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Stage II Maths Studies | Directed Investigation - CODES |

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Suppose all letters are given a number from 1 to 26. E.g. a=1, b=2, c=3 etc then “put out the cat” becomes 16 21 20 15 21 20 20 8 5 3 1 20. If we use the following randomly chosen matrix
A= 1483852321
-------------------------------------------------
as a multiplier for each of these then
‘put’ = B = 16 21 20 becomes 452 273 110 1. Verify this and ascertain the code for “out”, “the”, and “cat”. How can this message be coded by one matrix calculation?
PUT = 16 21 20 x 1483852321 = 16 x 14+21 x 8+20 x 316 x 8+21 x 5+20 x 2(16 x 3+21 x 2+20 x 1) = 224+168+60 128+105+40 (48+42+20)
-------------------------------------------------
= 452 273 110
OUT = 15 21 20 x 1483852321 = 15 x 14+21 x 8+20 x 315 x 8+21 x 5+20 x 2(15 x 3+21 x 2+20 x 1) = 210+168+60 120+105+40 (45+42+20)
-------------------------------------------------
= 438 265 107
THE = 20 8 5 x 1483852321 = 20 x 14+8 x 8+5 x 320 x 8+8 x 5+5 x 2(20x 3+8 x 2+5 x 1) = 280+64+15 160+40+10 (60+16+5)
-------------------------------------------------
= 359 210 81
CAT = 3 1 20 x 1483852321 = 3 x 14+1 x 8+20 x 33 x 8+1 x 5+20 x 2(3x 3+1 x 2+20 x 1) = 42+8+60 24+5+40 (9+2+20) = 110 69 31

This message can be coded by one matrix calculation by putting it vertically in a 3 x 4 matrix and then multiplying it by the randomly chosen matrix like so:
C
A
T
T
H
E
O
U
T
P
U
T
161520 321218 120155 20

1483852321 X 161520 321218 120155 20 = 452438359 110273265210 6911010781 31
-------------------------------------------------

2. Are there similar codes for the same letters?
U = 273, 265
T = 110, 107, 359, 31
-------------------------------------------------
No, there are no similar codes for the same letters as U equals to 273 and 265, while T equals to 110, 107, 359 and 31. 3. Compute the inverse of A and use it to decode your coded message.
According to the TI-84 Plus Texas Instruments Graphics Calculator:
A-1 = 1-21-25-41-46
-------------------------------------------------

1-21-25-41-46 X 452438359 110273265210 6911010781 31

O
U
T
P
U
T
C
A
T
T
H
E
= 161520 321218 120155 20

4. Calculate Det A and comment
Det A = 1483852321 = 145221 + 82813 + 38532 = 14(5 – 4) + 8(6 – 8) + 3(16 – 15) = 14(1) + 8(-2) + 3(1) = 17 – 16 = 1
-------------------------------------------------
The determinant of matrix A is 1. By having a whole number as a determinant, decoding and encoding messages is easier than if the determinant was a fraction. Also, it makes sure that a decoding matrix, or the inverse matrix, exists. It is not good if an encoded message cannot be decoded. 5. Explain why a coding matrix with a determinant of 1 would simplify the decoding process.
-------------------------------------------------
A coding matrix with a determinant of 1 would simplify the decoding process because it means the inverse has an integer component.
A matrix that is triangular has all 1s on the diagonal and zeros in the upper or lower half. The matrix below is triangular.
123012001
The transpose of a matrix (AT) has elements swapped aij = aji 6. Find the transpose of the matrix above

123012001T = 100210321
-------------------------------------------------

Here is a way to create a matrix that is suitable.
Let M be a matrix that is upper triangular and MT is its transpose.
Matrix M x MT has integer elements and has a determinant of 1. This is also true of its inverse. This matrix is the one we use.

7. Verify the process given produces an appropriate matrix.

Upper Triangular Matrix = 100210321
Find MT = 123012001

M x MT = 100210321 x 123012001

According to calculator = 1232583814

Determinant of M x MT = 158814 + 282143 + 32538

= 1(70 –64) + 2(24 – 28) + 3(16 – 15) = 1(6) + 2(-4) + 3(1) = 9 – 8 = 1

Calculator: Inverse = 6-41-45-21-21

Determinant of (M x MT)-1 = 65-2-21 + -4-2-411 + 1-451-2 = 6(5 –4) + -4(-2 – -4) + 1(8 – 5) = 6(1) + -4(2) + 1(3) = 6 – 8 + 3 = 9 - 8 = 1

8. Create your own matrix and use it to encode and decode ‘maths is funny’. Don’t forget to group the letters in threes.
MATHS IS FUNNY = MAT HSI SFU NNY
MAT = 13120

HSI = 8199

SFU = 19621

NNY = 141425

MATRIX M = 100810781

MATRIX MT = 187018001

MATRIX M x MT = 100810781 x 187018001

According to calculator = 18786564764114

MAT = 13120 x 18786564764114 = 16114492435
HSI = 8199 x 18786564764114 = 22318752298

SFU = 19621 x 18786564764114 = 21418862911

NNY = 141425 x 18786564764114
= 30126223844
Therefore the encrypted message would be:
-------------------------------------------------
161 1449 2435 223 1875 2298 214 1886 2911 301 2622 3844
MATRIX (M X MT)-1 = 3314-46457-46465-857-81 (on calculator)
Encrypted Codes = 16114492435 = 22318752298 = 21418862911 = 30126223844
-------------------------------------------------
To decode = Encrypted codes x Multiplier
16114492435 x 3314-46457-46465-857-81 = xyz
13120 = xyz x = 13 = M y = 1 = A
-------------------------------------------------
z = 20 = T
22318752298 x 3314-46457-46465-857-81 = xyz
8199 = xyz x = 8 = H y = 19 = S
-------------------------------------------------
z = 9 = I
21418862911 x 3314-46457-46465-857-81 = xyz
19621 = xyz x = 19 = S y = 6 = F
-------------------------------------------------
z = 21 = U

30126223844 x 3314-46457-46465-857-81 = xyz
141425 = xyz x = 14 = N y = 14 = N
-------------------------------------------------
z = 25 = Y
Therefore, the decrypted code is:
-------------------------------------------------
MATHS IS FUNNY 9. One major problem can be if the codes are mixed length. If codes produce 2 digit or 3 digit answers, how do you know where to separate them for decoding. One way is to make all codes 3 digits by adding multiples of 26. The letters work best when using modulo arithmetic putting a=0, b=1, c=2, etc.
When decoding we can use the modulo arithmetic to get our codes back.
Explain how this works for example in 8.
It is better to change all 2 digit numbers to 3 digit numbers as it would make it easier to the receiver to separate the long message. If a message was composed of both 2 digit and 3 digit numbers, the receiver would not know where to separate the message if it is sent as numbers. So it would make it difficult for the decipherer to decode the message.
Using question 8 as an example:
MATHS IS FUNNY
When encoded, the message will display:
161 1449 2435 223 1875 2298 214 1886 2911 301 2622 3844
In this example there are no 2 digit numbers, but instead there are 4 digit numbers. But this can also make it difficult for the receiver t o decipher the codes. The message would look like this:
1611449243522318752298214 1886291130126223844
In this form, it would be difficult to determine if these numbers were 3 digit or 4 digit numbers.

Therefore, to have 3 digit numbers you subtract multiples of 26. It would make it easier for the decipherer to read the message, knowing that the message is divided into groups of three numbers.
161 1449 2435 223 1875 2298 214 1886 2911 301 2622 3844
161 981 979 223 991 998 214 976 987 301 984 958
161 981 979 223 991 998 214 976 987 301 984 958
-------------------------------------------------
We add or subtract multiples of 26 because there are 26 letters of the alphabet. Every multiple of 26 repeats the letter it represents e.g. A=1, then by adding 26, we get 27. As the table shows, 27 = A. This means that the alphabet repeats again after 26.

10. Create your own code using at least a 3 x 3 matrix incorporating these hints. You should discuss ways of sending the coding and decoding information to your spy! Don’t forget to write about it and show and verify all the maths you have used to code and decode.
With technology these days, messages can be sent through emails and text messages. If a secret message was to be sent, information on how to read the code will be given. Modulo arithmetic will be used for the whole coding and decoding. All numbers in the message will be 2 digit numbers because when using modulo arithmetic, the maximum number is 26, a 2 digit number.
ENCRYPTION

To encrypt a message, a message is needed when using modulo arithmetic. A 3 x 3 matrix will be used and this would represent the first two numbers of the long message. So the first two numbers of the message will be 03 03.

MATRIX M = 12618016001

MATRIX MT = 10026101861

MATRIX M X MT = 1001134181343761861
This method was used to find a matrix with a determinant of 1.

If a number is greater than M, in this case, greater than 26, you divide the number by 26 and find the remainder.

Mathematically, the remainder that results after being divided by 26 replaces integers greater than 26.

Therefore,
1001/26 = 38 remainder 13
134/26 = 5 remainder 4
37/26 = 1 remainder 11

= 1341841161861 (mod 26)
Therefore the nine numbers are 13 04 18 04 11 06 18 06 01.
In this case, the message will be:
I AM BORING
This message has nine characters, a multiple of three.
The message is separated into groups of three letters and converted to numbers, where A=1, B=2 etc.
I A M B O R I N G
9 1 13 2 15 18 9 14 7
These groups will form matrices.
[9 1 13] [2 15 18] [9 14 7]
These matrices formed will be multiplied with the coding matrices. These will produce new set of numbers, which will be sent to the receiver.
IAM = 91131341841161861 = 355125181
-------------------------------------------------
= 172125
BOR = 215181341841161861 = 410281144
-------------------------------------------------
= 202114
ING = 91471341841161861 = 299232253
-------------------------------------------------
= 132419
-------------------------------------------------
Therefore the coded message is: 17 21 25 20 21 14 13 24 19
The last part of the message contains the decoding elements of the decryption matrix. This will save a lot of time for the receiver to figure out the inverse matrix for decryption.
To determine the inverse matrix, the Co-Matrix, that is, MATRIX T X TT has to be used as it was not yet changed. Determining the inverse by hand is difficult as it may take up to 10 steps. Instead, modern technology is able to calculate the inverse.

Calculator: Inverse = 1-26138-26677-3594138-359419081 = 108012082023 (mod 26)
This will be seen on the secret message as 01 00 08 00 01 20 08 20 23
Therefore, the whole message is:
03 03 13 04 18 04 11 06 18 06 0117 21 25 20 21 14 13 24 1901 00 08 00 01 20 08 20 23
However, the message looks continuous like this:
030090220026508200801050519240520152213010216000118000001
DECRYPTION
03 03 09 02 20 02 65 08 20 08 01 05 05 19 24 05 20 15 22 13 01 02 16 00 01 18 00 00 01
This message was received. We know that 03 03 represents the order of the matrix. To decrypt these numbers, a 3 x 3 matrix is used with elements
09 02 20 02 65 08 20 08 01.
922026582081
03 03 09 02 20 02 65 08 20 08 01 05 05 19 24 05 20 15 22 13 01 02 16 00 01 18 00 00 01
The last nine numbers are also a matrix, but for decoding, The matrix is used for decryption. The elements in this matrix are 01 02 16 00 01 18 00 00 01.
12160118001
03 03 09 02 20 02 65 08 20 08 01 05 05 19 24 05 20 15 22 13 01 02 16 00 01 18 00 00 01
This leaves the message part of the code: 05 05 19 24 05 20 15 22 13. Knowing a 3 x 3 matrix will be used, these numbers are grouped in to three as well.
5519 24520 152213
So, to decipher this message, the coding matrix and these groups must be multiplied together.

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