MDM4U Tools for Data Management Test 1 Instructions You must show how you arrived at your solution for past mark eligibility; the greater the detail, the greater the scope for part mark assessment. Marks: Knowledge and Understanding Application Communication Thinking / Inquiry / Problem Solving Total marks /11 /10 /5 /5 /31 Knowledge and Understanding 11 marks 1. Write the first six terms generated by the recursive formula 𝑡𝑛 = 2𝑡𝑛 − 2 + 𝑡𝑛2 − 1 given that 𝑡1 = 0 and 𝑡2 = 1. /4 Solution 𝑡3 = 2(0) + 12 = 1 𝑡5 = 2(1) + 32 = 11 ∴ 0, 1, 1, 3, 11, 127. 𝑡4 = 2(1) + 12 = 3 𝑡6 = 2(3) + 112 = 127 2. Write a non-recursive formula for the sequence formed by the recursive formula 𝑡𝑛 = 2(𝑡𝑛 − 1 + 1) − 𝑡𝑛 − 2 given that 𝑡1 = 0 and 𝑡2 = 3. /3 Solution 𝑡3 = 2(3 + 1) − 0 = 8 𝑡5 = 2(15 + 1) − 8 = 24 𝑡4 = 2(8 + 1) − 3 = 15 𝑡6 = 2(24 + 1) − 15 = 35 𝑡1 = 0 = 12 − 1 𝑡2 = 3 = 22 − 1 𝑡3 = 8 = 32 − 1 𝑡4 = 15 = 42 − 1 𝑡5 = 24 = 52 − 1 𝑡6 = 35 = 62 − 1 𝑡𝑛 = 𝑛2 − 1, 𝑛 ∊ ℕ 3. Utilize your solution to question 2 to evaluate 𝑡15 . Solution 𝑡15 = 152 − 1 = 224. /1 4. Given the communication network below, write the matrix representing the number of ways any two individuals can communicate with one another through at least 2 intermediates and at most 5 intermediates. /3 Solution 𝐴 𝐵 A=[ ] 𝐶 𝐷 [𝐴 0 1 [ 1 0 𝐵𝐶 0 0 0 1 1 0 0 0 𝐷] 0 0 ] - direct communication matrix 1 0 [𝐴 𝐵 𝐶 𝐷 ] 𝐴 𝐵 𝑀 = 𝐴3 + 𝐴4 + 𝐴5 + 𝐴6 = [ ] 𝐶 𝐷 0 4 [ 4 0 0 2 2 0 0 2 2 0 0 2 ] 2 0 Application 10 marks 5. Convert the number 124 which is in base 6 into its equivalent in base 8. /4 Solution 1246 = 1 ∙ 62 + 2 ∙ 6 1 + 4 ∙ 60 = 36 + 12 + 4 = 5210 . 52 = 6 ∙ 8 + 4 ∴ 5210 = 648 6. Given that A is 2×4-matrix, B is 2×4-matrix and C is 2×4-matrix, determine the dimensions of: Solution a) Matrix B + 2C does not exist. b) 2× 5 - matrix c) Matrix 𝐵2 does not exist. 7. What is the minimum number of colours required to colour in the map below? /3 Solution Two colours: Communication 4 marks 8. In what situation would you separate the bars in a bar graph? Answer: when the variable in the question is discrete. 9. Rewrite the following question in a biased manner with the intent of steering the reader towards a negative response: “What is your assessment of the Data Management course thus far?” /3 Answer: “Given its lackluster nature thus far, what is your assessment of DM4U1?” or “Given its limited mathematical content, what is your assessment of MDM4U1 thus far?” Thinking / Inquiry / Problem Solving 5 marks 10. About 30 minutes into his data management test Arash texts Chris a scrambled version of the message matrix 𝐻 𝐸 M = [ 𝐿 𝑃] 𝑀 𝐸 using the coding matrix 1 −1 0.5 C=[ ] 3 1 4 and the alpha numeric system: A B C D E F G H I J K L M N O P … X Y Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 … 24 25 26 (using the order C ∙ M). a) Create the coded message matrix using the message matrix and the coding matrix above. /2 Solution 1 CM = C ∙ M = [ 3 −1 1 8 0.5 ] ∙ [12 4 13 5 2.5 16] = [ 88 5 −8.5 ]. 51 b) Explain in mathematical detail, but do not solve, how Chris would go about decoding the coded message matrix generated by Arash in part a) to get back the message matrix including the way in which you mathematically generate any necessary formula(s) to do so. /3 Solution To decode the coded message matrix Chris should multiply it from left by inverse coding matrix: C −1 ∙ CM = (C −1 ∙ C) ∙ M = I ∙ M = M, where I is a unit 3× 3-matrix. Since C is 2× 3-matrix, C −1 must be 3× 2matrix. Now the question is how to determine C −1 . C −1 ∙ C = I ∴ 𝑢 [𝑤 𝑦 𝑣 𝑥 ] ∙ [1 3 𝑧 u + 3v = 1 −1 1 1 0.5 ] = [0 4 0 –u + v = 0 0 1 0 0 0] ∴ 1 0.5u + 4v = 0 No solution. Inverse matrix C −1 does not exist. Note: Only square matrices could have inverse.