Question

ENCRYPTION METHOD
The message to be encrypted is as follows: ‘Enter manual breathing mode’
Step 1: Convert secret message into sets of three (triplets).
ENT | ERM | ANU | ALB | REA | THI | NGM | ODE | NOW

Step 2: Assign each letter from the triplet sequence a value from the alphanumeric conversion table below and assign each triplet to a matrix of order 3×1. Triplet matrices are denoted by Cn, meaning the nth code triplet.
A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13

N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26

C_(1 = ) [■("5" @"14" @"20" )] 〖

Answer 1

ENCRYPTION METHOD
The message to be encrypted is as follows: ‘Enter manual breathing mode’
Step 1: Convert secret message into sets of three (triplets).
ENT | ERM | ANU | ALB | REA | THI | NGM | ODE | NOW

Step 2: Assign each letter from the triplet sequence a value from the alphanumeric conversion table below and assign each triplet to a matrix of order 3×1. Triplet matrices are denoted by Cn, meaning the nth code triplet.
A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13

N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26

C_(1 = ) [■("5" @"14" @"20" )] 〖 C〗_(2= ) [■("5" @"18" @"13" )] C_(3= ) [■("1" @"14" @"21" )] 〖 C〗_(4= ) [■("1" @"12" @"2" )] 〖 C〗_(5= ) [■("18" @"5" @"1" )] 〖 C〗_(6= ) [■("20" @"8" @"9" )] 〖 C〗_(7= ) [■("14" @"7" @"13" )] 〖 C〗_(8= ) [■("15" @"4" @"5" )] 〖 C〗_(9= ) [■("14" @"15" @"23" )]

Step 3: Create a cipher key matrix of order 3×3 utilising a birth date (i.e. 23 06 1995). Negative values were added in a checkerboard pattern for increased complexity. For memory and disguising purposes, the cipher key can be written as a letter-hyphen sequence using the alpha-numeric conversion table (0 representing a ‘?’) (The hyphens indicate a negative numeral). The two numerals on the end indicate the order of the cipher key matrix.
K = [■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )]" = ?-FA-BC-AJ-JE33"




Step 4: Pre-multiply coded triplet matrices by the cipher key matrix. Zn represents a matrix with values outside the range of the alpha-numeric conversion table numbers (i.e. 0 > Zn >26).Zn denotes the corresponding over-range matrix of the nth code triplet.
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("5" @"14" @"20" )]= [■(-64@12@19)]
KC1 = Z1
[-■("0" &"-6" &"1" @"2" &"3" &"-1" @"9" &"-9" &"5" )][■("1" @"12" @"22" )]= [■(@@)]
KC4 = Z4
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("14" @"7" @"13" )]= [■(@@)]
KC7 = Z7
[-■("0" &"-6" &"1" @"2" &"3" &"-1" @"9" &"-9" &"5" )][■("5" @"18" @"3" )]= [■(@@)]
KC2 = Z2
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("18" @"5" @"1" )]= [■(@@)]
KC5 = Z5
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("15" @"4" @"5" )]= [■(@@)]
KC8 = Z8
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("1" @"14" @"21" )]= [■(@@)]
KC3 = Z3
[-■("0" &"-6" &"1" @"2" &"3" &"-1" @"9" &"-9" &"5" )][■("20" @"8" @"9" )]= [■(@@)]
KC6 = Z6
[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )][■("14" @"15" @"23" )]= [■(@@)]
KC9 = Z9


Step 5: Reduce Z-matrices into the appropriate sized matrices with values between 0 and 26. A multiple of 26 will be subtracted from values in the Z-matrix which are over 26, and vice versa for values that are under.

[■(-64@12@19)]-[■("-3×26" @"0×26" @0×26)]=[■(14@12@19)]
Z1R1 = X1
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z4R4 = X4
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z7R7 = X7


[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z2R2 = X2
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z5R5 = X5
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z8R8 = X8


[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z3R3 = X3
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z6R6 = X6
[■(@@)]-[■("×26" @"×26" @×26)]=[■(@@)]
Z9R9 = X9

Step 6: the encrypted matrices (Xn) can now be converted from this numeric from do a letter-form.

C1 = [■(14@12@19)]"=" NLS
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="
C2 (Triplet 2) = [■(@@)]"="

Step 7: The encrypted message can now be written in the following sequence:
NLS |
DECRYPTION METHOD
Step 1: find the inverse of the key cipher matrix using expansion of the minors.
K = [■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )]= [■("a" _"11" &"a" _"12" &"a" _"13" @"a" _"21" &"a" _"22" &"a" _"23" @"a" _"31" &"a" _"32" &"a" _"33" )]
Therefore, the matrix of the cipher K is:
K-1 = "1" /|"K" | [■(|■("a" _"22" &"a" _"23" @"a" _"32" &"a" _"33" )|&|■("a" _"13" &"a" _"12" @"a" _"33" &"a" _"32" )|&|■("a" _"12" &"a" _"13" @"a" _"22" &"a" _"23" )|@|■("a" _"23" &"a" _"21" @"a" _"33" &"a" _"31" )|&|■("a" _"11" &"a" _"13" @"a" _"31" &"a" _"33" )|&|■("a" _"13" &"a" _"11" @"a" _"23" &"a" _"21" )|@|■("a" _"21" &"a" _"22" @"a" _"31" &"a" _"32" )|&|■("a" _"12" &"a" _"11" @"a" _"32" &"a" _"31" )|&|■("a" _"11" &"a" _"12" @"a" _"21" &"a" _"22" )| )]="1" /|"K" | [■(|■(3&"-1" @"-9" &"5" )|&|■(&@&)|&|■(&@&)|@|■(&@&)|&|■(&@&)|&|■(&@&)|@|■(&@&)|&|■(&@&)|&|■(&@&)| )]
The determinant of every 2x2 matrix within the above 3x3 matrix must be found. The formula for finding the determinant is as follows:
|■(&@&)|"=" |■("a" _"22" &"a" _"23" @"a" _"32" &"a" _"33" )|"= " [■("b" &"c" @"d" &"e" )]"= " |"be - cd" |"= " "a" _"22" "a" _"33" "- " "a" _"23" "a" _"32" "= "

Therefore, the the cipher key matrix only needs a determinant.
K-1 = "1" /|"K" | [■(&&@&&@&&)]



|K| is determined utilising the following process:
[■("a" _"11" &"a" _"12" &"a" _"13" @"a" _"21" &"a" _"22" &"a" _"23" @"a" _"31" &"a" _"32" &"a" _"33" )]"=" "a" _"11" [■("a" _"22" &"a" _"23" @"a" _"32" &"a" _"33" )]"-" "a" _"12" [■("a" _"21" &"a" _"23" @"a" _"31" &"a" _"33" )]"+" "a" _"13" [■("a" _"21" &"a" _"22" @"a" _"31" &"a" _"32" )]
(Upon simplifying) "=" "a" _"11" "a" _"22" "a" _"33" "+" "a" _"12" "a" _"23" "a" _"31" "+" "a" _"13" "a" _"21" "a" _"32" 〖"- a" 〗_"13" "a" _"22" "a" _"31" " –" 〖" a" 〗_"12" "a" _"21" "a" _"33" "- " 〖" a" 〗_"11" "a" _"23" "a" _"32"
|K|=[■("0" &"-6" &"1" @"-2" &"3" &"-1" @"9" &"-9" &"5" )]=("0×3×5" )"+ " ("-6×-1×9" )"+" ("1×-2×-9" )"- " ("1×3×9" )" - (6×-2×5)- (0×-9×-1)"
|K| = -15
The determinant is now known and thus the inverse of the matrix K can be found.
K-1 = 1/(-15) [■(&&@&&@&&)]=[■((-2)⁄5&(-7)⁄5&(-1)⁄5@(-1)⁄15&3⁄5&2⁄15@3⁄5&18⁄5&4⁄5)]
This matrix contains values that are do not correspond with those listed in the alpha-numeric conversion able, hence must be converted into a matrix that does:
[■((-2)⁄5&(-7)⁄5&(-1)⁄5@(-1)⁄15&3⁄5&2⁄15@3⁄5&18⁄5&4⁄5)] +[■("1×26" &"1×26" &"1×26" @"1×26" &"0×26" &"0×26" @0×26&"0×26" &"0×26" )]=[■(128⁄5&123⁄5&129⁄5@389⁄15&3⁄5&2⁄15@3⁄5&18⁄5&4⁄5)]

Therefore the inverse of the cipher key matrix K is [■(128⁄5&123⁄5&129⁄5@389⁄15&3⁄5&2⁄15@3⁄5&18⁄5&4⁄5)]