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.