CHAPTER 2 Matrices. Section 2.1 Operations with Matrices Section 2.2 Properties of Matrix Operations... 36

Transcription

1 CHAPER Matrices Section. Operations with Matrices... Section. Properties of Matrix Operations... 6 Section. he Inverse of a Matrix... 4 Section.4 Elementary Matrices Section.5 Markov Chains... 5 Section.6 More Applications of Matrix Operations... 6 Review Exercises Project Solutions... 75

2 CHAPER Matrices Section. Operations with Matrices. x, y 4. x + x + 6 y 8 4 x y 9 x 8 y + x 4 y (a) A B ( ) (b) A B ( ) (c) ( ) ( ) ( ) ( ) ( ) ( ) 6 6 A (d) A B ( ) (e) B + A (a) (b) (c) (d) (e) A + B A B () () ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 4 A A B B + A

3 Section. Operations with Matrices. (a) A + B is not possible. A and B have different sizes. (b) A B is not possible. A and B have different sizes. 6 (c) A 4 (d) A B is not possible. A and B have different sizes. (e) B + A is not possible. A and B have different sizes.. (a) c a b ( ) ( ) (b) c a b () ( ) Simplifying the right side of the equation produces w x 4 + y + w. y x + z + x By setting corresponding entries equal to each other, you obtain four equations. w 4 + y y + w 4 x + w x w y + z y z x + x x he solution to this linear system is: x, 4, z and w. y, 6. (a) (b) ( ) ( )( ) ( ) ( ) () ( )( ) () ( ) AB ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) BA (a) (b). (a) () ( )( ) () () ( )() ( ) ( ) ( )() ( ) () ( )( ) () () ( )() ( ) ( ) ( )() ( ) () ( ) ( )() () () ( )( ) ( ) () ( )( ) AB ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) BA () ( ) () () ( ) () ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) AB (b) BA is not defined because B is and A is.. (a) AB [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (b) BA [ ] ( ) + ( ) + ( ) + ( ) [ 4] 4. (a) AB is not defined because A is and B is. (b) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) BA

4 Chapter Matrices 6. (a) AB ( ) + ( ) + ( ) ( ) + ( ) + ( ) () + ( ) + 4 ( ) () + ( ) + () ( 4) + ( )( ) + ( )( ) ( ) + ( )( ) + ( )( ) () + ( )( ) + ( )( 4) () + ( )( ) + ( )() ( 4) ( ) ( )( ) ( ) ( ) ( )( ) () ( ) ( )( 4) () ( ) ( )() (b) BA is not defined because B is 4and A is. 8. (a) AB is not defined because A is 5and B is. (b) 6 4 BA ( ) + 6( 6) ( ) + 6 ( ) ( ) + 6( 8) ( ) + 6( 7) 4 ( ) + 6( ) 4 () + 6 ( ) 4 ( ) + ( ) 4 () + 8 ( ) 4( ) + ( 7) 44 ( ) + ( ) C + E is not defined because C and E have different sizes.. 4A is defined and has size 4because A has size BE is defined. Because B has size 4and E has size 4, the size of BE is. 6. D + C is defined and has size 4 because D and C have size As a system of linear equations, A x x + x + x + x4 x x + x4. x x + x 4 is Use Gauss-Jordan elimination on the augmented matrix for this system. Choosing x4 t, the solution is x, t x t, x t, and x4 t, where t is any real number. 4. In matrix form A x b, the system is x 5. 4 x Use Gauss-Jordan elimination on the augmented matrix. 5 4 x So, the solution is. x 4. In matrix form A x b, the system is 4 9 x. x Use Gauss-Jordan elimination on the augmented matrix x So, the solution is 5. x

5 Section. Operations with Matrices 44. In matrix form A x b, the system is x x. x Use Gauss-Jordan elimination on the augmented matrix. x So, the solution is x. x 46. In matrix form A x b, the system is 4x 7 x. 6 5x 4 Use Gauss-Jordan elimination on the augmented matrix x 4 So, the solution is x 5. x 48. In matrix form A x b, the system is x x x. x4 x 5 5 Use Gauss-Jordan elimination on the augmented matrix. 5 So, the solution is x x x. x4 x5 5. he augmented matrix row reduces as follows. 4 here are an infinite number of solutions. For example, x, x, x. So, b he augmented matrix row reduces as follows So, 5 b ( ) Expanding the left side of the equation produces a a A a a a a a a a a a a and you obtain the system a a a a a a a a. Solving by Gauss-Jordan elimination yields a, a, a, and a. So, you have A.

6 4 Chapter Matrices 56. Expanding the left side of the matrix equation produces a b a + b a + b 7. c d c + d c + d 4 You obtain two systems of linear equations (one involving a and b and the other involving c and d). a + b a + b 7, and c + d 4 c + d. Solving by Gauss-Jordan elimination yields a 48, b, c 7, and d AA 9 6. AB ( ) ( 5) Similarly, BA. 6. (a) (b) a b b b a b a b a b a b b b ab ab ab AB a b b b a b a b a b he ith row of B has been multiplied by aii, the ith diagonal entry of A. b b ba ab ab ab BA b b b a ab ab ab b b b a ab ab ab he ith column of B has been multiplied by aii, the ith diagonal entry of A. (c) If a a a, then AB a B BA. 64. he trace is the sum of the elements on the main diagonal he trace is the sum of the elements on the main diagonal ( ) 68. Let AB c ij, where c n n n n a b. hen, r( AB) cii aikbki. i i k ij ik kj k Similarly, if BA d ij, dij bik akj. k n n n n hen r( BA) dii bikaki r( AB). i i k cos α sin αcos β sin β cos α cos β sin α sinβ cos α sin β sin α cos β 7. AB sin α cos αsin β cos βsin α cos β + cos α sin β sin α( sin β ) + cos α cos β cos β sin βcos α sin αcos β cos α sin β sin α cos β ( sin α) sin β cos α BA sin β cos βsin α cos αsin β cos α + cos β sin α sin β( sin α) + cos β cos α So, you see that AB ( α + β ) sin( α + β ) ( α + β ) cos( α + β ) cos BA. sin ( )

7 Section. Operations with Matrices 5 a a 7. Let A and B a a b b. b b hen the matrix equation AB BA is equivalent to a ab b b ba a. a a b b b b a a his equation implies that ab + ab ba ba ab b a a b + a b b a b a a b b a which is impossible. So, the original equation has no solution. 74. Assume that A is an m nmatrix and B is a p q matrix. Because the product AB is defined, you know that n p. Moreover, because AB is square, you know that m q. herefore, B must be of order n m, which implies that the product BA is defined. 76. Let rows s and t be identical in the matrix A. So, asj aijfor j,, n. Let AB c ij, where c n a b. hen, c ij ik kj k are the same. n a b, and c sj sk kj k n a b. Because ask atk for k,, n, rows s and t of AB tj tk kj k 78. (a) No, the matrices have different sizes. (b) No, the matrices have different sizes. (c) Yes; No, BA is undefined (a) Multiply the matrix for by. his produces a matrix giving the information as percents of the total population. 9,6 5, ,95 4, A 7,799 7,75 4, , ,, Multiply the matrix for by. his produces a matrix giving the information as percents of the total population. 6,6 5, ,77 4, B 7,954 7,7 6, , ,4, (b) B A (c) he 65+ age group is projected to show relative growth from to over all regions because its column in B A contains all positive percents.

8 6 Chapter Matrices 84. AB (a) rue. he number of elements in a row of the first matrix must be equal to the number of elements in a column of the second matrix. See page 4 of the text. (b) rue. See page 45 of the text. Section. Properties of Matrix Operations. ( ) ( ) ( ) ( ) ( ) ([ 5 4 ] [ ] ) ( 8) + 9 [ ] ( ) ( 9) 4 + ( ) ( ) 8. A + B ( a b) B ( ( 4) ) ( ). ( ab) O ( )( 4) ( ) 4. (a) X A B (b) X A B 4 X X 7 5 X 5 7

9 Section. Properties of Matrix Operations 7 (c) X + A B X X 6 6 X (d) A + 4B X X 4 X X 6. ccb ( ) ( ) 8. CBC ( ). BC ( O) ( ) AB C 4. (a) ( ) (b) ABC ( ) BcA ( ) ( ) AB BA AB 8. AC BC 4 But A B. 4. AB 4 But A O and B O. O

10 8 Chapter Matrices. 4. A A + IA A I 4 A A I I. So, ( ) 8. In general, AB BAfor matrices. 4. D ( AB) B A 44. ( AB) B A (a) (b) 48. (a) (b) A A 4 4 AA A A AA () 7 ( ) 7 A 7 () 7 ( ) 7 ()

11 Section. Properties of Matrix Operations 9 5. () ( ) A () ( ) () 8, you have 7 () 54. Because A ( ) A. 56. (a) False. In general, for n nmatrices A and B it is not true that AB BA. For example, let A hen AB BA. (b) False. Let A, B, C. hen AB (c) rue. See heorem.6, part on page 57. AC, but B C., B. 58. ax + A( bb) b( AB + IB) ax + ( Ab) B b( AB + B) ax + bab bab + bb ( ) ( ) ax bab + bb + ( bab) ax bab + ( bab) + bb Original equation Associative property; property of the identity matrix Property of scalar multiplication; distributive property ax + bab + bab bab + bb + bab Add bab to both sides. ax bb Additive inverse Commutative property Additive inverse b X B Divide by a. a 6. f( A)

12 4 Chapter Matrices 6. ( cd) A ( cd) a ( cd) a c( da ) cda c( da) ij ij ij ij 64. ( ) ( ) ( ) c + d A c + d a c + d a ca + da ca + da ca + da ca + da 66. (a) o show that A( BC) ( AB) C, ij ij ij ij ij ij ij ij compare the ijth entries in the matrices on both sides of this equality. Assume that A has size n p, B has size p r, and C has size r m. hen the entry in the kth row and the jth column of BC is r bc. herefore, the entry in ith row and jth column of A(BC) is l kl lj p r aik bklclj aikbklclj k l k, l he entry in the ith row and jth column of (AB)C is column. p So, d a b for each l,, r. il k ik kl r p aikbklclj aikbklcij i k k, l.. r l dc, il lj So, the ijth entry of ( ) where dil is the entry of AB in the ith row and the lth AB Cis Because all corresponding entries of A(BC) and (AB)C are equal and both matrices are of the same size ( n m), you conclude that A( BC) ( AB) C. (b) he entry in the ith row and jth column of ( A + BC ) is ( + ) + ( + ) + + ( + ) ail bil cj ai bi c j ain bin cnj, whereas the entry in the ith row and jth column of AC + BC is ( ac ij + + ac in nj) + ( bc i j + + bc in nj), which are equal by the distributive law for real numbers. (c) he entry in the ith row and jth column of ( ) cabis cab ij + ab i j + + ab in nj. ( cai) bj + ( cai) bj + + ( cain) bnj Because these three expressions are equal, you have shown that cab ( ) ( cab ) AcB ( ). 68. () ( ) ( ) he corresponding entry for ( ) and the corresponding entry for A(cB) is ai( cbj) + ai( cb j) + + ain( cbnj). ij ij ij ij ji ji ji ji ij ij ji ji A + B a + b a + b a + b a + b A + B () ( ca) ( c a ) ca ca c a c( A ) (4) he entry in the ith row and jth column of ( ) ca B is AB is ajbi + ajb i + ajnbni. On the other hand, the entry in the ith row and jth column of B A is ba + b a + + ba, which is the same. i j i j ni jn 7. (a) Answers will vary. Sample answer: (b) Let A and B be symmetric. If AB BA, then ( ) AB B A BA AB and AB is symmetric. If ( AB) AB, then ( ) AB AB B A BAand AB BA. 7. Because A A, the matrix is symmetric. 74. Because A A, the matrix is skew-symmetric. 76. If A A and B B, skew-symmetric. then ( A B) A B A B ( A B), which implies that A + B is

13 Section. he Inverse of a Matrix Let A a a a an a a a an. an an an ann A A a a a an a a a an a a a an a a a an a a a a a a a a n n n nn n n n nn a a a a an an a a a a an an an an an an an an a a a a an an ( a a) a a an an ( an an) ( an an) ( an an) So, A A is skew-symmetric. Section. he Inverse of a Matrix AB BA AB BA AB BA Use the formula A d b, ad bc c a where a b A. c d So, the inverse is A. Use the formula A 4 4. ( ) ( )( ) 4 4 d b, ad bc c a where a b A. c d So, the inverse is A. ()( ) ( )( )

14 4 Chapter Matrices. Using the formula d b A, ad bc c a where a b A c d you see that ad bc ( )( ) ( )( ). So, the matrix has no inverse. 4. Adjoin the identity matrix to form [ A I] Using elementary row operations, reduce the matrix as follows I A 6. Adjoin the identity matrix to form [ A I] Using elementary row operations, reduce the matrix as follows. 4 7 I A herefore, the inverse is A 8. Adjoin the identity matrix to form [ A I] Using elementary row operations, you cannot form the identity matrix on the left side. herefore, the matrix has no inverse.. Adjoin the identity matrix to form [ A I] Using elementary row operations, you cannot form the identity matrix on the left side. herefore, the matrix has no inverse.. Adjoin the identity matrix to form [ A I] Using elementary row operations, reduce the matrix as follows I A herefore, the inverse is A Adjoin the identity matrix to form [ A I] 5 5 Using elementary row operations, you cannot form the identity matrix on the left side. herefore, the matrix has no inverse. 6. Adjoin the identity matrix to form [ A I]. Using elementary row operations, reduce the matrix as follows. I A herefore, the inverse is A.

15 Section. he Inverse of a Matrix 4 8. Adjoin the identity matrix to form [ A I] Using elementary row operations, reduce the matrix as follows. [ A I] herefore the inverse is A Adjoin the identity matrix to form [ A I] Using elementary row operations, reduce the matrix as follows I A..5.. herefore, the inverse is A A ()( ) ( )( ) ad bc 4 A A 5 ( )( ) ( ) ad bc A A ad ( )( ) ( )( ) bc A A ( A ) A ( A ) he results are equal. 4. A ( A ) A ( A ) he results are equal.

16 44 Chapter Matrices 4. (a) ( ) AB B A (b) ( A ) ( A ) A (c) ( ) 44. (a) ( ) A AB B A (b) ( A ) ( A ) (c) ( ) A A he coefficient matrix for each system is A and the formula for the inverse of a matrix produces 4 4 A. + (a) (b) 4 4 A x b 7 5 he solution is: x and y x A b he solution is: x and y. 48. he coefficient matrix for each system is A. Using the algorithm to invert a matrix, you find that the inverse is A. (a) A x b he solution is: x, x, and x. (b) A x b he solution is: x, x, and x. 5. Using a graphing utility or software program, you have Ax b x where A b x x A, x x, and b 4 x4 x he solution is: x, x, x, x4, and x 5.

17 Section. he Inverse of a Matrix Using a graphing utility or software program, you have Ax x b A b where x x x A, x, and x4 5 5 x5 4 6 x 6 b. 9 he solution is: x, x, x, x 4, x 5, and x he inverse of A is given by A x. x 4 Letting A A, you find that So, x. x 4 x 56. he matrix will be singular if 4 ( )( ) ( )( ). ad bc x 4, which implies that 4x 6or x. 58. First, find 4 A. ( A) A 4 + hen, multiply by 4 to obtain A 4( 4 A) Using the formula for the inverse of a matrix, you have d b A ad bc c a sec θ tan θ sec θ tan θtan θ sec θ sec θ tan θ. tan θ sec θ 6. Adjoin the identity matrix to form [ F I] Using elementary row operations, reduce the matrix as follows I F So, F and F w d ( ) ( ) ( ) ( ) So, ( ) ( ) A A A A I I and n n A A AA I I n n A A. I A I A I IA AI + 4A I 4A + 4A I 4A + 4A because A A I 66. ( )( ) I A I A. So, ( ) ( ) 68. Because ABC I, A is invertible and A BC. So, ABC A A and BC A I. So, B CA. 7. Let A A and suppose A is nonsingular. hen, A exists, and you have the following. ( ) ( ) A A A A A A A I A I

18 46 Chapter Matrices 7. (a) rue. See heorem.8, part on page 67. (b) False. For example, consider the matrix, which is not invertible, but. (c) False. If A is a square matrix then the system A x bhas a unique solution if and only if A is a nonsingular matrix. 74. A has an inverse if aii for all i nand A a a. a nn 76. A (a) A A + 5I ( ) ( ) 5 5 5( 5 ) ( I A ) A I (b) ( ) A I A A A I I Similarly, ( ) Or, ( ) I A A (c) he calculation in part (b) did not depend on the entries of A. directly. 78. Let C be the inverse of ( I AB), that is C ( I AB). hen CI ( AB) ( I ABC ) I. Consider the matrix I + BCA. Claim that this matrix is the inverse of I BA. o check this claim, show that ( I + BCA)( I BA) ( I BA)( I + BCA) I. First, show ( )( + ) + + ( ) + (( ) ) I BA I BCA I BA BCA BABCA I BA B C ABC A I BA B I AB C A Similarly, show ( I + BCA)( I BA) I. 8. Answers will vary. Sample answer: A or A BA + BA I 8. AA A A a b d b a b d b ad bc c dad bc c a ad bc c dc a ad bc ad bc d ba b ad bc ad bc c ac d ad bc ad bc Section.4 Elementary Matrices. his matrix is not elementary, because it is not square. 4. his matrix is elementary. It can be obtained by interchanging the two rows of I. 6. his matrix is elementary. It can be obtained by multiplying the first row of I by, and adding the result to the third row. 8. his matrix is not elementary, because two elementary row operations are required to obtain it from I 4.. C is obtained by adding the third row of A to the first row. So, E.

19 Section.4 Elementary Matrices 47. A is obtained by adding times the third row of C to the first row. So, E. 4. Answers will vary. Sample answer: Matrix Elementary Row Operation Elementary Matrix R R 6 ( ) R R ( ) R R 6 So,. 6. Answers will vary. Sample answer: Matrix Elementary Row Operation Elementary Matrix ( ) R + R R 4 4 R + R R ( ) 4 7 ( ) R R ( ) R + R R ( ) 7 R R 7 So,

20 48 Chapter Matrices 8. Matrix Elementary Row Operations Elementary Matrix R + ( ) R R R + R R 4 ( 4) ( ) R R R + ( 7) R R R4 ( ) R R + 6 ( 6) R R R4 ( 7) R R ( 8) R 4 R4 8 So,

21 Section.4 Elementary Matrices 49. o obtain the inverse matrix, reverse the elementary row operation that produced it. So, multiply the first row by to obtain 5 E 5.. o obtain the inverse matrix, reverse the elementary row operation that produced it. So, add times the second row to the third row to obtain. E 4. o obtain the inverse matrix, reverse the elementary row operation that produced it. So, add k times the third row to the second row to obtain E k. 6. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form. ( ) R R R R R E E Use the elementary matrices to find the inverse. A EE 8. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form. ( ) R R R + R R ( ) R R R E E E Use the elementary matrices to find the inverse. A EEE

22 5 Chapter Matrices For Exercises 6, answers will vary. Sample answers are shown below.. he matrix A is itself an elementary matrix, so the factorization is A.. Reduce the matrix A as follows. Matrix Elementary Row Operation Elementary Matrix Add times row one to row two. E Multiply row two by. E Add times row two to row one. E So, one way to factor A is A E E E. 4. Reduce the matrix A 5 6 as follows. 4 Matrix Elementary Row Operation Elementary Matrix Add times row one to row two. E 4 Add times row one to row three. E Add times row two to row three. E Add times row three to row one. E4 Add times row two to row one. E5 So, one way to factor A is A E E E E E 4 5.

23 Section.4 Elementary Matrices 5 6. Find a sequence of elementary row operations that can be used to rewrite A in reduced row-echelon form. ( ) R R 4 5 R4 R R4 ( 5) R4 R4 R R R ( ) R4 R R R4 R R + R R So, one way to factor A is 4 E E E E E E E A E E E E4 E5 E6 E (a) EA has the same rows as A except the two rows that are interchanged in E will be interchanged in EA. (b) Multiplying a matrix on the left by E interchanges the same two rows that are interchanged from In in E. So, multiplying E by itself interchanges the rows twice and E I n.

24 5 Chapter Matrices 4. a a a +. c c c A b b b ab 4. (a) False. It is impossible to obtain the zero matrix by applying any elementary row operation to the identity matrix. (b) rue. If A EE E k, where each Ei is an elementary matrix, then A is invertible (because every elementary matrix is) and A E E E k. (c) rue. See equivalent conditions () and () of heorem Matrix Elementary Matrix A 6 4 U E EA U A E U LU 46. Matrix Elementary Matrix A 7 U EEA U A E E U E E LU 48. Matrix Elementary Matrix A 6 6 U E E E A U A E E E U E E E LU y 4 y 4 Ly b : y 5 y4 y 4, y + y 4 y, y + y + y 5 y, and y4. x 4 x Ux y : x x4 x 4, x, x x x, and x. So, the solution to the system A x bis: x, x x x4.

25 Section.5 Markov Chains 5 5. A A. Because A A, Ais not idempotent. 5. A. Because A A, Ais not idempotent. 54. Assume A is idempotent. hen A ( A ) ( A A ) A A A which means that Now assume A A ( A A ) ( A ) A is idempotent. A is idempotent. hen A AA A which means that A is idempotent. 56. ( AB) ( AB)( AB) A( BA) B A( AB) B ( AA)( BB) AB So, ( AB) AB, and AB is idempotent. 58. If A is row-equivalent to B, then A E E E B k, where E,, Ek are elementary matrices. So, B E E Ek A, which shows that B is row equivalent to A. 6. (a) When an elementary row operation is performed on a matrix A, perform the same operation on I to obtain the matrix E. (b) Keep track of the row operations used to reduce A to an upper triangular matrix U. If A row reduces to U using only the row operation of adding a multiple of one row to another row below it, then the inverse of the product of the elementary matrices is the matrix L, and A LU. (c) For the system A x b, find an LU factorization of A. hen solve the system L y b for y and U x y for x. Section.5 Markov Chains. he matrix is not stochastic because every entry of a stochastic matrix satisfies the inequality a ij. 4. he matrix is not stochastic because the sum of entries in a column of a stochastic matrix is. 6. he matrix is stochastic because each entry is between and, and each column adds up to.

26 54 Chapter Matrices 8. 6% Gas (G) 4% Liquid (L) 7% 5% Solid (S) 5% % he matrix of transition probabilities is shown. From G L S.6 G P.4.7 L o..5 S he initial state matrix represents the amounts of the physical states is shown..(,) X.6(,) 6.(,) o represent the amount of each physical state after the catalyst is added, multiply P by X to obtain.6 PX So, after the catalyst is added there are molecules in a gas state, 6 molecules in a liquid state, and 8 molecules in a solid state.. X X X PX PX PX Form the matrix representing the given transition probabilities. A represents infected mice and B noninfected. From A B.. A P o.8.9 B he state matrix representing the current population is. A X..7 B (a) he state matrix for next week is.... X PX So, next week.( ) mice will be infected..... (b) X PX X PX In weeks,.( ) mice will be infected.

27 Section.5 Markov Chains Form the matrix representing the given transition probabilities. Let S represent those who swim and B represent those who play basketball. From S B..4 S P o.7.6 B he state matrix representing the students is.4 S X..6 B (a) he state matrix for tomorrow is X PX So, tomorrow.6( 5) 9 students will swim and.64( 5) 6 students will play basketball. (b) he state matrix for two days from now is X P X So, two days from now.64( 5) 9students will swim and.66( 5) 59 students will play basketball. (c) he state matrix for four days from now is X P X So, four days from now,.664( 5) 9students will swim and.666( 5) 59 students will play basketball. 6. Form the matrix representing the given transition probabilities. Let A represent users of Brand A, B users of Brand B, and N users of neither brands. From A B N A P B o N he state matrix representing the current product usage is X A B 5 N (a) he state matrix for next month is X P X So, next month the distribution of users will be.7, 4,5 for Brand A,.9, 4, for Brand B, and.7, 4,5 for neither. (b) (c) X.5 P X..5 In months, the distribution of users will be.5, 7,65 for Brand A,., 6,65 for Brand B, and.5, 5,75 for neither. X.9 P X In 8 months, the distribution of users will be.9, 4,5 for Brand A,.8, 4,88 for Brand B, and.5,,66 for neither.

28 56 Chapter Matrices 8. he stochastic matrix. P.7 is regular because P has only positive entries. PX X.x x.7x x.x x x +.7x x Because x + x, the system of linear equations is as follows. x +.x x.x x + x he solution to the system is x and x. So, X.. he stochastic matrix. P.8 is not regular because every power of P has a zero in the second column.. x x PX X.8 x x.x x.8x + x x Because x + x, the system of linear equations is as follows..8x.8x x + x he solution of the system is x and x. So, X.. he stochastic matrix 7 5 P 5 is regular because P has only positive entries. PX X 7 5 x x x x 5 7 x 5 + x x x + x 5 x Because x + x, the system of linear equations is as follows. 7 x + x 5 7 x 5 x x + x 6 he solution of the system is x and 6 7 x. So, 7 X he stochastic matrix 9 4 P is regular because P has only positive entries. PX X x x 9 4 x x 4 x x 9 4 x 9 + x 4 + x x x + x + x x 4 x 9 + x 4 + x x Because x + x + x, the system of linear equations is as follows. 7 x x x 9 4 x x x 4x 9 x 4 x x x x he solution of the system is x., x.4, and x So,.7 X.4..

29 Section.5 Markov Chains he stochastic matrix 5 P is regular because P has only positive entries. PX X x x 5 x x 5 x x 6 5 x + x 5 + x x x + x 5 x x 6 + x 5 x Because x + x + x, the system of linear equations is as follows. x + x 5 + x x 4x 5 x 6 + x 5 x x + x + x he solution of the system is x, x, and ( ) x. So, X he stochastic matrix.. P is not regular because every power of P has two zeros in the second column. PX X..x x.7. x x..4 x x.x +.x x.7x + x +.x x.x +.4x x Because x + x + x, the system of linear equations is as follows..9x +.x.7x.x +.x.6x x + x + x he solution of the system is x, x, and x. So, X.. he stochastic matrix P is not regular because every power of P has three zeros in the first column. PX X x x x x x x x4 x4 x x x x x x x x 4 4 Because x + x + x + x4, the system of linear equations is as follows. x + x x x x + x + x + x 4 Let x s and x4 t. he solution of the system is x4 t, x s, x s, and x s t, where s, t, and s + t.

30 58 Chapter Matrices. Exercise : o find X, let X x. hen use the x x matrix equation PX X to obtain..6.5x x..6.5x x..6.5 x x or.x +.6x +.5x x.x +.6x +.5x x.x +.6x +.5x x Use these equations and the fact that x + x + x to write the system of linear equations shown..6x +.6x +.5x.x.x +.5x.x +.6x +.5x x + x + x he solution of the system is 6 x x,, and x So, the steady state matrix is 6 X. 4 Exercise 5: o find X, let 4. matrix equation PX X to obtain x x x x x x x4 x 4 or x x x x x x. x x 4 4 x x X. hen use the x x 4 Use these equations and the fact that x + x + x + x4 to write the system of linear equations shown. x + x + x + x4 Let x r, x s, and x4 t, where r, s, and t are real numbers between and. he solution of the system is x r s t, x r, x s, and x4 t, where r, s, and t are real numbers such that r, s, t, and r + s + t. So, the steady state matrix is r s t r X. s t x x Exercise 6: o find X, let X. hen use the x x 4 matrix equation PX 4 x x x x 4x x x x4 or X to obtain x + x + x + 4 x x x + x + x x 5 4 x x x 9 + x 4 + x 5 4 x x 6 + x 9 + x 4 + x 5 4 x4 Use these equations and the fact that x + x + x + x4 to write the system of equations shown. 4 x x x x x x x x x x 6 9 x 4 x 5 4 x 4 6 x 9 x 4 x 5 4 x x x x he solution of the system is x, x, x, and x 4 So, the steady state matrix is X

31 Section.5 Markov Chains Form the matrix representing the given transition probabilities. Let A represent those who received an A and let N represent those who did not. From A N.7. A P o..9 N o find the steady state matrix, solve the equation PX X, where X to write a system of equations..7x +.x x.x +.x.x +.9x x.x.x x + x x + x x and use the fact that x + x x he solution of the system is x and 4 x 4. So, the steady state matrix is X 4. his indicates that eventually 4 4 of the students will receive assignment grades of A and of the students will not Form the matrix representing transition probabilities. Let A represent heatre A, let B represent heatre B, and let N represent neither theatre. From A B N P..6. A B o B x o find the steady state matrix, solve the equation PX X where X + + x and use the fact that x x x x to write a system of equations..x +.6x +.x x.9x +.6x +.x.5x +.8x +.4x x.5x.9x +.4x.85x +.86x +.97x x.85x +.86x.x x + x + x x + x + x he solution of the system is x, x 9, and x So, the steady state matrix is X. his indicates 9 9 that eventually.4% of the people will attend heatre A, 4.% of the people will attend heatre B, and % of the people will attend neither theatre on any given night. 8. he matrix is not absorbing; he first state S is absorbing, however the corresponding Markov chain is not absorbing because there is no way to move from S or S to S. 4. he matrix is absorbing; he fourth state 4 S is absorbing and it is possible to move from any of the states to 4 S in one transition.

32 6 Chapter Matrices 4. Use the matrix equation PX X, or. x x. x x.7 x x along with the equation x + x + x to write the linear system.9x.x..7x x + x + x he solution of this system is x, x t, and x t, where t is a real number such that t. So, the steady state matrix is X t, where t t. 44. Use the matrix equation PX X or.7..x x..5.6x x..x x... x4 x 4 along with the equation x + x + x + x4 to write the linear system.x +.x +.x4.x +.5x +.6x4.8x +.x4..x +.x.9x4 x + x + x + x 4 he solution of this system is x, x, x, and x 4. So, the steady state matrix is X. 46. Let S n be the state that Player has n chips. From S S S S S4.7 S.7 S P..7 S o and X. S. S 4 So, PX n P X So, the probability that Player reaches S4 and wins the tournament is (a) o find the nth state matrix of a Markov chain, n compute X n P X, where X is the initial state matrix. (b) o find the steady state matrix of a Markov chain, n determine the limit of P X, as n where, X is the initial state matrix. (c) he regular Markov chain is PX, P X, P X,, where P is a regular stochastic matrix and X is the initial state matrix. (d) An absorbing Markov chain is a Markov chain with at least one absorbing state and it is possible for a member of the population to move from any nonabsorbing state to an absorbing state in a finite number of transitions. (e) An absorbing Markov chain is concerned with having an entry of and the rest in a column, whereas a regular Markov chain is concerned with the repeated multiplication of the regular stochastic matrix.

33 Section.5 Markov Chains 6 5. (a) When the chain reaches S or S4, it is certain in the next step to transition to an adjacent state, S or S, respectively, so S and S4 reflect to S or S. (b).4. P.6.7 (c) (d) P P Other high even or odd powers of P give similar results where the columns alternate. X Half the sum entries in the corresponding columns of n n P and P + approach the corresponding entries in X. 5. (a) Yes, it is possible. (b) Yes, it is possible. Both matrices X satisfy PX X. he steady state matrix depends on the initial state matrix. In general, the steady state matrix is X 6 t 5 5 t 6, 5 t 6 t 54. Let 6 where t is any real number such that t. In part (a) t and in part (b), t 6. a b P a b be a stochastic matrix, and consider the system of equations PX X. a b x x a bx x You have ax + bx x ( ax ) + ( bx ) x or ( a ) x + bx ( ax ) bx. Letting x b and x a, you have the state matrix X satisfying PX X X b. a 56. Let P be a regular stochastic matrix and X be the initial state matrix. n n lim P X lim P x + x + + x n ( ) k n n n n lim lim lim n n n P x + P x + + P x Px + Px + + Pxk Px ( + x + + xk ) PX X, where X is a unique steady state matrix. k

34 6 Chapter Matrices Section.6 More Applications of Matrix Operations. Divide the message into groups of four and form the uncoded matrices. H E L P _ I S _ C O M I N G [ 8 5 6] [ 9 9 ] [ 5 9] [ 4 7 ] Multiplying each uncoded row matrix on the right by A yields the coded row matrices [ 8 5 6] A [ ] [ ] [ 9 9 ] A [ ] [ 5 9] A [ 7 7 ] [ 4 7 ] A [ ]. So, the coded message is 5,,, 4, 8,, 8, 47, 7,, 7,, 5, 49, 7, Find A, and multiply each coded row matrix on the right by A to find the associated uncoded row matrix. 4 [ 85 ] [ 5], O [ 6 8] A [ ] _, B [ 5] A [ 5 ] E, _ [ 84 7] A [ 5 8] O, R [ 4 56] A [ 4 ] _, N [ 9 5] A [ 5 ] O, [ 6 8] A [ ] _, [ 45] A [ 5 ] O, _ [ 9 6] A [ 5] B, E So, the message is O_BE_OR_NO_O_BE. 6. Find A 8 4, 8 6 and multiply each coded row matrix on the right by A to find the associated uncoded row matrix H, A, V [ 9 5 ] A [ 5 ] E, _, A [ ] A [ 7 8 ] _, G, R [ ] A [ 5 ] E, A, [ 8] A [ 5 ] _, W, E [ 5 6] A [ 5 5] E, K, E 4 48 A 4 4 N, D, _ [ ] A [ ] [ ] [ ] [ ] he message is HAVE_A_GREA_WEEKEND_.

35 Section.6 More Applications of Matrix Operations 6 a b 8. Let A and find that c d _ S [ 9 a b 9] c d [ 9] U E [ 7 a b 6] [ c d 5 ]. his produces a system of 4 equations. 9a 9c 9b 9d 9 7a + 6c 7b + 6d 5. Solving this system, you find a, b, c, and d. So, A. Multiply each coded row matrix on the right by A to yield the uncoded row matrices. [ ], [ 4 ], [ 5 ], [ 5 ], [ 8 4 ], [ 5 8 ], [ 9 ], [ 9 ], [ 5 ]. his corresponds to the message CANCEL_ORDERS_SUE.. You have [ 45 w x 5] y z [ 5] and [ 8 w x ] y z [ 8 4 ]. So, 45w 5y and 45x 5z 5 8w y 8 8x z 4. Solving these two systems gives w y, x, and z. So, A. (b) Decoding, you have: [ 45 5] A [ 5] J, O [ 8 ] A [ 8 4] H, N [ 8 8] A [ 8 ] _, R [ 5 ] A [ 5 ] E, [ 8 6] A [ 8] U, R [ 4 8] A [ 4 ] N, _ [ 75 55] A [ 5], O [ ] A [ ] _, B [ ] A [ 9] A, S 5 A 5 E, _ [ ] [ ] he message is JOHN_REURN_O_BASE_.. Use the given information to find D. User A B.. A D Supplier.4.4 B he equation X DX + E may be rewritten in the form ( I D) X E, that is.7., X..4.6, Solve this system by using Gauss-Jordan elimination to obtain 9,4 x. 5,94

36 64 Chapter Matrices 4. From the given matrix D, form the linear system X DX E, I D X E that is + which can be written as ( ), X Solving this system,,875 X 7,. 4,5 6. (a) he line that best fits the given points is shown in the graph. y (, ) (b) Using the matrices X and Y, you have 4 (, ) X X (, ) (, ) 4, X Y 4, and 6 ( ) 4 4 A X X X Y. 6 So, the least squares regression line is y x +. (c) Solving Y XA + E for E, you have.. E Y XA... So, the sum of the squares error is E E.. x 8. (a) he line that best fits the given points is shown in the graph. 4 y (b) Using the matrices X and Y, you have (, ) (, ) X X, X Y, and A 4 ( X X) ( X Y). So, the least squares regression line is y x (c) Solving Y XA + E for E, you have E Y XA and the sum of the squares error is E E.5.. Using the matrices (5, ) (4, ) (6, ) (4, ) x 5 6 (, ) (, ) X and Y, 5 6 you have 9 X X, X Y, and A ( X X) ( X Y). So, the least squares regression line is y x.

37 Section.6 More Applications of Matrix Operations 65. Using matrices 4 X and Y, you have X X 4 8, X Y, and A ( X X) ( X Y)..8 4 So, the least squares regression line is y.8x Using matrices 4 X and Y, you have X X 4, X Y 7, and 7 7 A 4 4 ( X X) ( X Y). So, the least squares regression line is y.65x Using matrices 6 4 X 5 and Y, you have X X 5, X Y, and A ( X X) ( X Y) So, the least squares regression line is y 75 x Using matrices X and Y.7,.4.6 you have X X and X Y ( ).48 A X X X Y.8 So, the least squares regression line is y.8x.48.. (a) o encode a message, convert the message to numbers and partition it into uncoded row matrices of size n. hen multiply on the right by an invertible n n matrix A to obtain coded row matrices. o decode a message, multiply the coded row matrices on the right by A and convert the numbers back to letters. (b) A Leontief input-output model uses an n n matrix to represent the input needs of an economic system, and an n matrix to represent any external demands on the system. (c) he coefficients of the least squares regression line are given by ( ) A X X X Y.

38 66 Chapter Matrices Review Exercises for Chapter ( ) + ( ) ( ) + ( ) ( ) + ( ) ( ) ( ) ( ) ( ) ( ) ( ) x 8. Letting A, x, and b x system can be written as Ax b x 5. x 4 x x. 4 x 5, the 4 Using Gaussian elimination, the solution of the system is 6 7 x. 7 x. Letting A, x x, and b, 4 x the system can be written as Ax b. Using Gaussian elimination, the solution of the system is 5 x. 6 A A A AA A A A [ ] AA [ ] [ 4] 6. From the formula d b A, ad bc c a you see that ad bc 4 ( ) ( )( 8), and so the matrix has no inverse. 8. Begin by adjoining the identity matrix to the given matrix. [ A I] his matrix reduces to I A. So, the inverse matrix is A.

39 Review Exercises for Chapter A x b x 4x Because equation A x A 5 5 A, solve the bas follows. x b A x b x x x Using Gauss-Jordan elimination, you find that A Solve the equation A x bas follows A x b A x b x 5 4y 4 4 A Because, solve the equation Ax bas follows A 9 x b 6. A x b x y 4 4z 7 Using Gauss-Jordan elimination, you find that A Solve the equation Ax bas follows A 8 4 x b A, you can use the formula for the inverse of a matrix to obtain A. 8. Because ( ) 4 So, 4 4 A. 4. he matrix x will be nonsingular if 4 ( )( ) ( )( ) ad bc 4 x, which implies that x 8.. Because the given matrix represents 6 times the second row, the inverse will be times the second row. 6 6 For Exercises 4 and 6, answers will vary. Sample answers are shown below. 4. Begin by finding a sequence of elementary row operations to write A in reduced row-echelon form. Matrix Elementary Row Operation Elementary Matrix 4 Interchange the rows. E 4 Add times row to row. E Add 4 times row to row. E hen, you can factor A as follows. 4 A E E E 4

40 68 Chapter Matrices 6. Begin by finding a sequence of elementary row operations to write A in reduced row-echelon form. Matrix Elementary Row Operation Elementary Matrix Multiply row one by. E Add times row one to row three. E Add times row three to row one. E Multiply row two by. E4 So, you can factor A as follows. 4 A E E E E a b 8. Letting A, you have c d A a ba b a + bc ab + bd. c dc d ac + dc cb + d So, many answers are possible.,, etc. 4. here are many possible answers. A, B AB But, BA.

41 Review Exercises for Chapter 69 ( ) A + B A A + B B I 4. Because ( A + B )( A + B ) I, if ( A + B ) exists, it is sufficient to show that ( ) ( ) for equality of the second factors in each equation. ( ) ( ) ( ) ( ( ) ) ( ( ) ) A + B A A + B B A A A + B B + B A A + B B ( ) ( ) A A A + B B + B A A + B B ( ) ( ) I A + B B + B A A + B B ( ) ( ) ( ) ( ) I + B A A + B B ( ) ( ) B B + B A A + B B ( )( ) B B A A B B + + ( )( ) B A B A B B + + B B I IB herefore, ( + ) ( + ) B A B A A B B. 44. Answers will vary. Sample answer: Matrix Elementary Matrix A U E 4 4 EA U 4 4 A E U LU 46. Matrix Elementary Matrix A U E E E EEEA U A E E E U LU

42 7 Chapter Matrices 48. Matrix Elementary Matrix A U E EA U A LU y 7 7 y Ly b: y y y 8 4 x 7 4 x Uxy: x x x (a) In matrix B, grading system counts each midterm as 5% of the grade and the final exam as 5% of the grade. Grading system counts each midterm as % of the grade and the final exam as 6% of the grade. (b) AB (c) wo students received an A in each grading system. 54. f( A) he matrix is not stochastic because the sum of entries in columns and do not add up to. 58. his matrix is stochastic because each entry is between and, and each column adds up to. 6. X PX.7.69 X PX X PX X PX X PX X PX

43 Review Exercises for Chapter Begin by forming the matrix of transition probabilities. From Region P..8. o Region (a) he population in each region after year is given by PX , Region So,,., Region.. 9, Region (b) he population in each region after years is given by PX ,5 Region So,,., Region ,875 Region 66. he stochastic matrix 4 7 P 7 is not regular because for all powers. o find X, begin by letting matrix equation PX n P has a zero in the first column x X. hen use the x X to obtain 4 7 x x. x x 7 Use these matrices and the fact that x + x to write the system of linear equations shown. 4 x 7 4 x 7 x x + he solution of the system is x and x So, the steady state matrix is X. 68. he stochastic matrix. P is regular because P has only positive entries. o find X, let X [ x x x ]. hen use the matrix equation PX X to obtain..x x.5.9 x x..5..8x x Use these matrices and the fact that x + x + x to write the system of linear equations shown. x +.x.5x.x.5x +.x.x x + x + x, 5, he solution of the system is x x and x 5. So, the steady state matrix is 5 X. 5

44 7 Chapter Matrices 7. Form the matrix representing the given probabilities. Let C represent the classified documents, D represent the declassified documents, and S represent the shredded documents. From C D S P.7. C..75 D o..5 S x Solve the equation PX X, where X + + to write a system of equations. x and use the fact that x x x x.7x +.x x.x +.x.x +.75x x.x.5x.x +.5x + x x.x +.5x x + x + x x + x + x So, the steady state matrix is X. his indicates that eventually all of the documents will be shredded. 7. he matrix.8 P..7.6 is absorbing. he first state S is absorbing and it is possible to move from S to S in two transitions and to move from S to S in one transition. 74. (a) False. See Exercise 65, page 6. (b) False. Let A and B. hen A + B. A + B is a singular matrix, while both A and B are nonsingular matrices. 76. (a) rue. See Section.5, Example 4(b). (b) False. See Section.5, Example 7(a). 78. he uncoded row matrices are B E A M _ M E _ U P _ S C O Y _ [ 5 ] [ ] [ 5 ] [ 6 9] [ 5 ] [ 5 ] Multiplying each matrix on the right by A yields the coded row matrices. [ 7 6 ] [ ] [ 6 4] [ 6 7] [ ] [ ] So, the coded message is 7, 6,,,,,, 6, 4, 6,, 7,,,, 5, 45, 55.

45 Review Exercises for Chapter 7 8. Find A to be 4 A and the coded row matrices are [ 5 ], [ 8 9 ], [ 9 ], [ 5 ], [ 56 ], [ 5 ], [ 7 ], [ 9 4 ], [ 5 ]. Multiplying each coded row matrix on the right by A yields the uncoded row matrices. S H O W _ M E _ H E _ M O N E Y _ [ 9 8] [ 5 ] [ ] [ 5 ] [ 8] [ 5 ] [ 5] [ 4 5] [ 5 ] So, the message is SHOW_ME_HE_MONEY_. 8. Find A to be 4 A 8 9, 5 4 and multiply each coded row matrix on the right by A to find the associated uncoded row matrix. 4 A [ ] [ ] [ ] Y, A, N A 5 5 K, E, E A 9 S, _, W A 9 4 I, N, _ A 5 8 W, O, R 4 A 4 L, D, _ 66 5 A S, E, R A I, E, S [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] A [ ] [ ] A [ ] [ ] A [ ] _, I, N _, S, E V, E, N he message is YANKEES_WIN_WORLD_SERIES_IN_SEVEN. 84. Solve the equation X DX + E for X to obtain I D X E which corresponds to solving the ( ), augmented matrix he solution to this system is, X,. 5, 86. Using the matrices X 4 and Y, you have X X 5, 9 X Y 4, and 6 A X X X Y So, the least squares regression line is y.7 x. ( )

46 74 Chapter Matrices 88. Using the matrices 4 X and Y, you have X X 5, X Y, and 8.4 A X X X Y. 8.8 ( ) 5 So, the least squares regression line is y.8x +.4, or y 9 x + 9. (a) Using the matrices X X and X Y X and Y, you have Now, using ( X X) to find the coefficient matrix A, you have 97 ( ) A X X X Y So, the least squares regression line is y.86x +.7. (b) Using a graphing utility, the regression line is y.86x +.7. (c) Year Actual Estimated he estimated values are close to the actual values.

47 Project Solutions for Chapter 75 Project Solutions for Chapter Exploring Matrix Multiplication. est seems to be the more difficult test. he averages were: est average 75 est average Anna, David, Chris, Bruce. Answers will vary. Sample answer: M represents scores on the first test. M represents scores on the second test.. 4. index ; index index ; index ; index 4 4. Answers will vary. Sample answer: [ ] M represents Anna s scores. [ ] M represents Chris s scores. 5. Answers will vary. Sample answer: M represents the sum of the test scores for each student, and M represents each students average. 6. [ M ] represents the sum of scores on each test; [ M 4 ] represents the average on each test. M represents the overall points total for all students on all tests. 7. [ ] 8. [ ] 9. 8 M 8.5. M. 5. k 6. No. If A is nilpotent and invertible, then A k some k and A O. So, ( ) O for k k k A A I O A A A A A IA O which is impossible. k 7. If A is nilpotent, then ( A ) ( A ) O. k k ( ) ( ) k But A A O, which shows that A is nilpotent with the same index. 8. Let A be nilpotent of index k. hen k k I A A + A + + A + A + I k I A I, ( )( ) which shows that k k ( A + A + + A + A + I) is the inverse of I A., Nilpotent Matrices. A and A, so the index is.. (a) Nilpotent of index (b) Not nilpotent (c) Nilpotent of index (d) Not nilpotent (e) Nilpotent of index (f) Nilpotent of index