogeneous was used in the proof of Theorem. to show that the system is consistent (because it has the trivial solution). However, nonhomogeneous systems can be inconsistent, and so the proof of Theorem. does not work for nonhomogeneous systems. (8) (a) R(A) and A are row equivalent, and hence have the same reduced row echelon form (by Theorem.), and the same rank. (b) The reduced row echelon form of A cannot have more nonzero rows than A. (c) If A has k rows of zeroes, then rank(a) = m least k rows of zeroes, so rank(ab) m k. k. But AB has at (d) Let A = R n ( (R (R (D))) ), where D is in reduced row echelon form. Then rank(ab) = rank(r n ( (R (R (D))) )B) = rank(r n ( (R (R (DB))) )) (by part () of Theorem.) = rank(db) (by repeated use of part (a)) rank(d) (by part (c)) = rank(a) (by de nition of rank). (9) (a) As in the abridged proof of Theorem.8 in the text, let a ; : : : ; a m represent the rows of A, and let b ; : : : ; b m represent the rows of B. For the Type (I) operation R : hii chii: Now b i = 0a + 0a + + ca i + 0a i a m, and, for k = i, b k = 0a + 0a + + a k +0a k+ + +0a m. Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (II) operation R : hii chji + hii: Now b i = 0a + 0a + + ca j + 0a j+ + + a i + 0a i+ + : : : + 0a m, where our notation assumes i > j. (An analogous argument works for i < j.) And, for k = i, b k = 0a +0a + +a k +0a k+ + +0a m. Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. For the Type (III) operation R : hii! hji: Now, b i = 0a + 0a + + a j + 0a j a m, b j = 0a + 0a + + a i + 0a i a m, and, for k = i, k = j, b k = 0a + 0a + + a k + 0a k a m. Hence, each row of B is a linear combination of the rows of A, implying it is in the row space of A. (b) This follows directly from part (a) and Lemma.. (0) Let k be the number of matrices between A and B when performing row operations to get from A to B. Use a proof by induction on k. Base Step: If k = 0, then there are no intermediary matrices, and Exercise 9

34 Andrilli/Hecker - Answers to Exercises Section. 9 shows that the row space of B is contained in the row space of A. Inductive Step: Given the chain A! D! D!! D k! D k+! B; we must show that the row space B is contained in the row space of A. The inductive hypothesis shows that the row space of D k+ is in the row space of A, since there are only k matrices between A and D k+ in the chain. Thus, each row of D k+ can be expressed as a linear combination of the rows of A. But by Exercise 9, each row of B can be expressed as a linear combination of the rows of D k+. Hence, by Lemma., each row of B can be expressed as a linear combination of the rows of A, and therefore is in the row space of A. By Lemma. again, the row space of B is contained in the row space of A. () (a) Let x ij represent the jth coordinate of x i. The corresponding homogeneous system in variables a ; : : : ; a n+ is 8 >< a x + a x + + a n+ x n+; = ; >: a x n + a x n + + a n+ x n+;n = 0 which has a nontrivial solution for a ; : : : ; a n+, by Corollary.. (b) Using part (a), we can suppose that a x + + a n+ x n+ = 0, with a i = 0 for some i. Then x i = a a a i x i a a i x i+ i a i x i+ a n+ a i x n+. Let b j = aj a i for j n +, j = i. () (a) T (b) T (c) F (d) F (e) F (f) T Section. () The product of each given pair of matrices equals I. () (a) Rank = ; nonsingular (b) Rank = ; singular (c) Rank = ; nonsingular (d) Rank = ; nonsingular (e) Rank = ; singular () No inverse exists for (b), (e) and (f). (a) " 0 0 # (c) " 8 8 # (d) " # 0

35 Andrilli/Hecker - Answers to Exercises Section. () No inverse exists for (b) and (e). (a) 0 (c) () (a) (b) 0 8 a 0 0 a a a a () (a) The general inverse is cos sin " p When =, the inverse is (d) (f) (c) " p When =, the inverse is When =, the inverse is (b) The general inverse is When =, the inverse is When =, the inverse is When =, the inverse is () (a) Inverse = " # sin. cos # p 0 0 p p p. cos sin 0 sin cos p a a a nn. #.. 0 p p p. p 0 p ; solution set = f(; )g.

36 Andrilli/Hecker - Answers to Exercises Section. (b) Inverse = (c) Inverse = (8) (a) Consider ; solution set = f( ; ; )g ; solution set = f(; 8; ; )g (b) Consider (c) A = A if A is involutory. (9) (a) A = I, B = I, A + B = O (b) A =, B = 0 0 0, A + B = 0 0 (c) If A = B = I, then A + B = I, A = B = I, A + B = I, and (A + B) = I, so A + B = (A + B). (0) (a) B must be the zero matrix. (b) No. AB = I n implies A exists (and equals B). Multiply both sides of AC = O n on the left by A. () : : : ; A 9 ; A ; A ; A ; A ; A ; : : : () B A is the inverse of A B. () (A ) T = (A T ) (by Theorem., part ()) = A. () (a) No step in the row reduction process will alter the column of zeroes, and so the unique reduced row echelon form for the matrix must contain a column of zeroes, and so cannot equal I n. (b) Such a matrix will contain a column of all zeroes. Then use part (a). (c) When pivoting in the ith column of the matrix during row reduction, the (i; i) entry will be nonzero, allowing a pivot in that position. Then, since all entries in that column below the main diagonal are already zero, none of the rows below the ith row are changed in that column. Hence, none of the entries below the ith row are changed when that row is used as the pivot row. Thus, none of the nonzero entries on the main diagonal are a ected by the row reduction steps on previous columns (and the matrix stays in upper triangular form throughout the process). Since this is true for each main diagonal position, the matrix row reduces to I n.

37 Andrilli/Hecker - Answers to Exercises Section. (d) If A is lower triangular with no zeroes on the main diagonal, then A T is upper triangular with no zeroes on the main diagonal. By part (c), A T is nonsingular. Hence A = (A T ) T is nonsingular by part () of Theorem.. (e) Note that when row reducing the ith column of A, all of the following occur:. The pivot a ii is nonzero, so we use a type (I) operation to change it to a. This changes only row i.. No nonzero targets appear below row i, so all type (II) row operations only change rows above row i.. Because of (), no entries are changed below the main diagonal, and no main diagonal entries are changed by type (II) operations. When row reducing [AjI n ], we use the exact same row operations we use to reduce A. Since I n is also upper triangular, fact () above shows that all the zeroes below the main diagonal of I n remain zero when the row operations are applied. Thus, the result of the row operations, namely A, is upper triangular. () (a) Part (): Since AA = I, we must have (A ) = A. Part (): For k > 0, to show (A k ) = (A ) k, we must show that A k (A ) k = I. Proceed by induction on k. Base Step: For k =, clearly AA = I. Inductive Step: Assume A k (A ) k = I. Prove A k+ (A ) k+ = I. Now, A k+ (A ) k+ = AA k (A ) k A = AIA = AA = I. This concludes the proof for k > 0. We now show A k (A ) k = I for k 0. For k = 0, clearly A 0 (A ) 0 = I I = I. The case k = is covered by part () of the theorem. For k, (A k ) = ((A ) k ) (by de nition) = ((A k ) ) (by the k > 0 case) = A k (by part ()). (b) To show (A A n ) = An A, we must prove that (A A n )(An A ) = I. Use induction on n. Base Step: For n =, clearly A A = I. Inductive Step: Assume that (A A n )(An A ) = I. Prove that (A A n+ )(An+ A ) = I. Now, (A A n+ )(An+ A ) = (A A n )A n+ A n+ (A n A ) = (A A n )I(A n A ) = (A A n )(A n A ) = I. () We must prove that (ca)( c A ) = I. But, (ca)( c A ) = c c AA = I = I. () (a) Let p = s, q = t. Then p; q > 0. Now, A s+t = A (p+q) = (A ) p+q = (A ) p (A ) q (by Theorem.) = A p A q = A s A t.

38 Andrilli/Hecker - Answers to Exercises Section. (b) Let q = t. Then (A s ) t = (A s ) q = ((A s ) ) q = ((A ) s ) q (by Theorem., part ()) = (A ) sq (by Theorem.) = A sq (by Theorem., part ()) = A s( q) = A st. Similarly, (A s ) t = ((A ) s ) q (as before) = ((A ) q ) s (by Theorem.) = (A q ) s = (A t ) s. (8) First assume AB = BA. Then (AB) = ABAB = A(BA)B = A(AB)B = A B. Conversely, if (AB) = A B, then ABAB = AABB =) A ABABB = A AABBB =) BA = AB. (9) If (AB) q = A q B q for all q, use q = and the proof in Exercise 8 to show BA = AB. Conversely, we need to show that BA = AB ) (AB) q = A q B q for all q. First, we prove that BA = AB ) AB q = B q A for all q. We use induction on q. Base Step (q = ): AB = A(BB) = (AB)B = (BA)B = B(AB) = B(BA) = (BB)A = B A. Inductive Step: AB q+ = A(B q B) = (AB q )B = (B q A)B (by the inductive hypothesis) = B q (AB) = B q (BA) = (B q B)A = B q+ A. Now we use this lemma (BA = AB ) AB q = B q A for all q ) to prove BA = AB ) (AB) q = A q B q for all q. Again, we proceed by induction on q. Base Step (q = ): (AB) = (AB)(AB) = A(BA)B = A(AB)B = A B. Inductive Step: (AB) q+ = (AB) q (AB) = (A q B q )(AB) (by the inductive hypothesis) = A q (B q A)B = A q (AB q )B (by the lemma) = (A q A)(B q B) = A q+ B q+. (0) Base Step (k = 0): I n = (A I n )(A I n ). Inductive Step: Assume I n +A+A + +A k = (A k+ I n )(A I n ) ; for some k: Prove I n +A+A + +A k +A k+ = (A k+ I n )(A I n ) : Now, I n +A+A + +A k +A k+ = (I n +A+A + +A k )+A k+ = (A k+ I n )(A I n ) + A k+ (A I n )(A I n ) (where the rst term is obtained from the inductive hypothesis) = ((A k+ I n ) + A k+ (A I n ))(A I n ) = (A k+ I n )(A I n ) : () (a) Suppose that n > k. Then, by Corollary., there is a nontrivial X such that BX = O. Hence, (AB)X = A(BX) = AO = O. But, (AB)X = I n X = X. Therefore, X = O, which gives a contradiction. (b) Suppose that k > n. Then, by Corollary., there is a nontrivial Y such that AY = O. Hence, (BA)Y = B(AY) = BO = O. But, (BA)Y = I k Y = X. Therefore, Y = O, which gives a contradiction. (c) Parts (a) and (b) combine to prove that n = k. Hence A and B are both square matrices. They are nonsingular with A = B by the de nition of a nonsingular matrix, since AB = I n = BA.