Sol-Worksheet-Unit-2(Matrice-System equ).pdf - Solution...

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Unformatted text preview: Solution Guide Open University of Hong Kong MATH S201F Finite Mathematics for Business Solutions for Worksheet Unit 2: Matrices and Systems of Linear equations All solutions are subject to error. Section 1: Basic Operation and Matrix Inverse Problem 1 Solution 1. The dimension of A and B are both (2 A+B 3); hence, they can be added. Now = 4 1 3 5 = 4+8 1+9 8 7 8 9 + 0 7 2 3 3+0 5+7 8 + ( 2) ( 7) + ( 3) 2 3 4 1 12 10 = 3 12 6 10 and B+A = 8 0 9 7 = 8 + 4 0 + ( 3) ( 2) + 8 9+1 7+5 ( 3) + ( 7) + 3 5 8 7 12 10 = 3 12 6 10 A and B are commutative, hence A+B =B+A 2. A + (B + C) = 4 1 3 5 8 7 + 8 9 0 7 2 3 = 4 1 3 5 8 7 + 8+1 0 + ( 1) 9 + ( 3) 7+2 = 4 1 3 5 8 7 + 9 6 = 4 + 9 ( 3) + ( 1) 1+6 5+9 8 + ( 4) ( 7) + 2 = 13 7 4 14 4 5 = 4 1 0 7 + 1 3 1 2 2 5 = 4+8 1+9 = 12 10 = 12 + 1 ( 3) + ( 1) 10 + ( 3) 12 + 2 1 9 1 3 + 1 2 2 5 ( 2) + ( 2) ( 3) + 5 4 2 and (A + B) + C 3 5 8 7 + 8 9 2 3 ( 3) + 0 8 + ( 2) 5+7 ( 7) + ( 3) 3 12 6 10 + 1 3 1 2 + 1 2 2 5 = 13 7 4 14 2 5 6 + ( 2) ( 10) + 5 A, B and C are associative, hence, we have A + (B + C) = (A + B) + C Page 1 1 3 4 5 Solution Guide Problem 2 Solution 1. A + DT = 4 1 3 5 8 7 = 4 1 3 5 8 7 2 4 = 0 3 2 3T 3 8 5 1 6 +4 3 0 6 3 + 3 8 0 1 8 6 2. DT + B 2 3T 3 8 5 + 1 6 = 4 3 0 6 3 = 3 8 0 1 8 9 0 7 2 3 8 0 9 7 + 2 3 2 12 = 3 1 2 2 3. AT + O + B T 4 1 = = 3 5 2 4 4 3 8 8 7 3 T 2 3 0 T 8 0 2 0 5+ 9 7 3 0 3 2 3 2 0 8 9 12 0 5+4 0 7 5=4 3 0 2 3 6 0 +4 0 0 2 1 0 5 5+4 0 7 0 3 10 12 5 10 4. D + AT + C T 2 6 = 4 3 0 2 6 = 4 3 0 2 1 = 4 1 6 3 3 8 5+ 1 3 2 3 8 5+4 1 3 1 1 5 1 4 1 3 5 4 3 8 8 7 3 T 1 3 + 2 1 1 5 5+4 1 7 2 1 2 3 2 5 3 2 5 5 Problem 3 Solution 1. A + BT 2 3:87 = 4 9:7 0 2 3:87 = 4 9:7 0 2 25: 37 : 84 = 4 3: 07 3 2 14:6 2:8 5 + 4 4:8 3 2 0:32 14:6 35:74 2:8 5 + 4 11:53 4:8 3 24: 48 23: 07 36: 63 34: 55 5 16: 34 2: 03 0:32 35:74 11:53 Page 2 21:5 24:8 8:47 21:5 8:86 3:07 8:86 0:89 37:35 24:8 0:89 4:81 3T 3:07 4:81 5 6:83 3 8:47 37:35 5 6:83 T Solution Guide 2. B + AT 2 21:5 = 4 24:8 8:47 2 21:5 = 4 24:8 8:47 2 25: 37 = 4 24: 48 23: 07 3T 3 2 3:87 0:32 14:6 3:07 2:8 5 4:81 5 + 4 9:7 35:74 0 11:53 4:8 6:83 3 3 2 3:87 9:7 0 8:86 3:07 0:89 4:81 5 + 4 0:32 35:74 11:53 5 14:6 2:8 4:8 37:35 6:83 3 : 84 3: 07 36: 63 16: 34 5 34: 55 2: 03 8:86 0:89 37:35 3. (A + B) = = Hence, 2 3 2 3:87 0:32 14:6 21:5 8:86 4 9:7 35:74 2:8 5 + 4 24:8 0:89 0 11:53 4:8 8:47 37:35 2 3 25: 37 8: 54 11: 53 4 15: 1 36: 63 2: 01 5 8: 47 48: 88 2: 03 3T 2 11:53 25:37 15:1 2:01 5 = 4 8:54 36:63 2:03 11:53 2:01 3 8:47 48:88 5 2:03 3T 2 3:87 0:32 14:6 21:5 8:86 4 9:7 35:74 2:8 5 + 4 24:8 0:89 0 11:53 4:8 8:47 37:35 2 3 2 3:87 9:7 0 21:5 24:8 4 0:32 35:74 11:53 5 + 4 8:86 0:89 14:6 2:8 4:8 3:07 4:81 2 3 25: 37 15: 1 8: 47 4 8: 54 36: 63 48: 88 5 11: 53 2: 01 2: 03 3T 3:07 4:81 5 6:83 3 8:47 37:35 5 6:83 2 25:37 8:54 (A + B)T = 4 15:1 36:63 8:47 48:88 4. AT + B T = = = 3 3:07 4:81 5 6:83 2 Remark: We observe that (A + B)T = AT + B T In fact, this is true for any two matrices A and B which are of the same size. Problem 4 Solution 1. O 3B = O + ( 3)B 2 3 2 0 0 6 0 0 7 6 7 6 = 6 4 0 0 5+4 0 0 2 3 2 0 0 6 0 0 7 6 7 6 = 6 4 0 0 5+4 0 0 ( 3) ( 6) ( 3) ( 3) ( 3) 0 ( 3) 8 ( 3) ( 3) ( 3) ( 2) ( 3) 3 ( 3) 9 3 2 18 9 18 9 7 6 0 24 7 6 0 24 = 9 6 5 4 9 6 9 27 9 27 Page 3 3 7 7 5 3 7 7 5 Solution Guide 2. B 6 6 0 6 4 3 3 2 6 6 0 6 4 3 3 2 6 6 0 6 4 3 3 4A = = = 3. CT 3 2 3 2 3 3 6 5 7 8 7 7 7 + ( 4) 6 4 4 5 6 7 5 2 5 3 9 3 2 ( 4) 2 ( 4) 3 6 ( 4) ( 4) ( 4) ( 8 7 7+6 ( 4) 2 5 4 ( 4) 6 ( 4) 5 ( 4) 9 3 2 3 2 14 8 12 3 7 6 16 6 16 20 8 7 7=6 7+6 28 5 4 27 2 5 4 24 17 20 12 9 2 2A 5 4 = 6 3 2 5 4 6 6 3 = 6 4 7 2 8 1 2 5 4 6 6 3 = 6 4 7 2 8 1 2 1 6 14 = 6 4 5 2 7 2 8 1 T 2 2 6 4 + ( 2) 6 4 6 5 3 3 5) 7 7 7 5 3 3 15 28 7 7 30 5 3 3 3 5 7 7 7 5 3 3 ( 2) 2 ( 2) 3 7 6 ( 2) ( 4) ( 2) ( 5) 7 7 7+6 5 4 ( 2) 6 ( 2) 7 5 ( 2) 5 ( 2) 3 3 3 2 4 6 7 7 6 8 10 7 7+6 5 4 12 14 5 10 6 3 2 13 7 7 12 5 5 2 3 Problem 5 Solution 1. The additive inverse of 4C is negative of ( 4C). That is 4C = 4 5 4 6 3 7 2 8 1 = ( 4C) = 4C: Now, 20 16 24 12 28 8 32 4 We can check that ( 4C) + (4C) = 0 = 4 5 4 6 3 7 2 8 1 + 20 16 24 12 2. We may compute 6A + 2C T …rst. 6A + 2C T 28 8 32 4 2 2 6 4 = 66 4 6 5 2 12 6 24 6 = 4 36 30 2 12 6 24 = 6 4 36 30 = 0 0 0 0 3 3 5 7 7+2 7 5 3 3 18 30 7 7+ 42 5 18 3 2 18 6 30 7 7+6 5 4 42 18 Page 4 0 0 0 0 5 4 : 6 3 7 2 8 1 14 4 T 16 2 T 10 8 12 6 10 12 14 16 3 2 8 22 6 12 6 7 7=6 4 5 4 50 2 46 3 26 24 7 7 46 5 20 Solution Guide Hence, the additive inverse of 6A + 2C T is 3 26 24 7 7 46 5 20 2 22 6 12 6 4 50 46 3. We compute (B T + 5C)T …rst. (B T + 5C)T = (B T )T + 5C T = B + 5C T 2 6 3 6 0 8 = 6 4 3 2 3 9 2 6 3 6 0 8 = 6 4 3 2 3 9 2 6 3 6 0 8 = 6 4 3 2 3 9 Hence, the additive inverse of (B T + 5C)T is 2 3 5 4 3 25 20 7 7+5 5 7 7+ 5 3 30 15 2 25 7 6 30 7+6 5 4 35 40 7 2 35 10 T 8 1 T 40 5 3 2 19 20 6 30 15 7 7=6 10 5 4 32 43 5 3 17 23 7 7 8 5 14 3 17 23 7 7 8 5 14 19 30 32 43 6 6 4 6 3 It can be seen that for any matrix A, we have (AT )T = A Problem 6 Solution 1. We observe that A is 2 3; B is 2 3, and hence B T is 3 2: A has three columns, which is equal to the number of rows in B T : Hence, it is legal to compute the matrix product AB T : We have AB T 3 4 = 4 3 3 4 = 4 3 3 = ( 4) 22 20 = 5 2 5 2 2 3 0 2 1 3 0 1 5 2 3 4 2 1 3+4 2+5 1 3 + ( 3) 2 + ( 2) 1 2 T 3 1 ( 4) 0 + 4 ( 1) + 5 ( 2) 0 + ( 3) ( 1) + ( 2) ( 2) 14 7 2. B is 2 3 and C is 3 2: The number of columns in B equals the number of rows in C: It is reasonable to complete the matrix product 3BC which 2 2 is 3 2 2 4 3 2 1 4 0 3 5 3BC = 3 0 1 2 1 6 = 3 = 3 0 3 2+2 2 + ( 1) 5 2 12 9 = 0 + 1 ( 1) 0 + ( 2) ( 1) 15 36 6 27 Page 5 0 3 4+2 4 + ( 1) 3 + 1 ( 6) 3 + ( 2) ( 6) Solution Guide 3. We observe that B is 2 3 and C T is also 2 3 (since C is 3 2). B has three columns, which does not equal the number of rows in C T : This shows that BC T is not de…ned. 4. Scienti…c Notebook exercise. Problem 7 Solution AC = 3 4 (AC)B = 1 6 4 3 2 2 4 0 1 5 2 Hence, On the other hand, we have 6 13 3 4 3 5 6 2 2 CB = 4 0 1 Hence, A(CB) = 3 4 3 0 4 3 5 2 This veri…es that (AC)B = A(CB): 3 0 2 1 1 2 2 1 1 2 6 4 0 3 0 3 4 2 3 4 3 5= 6 = 1:0 6:0 6:0 13:0 3:0 18:0 2 6:0 =4 0 3:0 3 6 6 5= 11 8:0 1:0 13:0 20:0 0 3:0 4:0 3 6:0 6:0 5 11:0 3:0 18:0 8:0 13:0 1:0 20:0 Problem 8 Solution We have and 2 3 BT C T = 4 2 1 C T BT = 3 0 1 5 2 2 4 0 3 2 0 4 3 1 6 Clearly, B T C T 6= C T B T : They are di¤ erent sizes. 1 6 2 3 4 2 1 2 6:0 =4 0 6:0 3 0 1 5= 2 0 3:0 6:0 5:0 12:0 3 3:0 4:0 5 11:0 2:0 9:0 Problem 9 Solution 1. AT B 3DDT 2 3 3 2 3 4 2 3 2 1 3 5 = 4 4 34 3 5 0 1 2 5 2 4 2 3 3 + ( 4) 0 3 2 + ( 4) ( 1) 3 = 4 4 3 + ( 3) 0 4 2 + ( 3) ( 1) 4 5 3 + ( 2) 0 5 2 + ( 2) ( 1) 5 3 2 2 2 2 ( 3) 2 4 3 4 ( 3) 2 ( 3) ( 3) ( 3) 4 5 4 2 4 ( 3) 4 4 2 3 2 3 9 10 11 4 6 8 9 12 5 = 4 12 11 10 5 3 4 6 15 12 9 8 12 16 2 3 2 3 2 9 10 11 12 18 24 27 36 5 = 4 = 4 12 11 10 5 + 4 18 15 12 9 24 36 48 Page 6 2 3 4 1 + ( 4) 1 + ( 3) 1 + ( 2) 3 ( 2) ( 2) 5 ( 2) 3 30 9 3 13 46 5 39 28 16 48 Solution Guide 2. DEC T 2 3 2 4 3 5 3 7 4 2 3 6 14 2 4 9 21 5 4 12 28 2 6 2 + 14 4 4 ( 9) 2 + ( 21) 12 2 + 28 4 2 68 42 90 4 102 63 135 136 84 180 = = = = 3. We compute (A (A (A D) D) E as. D) = 3 4 E = 14 7 = 42 21 4 3 2 4 0 1 0 3 1 6 3T 4 3 5 6 4 3 5 3 2 4 3 5= 4 7 3 6 ( 1) + 14 ( 6) ( 9) ( 1) + ( 21) ( 6) 5 12 ( 1) + 28 ( 6) 6 0 + 14 3 ( 9) 0 + ( 21) 3 12 0 + 28 3 2 5 2 3 2 3 ( 4) 2 + 4 ( 3) + 5 4 2 + ( 3) ( 3) + ( 2) 98 49 We checked that these results agree with those in parts 1, 2 and 3. Problem 10 Solution 1. We assume a c X= b d Then we have 4 2 1 5 a c b d = 1 0 0 1 That is, we have 4a c 2a + 5c 4b d 2b + 5d 1 0 = To …nd a; b; c and d, we have to solve the following system 8 4a c = > > < 4b d = > 2a + 5c = > : 2b + 5d = of linear equations: 1 0 0 1 which can be split into two systems: 4a c = 1 2a + 5c = 0 and 4b d = 0 2b + 5d = 1 5 1 The …rst system has the solution fa = ;c= g: 22 11 1 2 The second system has the solution fb = ;d= g: 22 11 So the matrix X= 5 22 1 11 Page 7 0 1 1 22 2 11 4 = 14 7 Solution Guide 2. We assume a c Y = b d Then we have a c b d 3 0 7 7 1 0 = 0 1 That is, we have 3a 3c 7a + 7b 7c + 7d 1 0 = 0 1 We have to solve the following two systems: 3a = 1 7a + 7b = 0 and 3c = 0 7c + 7d = 1 It can be easily seen that the solutions are: 1 1 1 ; b = g and fc = 0; d = g: 3 3 7 fa = So the matrix 1 3 1 7 1 3 Y = 0 3. AB = 4 2 1 5 3 0 7 7 12 6 = 21 49 and det(AB) 12 21 6 49 = det = ( 12) = 49 21 588 + 126 = ( 6) 462 4. 3 0 BA = det(BA) = det = 2 7 7 4 2 1 5 2 38 14 35 35 38 14 = 2 38 14 35 = 462 Problem 11 Solution 1. AT B = 4 1 2 5 3 0 7 7 12 3 = 42 28 Then det AT B = det 12 3 42 28 = 462 Since det(AT B) 6= 0; the inverse of AT B exists. The inverse of AT B; is (AT B) 1 = 2 33 1 154 1 11 2 77 Check: we may con…rm that this is the right answer since 2 33 1 154 1 11 2 77 12 3 Page 8 42 28 = 1 0 0 1 = I2 Solution Guide 2. BT A = 3 7 0 7 4 2 1 5 12 42 = 3 28 Then det 12 42 3 28 = 462 Since det(B T A) 6= 0; the matrix B T A has an inverse which can be written as (B T A) and we have 2 33 1 11 1 2 33 1 11 = 1 154 2 77 12 42 1 154 2 77 3 28 1 0 0 1 12 20 21 35 = 3. CT D = 3 5 4 7 12 20 21 35 = and det =0 Since det(C T D) = 0; the matrix C T D does not have an inverse. 4. DT C = 4 7 3 5 12 21 = 20 35 then det 12 21 20 35 =0 Since det(DT C) = 0; the matrix DT C does not have an inverse. Problem 12 Solution 1. A2 A is 2 3 matrix. It cannot be multiplied with another 2 rows of A. Hence A 2. A 3 and B is 3 AB = 1 0 4 1 3; so A B can be multiplied: 3 2 1 4 7 6 4 1 + 32 + 0 8 2 1 5= 7 0+8+0 0 1 6 4 + 8 + 6 7 + 4 + 36 0 + 2 + 1 0 + 1 + 42 = 33 8 18 9 47 43 A B is 3 4. A A is not de…ned. B A is 2 3. B 3 matrix, it is because the column of A 6= the 3 and A is 2 3 so B 3 and C is 3 2 so A A is not de…ned, because the columns of matrix B 6= the rows of A. C A is 2 AC = 1 0 4 1 C can be multiplied to give: 3 2 4 2 6 4 4 + 0 + 12 0 0 5= 7 0 + 0 + 14 2 9 Page 9 2 + 0 + 54 0 + 0 + 63 = 16 14 56 63 Solution Guide 5. C A C is 3 2 and A is 2 2 3 so C 3 4 2 CA = 4 0 0 5 2 9 A can be multiplied to give: 2 4 + 0 16 + 2 1 4 6 =4 0+0 0+0 0 1 7 2+0 8+9 3 2 24 + 14 4 0+0 5=4 0 12 + 63 2 18 0 17 3 38 0 5 75 B2 6. A 3. A square matrix B 2 can be formed as 3 2 32 2 1 + 32 + 0 1 4 7 1 4 7 B 2 = 4 8 2 1 5 4 8 2 1 5 = 4 8 + 16 + 0 0+8+0 0 1 6 0 1 6 B is 3 3 and B 2 is 3 3. AB 2 can be multiplied, then A 2 3 33 19 53 1 4 6 4 24 37 64 5 = AB 2 = 0 1 7 8 8 37 and A is 2 7. B 3 2 33 7 + 4 + 42 56 + 2 + 6 5 = 4 24 8 0 + 1 + 36 4+8+7 32 + 4 + 1 0+2+6 B 2 can form a 2 177 80 19 37 8 3 53 64 5 37 3 matrix. 215 531 93 323 C B is 3 8. C 3 and C is 3 2. BC to form a 3 2 matrix. 2 32 1 4 7 4 BC = 4 8 2 1 5 4 0 0 1 6 2 3 2 2 18 0 5 = 4 34 9 12 3 65 25 5 54 B C is 3 2 and B is 3 3. CB is not de…ned, , because the columns of matrix C 6= the rows of B. 9. C 2 C is 3 2. It cannot be multiplied with a 3 C 6= the rows of C. 2 matrix, so C 2 is not de…ned, because the columns of matrix Problem 13 Solution 7 5 1. 5(A + B + C) = 5 2. A 1 = 1 7 3 3. (A + B + C) 4 1 4 3 3 5 5 6 5 + 4 7 2 3 3 5 = , where A + B + C = 17 15 4 5 + 1 = 1 17 4. AT B T , where AT = So AT B T = 7 4 5 3 9 7 5 8 4 3 6 2 9 15 15 T 7 4 = 5 3 = 5 3 52 30 17 15 =5 8 9 = 5 3 : 4 7 8 9 so (A + B + C) 2 3 8 17 2 6 =4 ; BT = 50 29 3 11 5 11 6 5 : Page 10 2 3 3 8 33 7 17 5 33 T = 6 2 85 75 40 45 Solution Guide T 7 5 T 5. (AB) C; where (AB) = = 62 45 = 473 199 T 26 19 4 5 2 3 = 4 3 6 5 62 26 T 2 3 45 19 4 5 4 5 2 3 1 2 0 0 2 3 259 109 B)T C; where 6. (A 7 5 B)T = (A 4 3 6 5 B)T C = 1 2 7. det(A) = 7 5 4 3 =7 3 4 5 = 1: 8. det(B) = 6 5 2 3 =6 3 2 5 = 8: So (A 9. det A det B = 1 0 0 T 2 3 4 5 T = 1 0 2 0 = 4 8 2 4 2 3 = 8=8 10. det(AB) = det A det B = 1 8=8 11. det(BC) = det B det C = 8 2 = 16 12. det(AB)T = det(AB) = 8 13. det(AB) 1 = 1 1 = det(AB) 8 14. det(5AB) = 52 det(AB) = 200: [Use Scienti…c Notebook to check your solution.] Problem 14 Solution 1. E 1 3 2 := 1 2 5 = 1 det(E) 5 2 2 3 2 3 = , where det (E) = 15 + 4 = 19: Hence 2. E 3 2 2 5 1 = 1 19 5 2 5 19 2 19 2 19 3 19 G = F: 3 2 2 5 a c b d = a c b d = a c b d = 0 3 " 3 2 2 5 2 5 5 19 2 19 Page 11 =) 1 2 # 19 3 19 0 3 0 3 2 5 2 5 =) = 6 19 9 19 0 1 Solution Guide 0 3 2 5 = 6 6 19 9 19 0 1 = 6 19 det (F ) det (G) = 6 and F G= 3. det(F ) = det det(G) = det 0 3 6 19 2 5 18 19 27 19 so det(F G) = det 2 5 = 36 . 19 6 19 9 19 0 1 = Hence we veri…ed that det (F ) 18 19 27 19 = 2 5 36 19 : det(G) = det (F G). Problem 15 Solution 1. det (D) = = 4(3 2 = 4( 3) 4 5 5 2 3 3 1 3 2 3 3) 2(5 D 2 2 2( 5) + 1(0) = and 5 det (D) = 5 2. 20 det 3 3 =4 = 20 23 3 2 +1 5 5 5) + 1(5 3 3 5) 1 3 2 3 3 7 5 2 2 3 3 3 2: ( 2) = det D = 5 5 3 2 10: 20 23 ( 2) = 3. det( 2D3 ) = ( 2)3 det(D3 ) = 8 5 ( 2)3 = 64: 2 3 3 5 0 3 2 5 4. M in(D) = 4 1 3 7 2 2 3 2 3 5 0 3 2 5 ; Adj(D) = 4 5 Cof (D): 4 1 3 3 7 2 0 D 1 2 3 1 4 1 5 Adj(D) = = det(D) 2 0 1 3 2 3 2 3 3 2 7 5 = 4 52 2 0 1 2 3 2 3 2 7 2 1 1 ( 5) 4 3 5 3 5 [Use Scienti…c Notebook to check your solution.] Problem 16 Solution 9 4 3 5 1. det(E) = 6 8 6 5 7 3 7 0 =9 8 6 5 3 7 0 2 6 1 where 8 6 5 3 7 0 2 6 1 5 2 6 1 4 3 5 6 = 8 7 0 3 7 0 6 1 2 6 1 4 3 5 +7 ( 3) 6 5 Page 12 6 1 8 6 5 2 6 1 +2 6 5 7 0 = 234 8 6 5 3 7 0 Solution Guide 4 3 5 Similar 3 7 0 2 6 1 Hence det(E) = 9(234) 2. det( 2E T 4 = = ( 2) det(E ) 4 3 5 8 6 5 2 6 1 4 3 5 = 18; 8 6 5 3 7 0 = 6( 179) + 7(18) + 5( 275) = 1931: 3I4 ) = det( 2E T ) T 179; det(3I4 ) 4 3 det(I4 ) = ( 2)4 (1931) 34 = 2502 576: since det(E T ) = det(E) = 1931, and det(I4 ) = 1: 4 3. det(3E 1 ) == 4 (3) (3) = = det(E) 1931 4. Cofactor of E is 2 234 6 469 cof (E) 6 4 33 430 179 301 50 593 18 261 146 264 81 1931 : 2 3 275 6 840 7 7 ; Adj(E) = 6 4 5 85 1116 2 234 179 18 275 234 1 1 6 179 T 6 E = Adj(E) = det(E) 1931 4 18 275 [Use Scienti…c Notebook to check your solution.] Page 13 469 301 261 840 469 301 261 840 33 50 146 85 33 50 146 85 3 430 593 7 7: 264 5 1116 3 430 593 7 7 264 5 1116 275 Solution Guide Section 2 Applications of Matrices Problem 17 Solution We can organize these data into a 3 4 matrix in which each chain store is represented by a row and each kind of product is represented by a column. KT=Kowloon Tong, HT=Homantin, TW=Tsuen Wan. 2A 35 4 28 42 KT HT TW B 61 55 28 C 111 96 75 D3 48 36 5 60 The total value in each store can be found by using the matrix product 3 2 2 3 12 35 61 111 48 6 7 4 28 55 96 36 5 6 8 7 4 15 5 42 28 75 60 20 3 3 2 2 = total value in KT 3533 (35 12) + (61 8) + (111 15) + (48 20) = 4 (28 12) + (55 8) + (96 15) + (36 20) 5 = 4 2936 5 = total value in HT = total value inTW 3053 (42 12) + (28 8) + (75 15) + (60 20) [The entries will change if either the order of chain store or the order of the products has been changed.] Problem 18 Solution The data concerning the stock of the …ve kinds of medicine can be organized in a 1 of …ve entries) as follows (in the order A, B, C, D and E): 5 matrix (or a row vector Q = [300; 260; 400; 580 760] The data concerning the prices can be organized into a 5 2 20 6 38 6 P =6 6 16 4 120 77 which may be called the price vector. 1 matrix or a column vector of …ve entries as 3 25 25 7 7 18 7 7 100 5 78 The total value of medicine is 300 260 = (300 (300 = 150 400 400 580 20) + (260 25) + (260 X 760 2 6 6 6 6 4 38) + (400 35) + (400 20 38 16 120 77 (in HK$) Problem 19 Solution 2 1 1. The matrix equation is 4 2 3 1 2 1 3 7 7 7 7 5 16) + (580 18) + (580 Y 141 080 25 35 18 100 78 32 3 2 3 1 x 55 1 5 4 y 5 = 4 95 5 2 z 80 Page 14 120) + (760 100) + (760 77) ; 78) Solution Guide 2. det(A) = 2 1 1 2 2 6 6 6 3. The cofactor 6 6 6 4 2 3 2 1 1 1 1 2 1 2 1 2 1 1 1 2 2 3 2 1 2 3 1 3 1 2 1 2 1 2 1 1 + 2 The adjugate A =cof(A)T = 4 3 1 4 = (4 1) 2 3 1 3 1 2 3 1 1 2 2 1 1 5 0 3 1 The inverse of A is det(A) AdjA = 12 4 1 4 4. The 2 x 4 y z The solution set (x; y; z) is x = A 3 2 3 3 2 1 1 1 1 2 1 (4 3) + (2 3 2 1 1 1 1 2 7 2 7 3 7 7=4 1 7 7 1 5 6) = 1 1 1 3 2 3 1 2 1 5 = 4 12 0 2 1 2 1 2 1 2 3 4 2 5 0 1 2 1 2 0 3 5 c: 3 2 3 55 5 2 2 2 1 5 1 5=4 1 4 95 5 = 4 35 5 2 2 2 2 1 0 80 15 manufacturer should produce 5 units of x, 35 units of y and 15 units of z. Problem 20 Solution The augment matrix form is Solve by Gaussian 3 Elimination: 2 4 4 7 9 R1 4 2 5 1 6 5 R2 2 4 4 1 R 2 3 4 4 7 R1 -2R2 4 0 6 5 Step 1: R1 -2R2 0 12 15 2 4 4 4 0 6 Step 2: R2a + 21 R3a 0 0 2 4 4 2 2 3 5 4 7 1 4 3 j 9 j 6 5 j 1 3 9 R1a 3 5 R2a 7 R3a 3 7 9 5 3 5 25 2 1 2 The above matrix is now in upper triangular form. The solution set is 25 1 1 Row 3: Solution is: z = 25 2 z = 2, Row 2: 6y 5z = 3 8 1 6y 5 3, : Solution is: y = 15 25 = Row 1: 4x + 4y 7y = 9 8 1 247 4x + 4 15 7 Solution is: 150 25 = 9, Problem 21 Solution 1. The cost of these materials for each style of house is 2 cost 3 20 6 180 7 10 2 0 2 6 7 Q R= 4 60 5 50 1 20 2 24 cost = 608 2428 Sp Con Page 15 Solution Guide 2. The requirement of each of four kinds of material must be ordered for three models: Sp Con 3 2 0 30 Q = 4 10 20 5 20 20 P C 1500 = 4 1100 1200 C L 10 2 50 1 B S 0 2 20 2 L B S 3 30 600 60 40 400 60 5 60 400 80 2 Sp , Con Model A Model B Model C Answer: Need to order: (1500 + 1100 + 1200) = 3800 yd3 of concrete, (30 + 40 + 60) 1000 = 130 000 board feet of lumber, (600 + 400 + 400) 100 = 20 000 ft2 of metal shingles. (60 + 60 + 80) 3. The total cost for 2 1500 30 600 4 1100 40 400 1200 60 400 2 3 72 840 = 4 54 640 5 60 720 1000 = 1400 000 bricks, and materials 3 2 3 20 60 6 180 7 7 60 5 6 4 60 5 80 24 Model A Model B Model C The total cost $72 840 for Model A, $54 640 for Model B, and $60 720 for Model C. Problem 22 Solution 1. The answer to this problem involves 6 quantities, two for each client. In this problem we take an arbitrary client b1 for the total investment and b2 for the annual return. There are three set of (b1 and b2 ): b1 b2 = Client 1 Client 2 20000 2400 50000 7500 ; Client 3 and 10000 1300 Let x = amount invested in Plan A by a given client, and let y = amount invested in Plan B by a given client. The system of equations is x + y = b1 0:1x + 0:2y = b2 Total invest equation Annual return equation A The matrix equations is det(A) = 0:2 1 0:1 1 0:2 1 0:1 x B x y b1 b2 = ; 0:1 = 0:1 1 = 2 1 10 10 Hence, the solution set is x = A Client 1: x = y 1 0:2 2 1 10 10 20000 2400 1 B: = 16 000 4000 Invest x = $16000 in Plan A, y = $4000 in Plan B. Client 2: Page 16 Solution Guide x y 2 1 = 10 10 50000 7500 = 25 000 25 000 Invest x = $25000 in Plan A, y = $25000 in Plan B. Client 3: x = y 2 1 10 10 10000 1300 = 7000 3000 Invest x = $7000 in Plan A, y = $3000 in Plan B. 2. Now change the percentage as A pays 8% and plan B pays 24%. The matrix A change to A 1 0:8 A 1 = 1 0:24 1 0:08 1 0:24 Client 1: x 1: 5 = y 0:5 x B x y = b1 b2 = 1: 5 0:5 1 6: 25 6: 25 20000 2400 6: 25 6: 25 15000 5000 = Invest x = $15000 in Plan A, y = $5000 in Plan B. Client 2: x 1: 5 = y 0:5 6: 25 6: 25 50000 7500 28125 21875 = Invest x = $28125 in Plan A, y = $21875 in Plan B. Client 3: x 1: 5 = y 0:5 6: 25 6: 25 10000 1300 6875 3125 = Invest x = $6875 in Plan A, y = $3125 in Plan B. Problem 23 Solution This is a similar problem as the previous one. The answer to this problem involves 6 quantities, two for each park. We take an arbitrary client b1 be the Number of workers and b2 be the Total weekly wages. There are three set of (b1 and b2 ): b1 b2 = Park 1 Park 2 120 28400 120 27200 ; Park 3 120 28000 and Let x be the number of experienced workers; and let y be the number of inexperienced workers. The system of equations is x + y = b1 Total investment equation 240x + 220y = b2 Annual return equation The matrix equations is A 1 240 1 220 x B x y b1 b2 Page 17 = ; Solution Guide det(A) = 220 240 = 20 1 240 Hence, the solution set is x = A 1 1 220 1 11 12 1 20 1 20 120 28400 = 100 20 = B In Park 1: x y 1 20 1 20 11 12 = Park 1: x = 100 experienced workers. In Park 2: x y = 11 12 1 20 1 20 120 27200 = 40 80 11 12 1 20 1 20 120 28000 = 80 40 Park 2:x = 40 experienced workers. In Park 3: x y = Park 3: x = 80 experienced workers. Problem 24 Solution 1. Let x be the type of Compact car model, and let y be the type of full-size car model. The system of equation is x + y = 800 10000x + 24000y = 12000000 total car purchase equation budget equation The matrix equations is A 1 10000 A 1 = 1 24...
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