We consider matrices $M$ of the form $M = \left( \begin{array} { l l } a & b \\ 5 & 3 \end{array} \right)$ where $a$ and $b$ are integers. The number $3 a - 5 b$ is called the determinant of $M$. We denote it $\operatorname { det } ( M )$. Thus $\operatorname { det } ( M ) = 3 a - 5 b$.
In this question we assume that $\operatorname { det } ( M ) \neq 0$ and we set $N = \frac { 1 } { \operatorname { det } ( M ) } \left( \begin{array} { c c } 3 & - b \\ - 5 & a \end{array} \right)$. Justify that $N$ is the inverse of $M$.
We consider the equation $( E ) : \quad \operatorname { det } ( M ) = 3$. We wish to determine all pairs of integers ( $a ; b$ ) that are solutions of equation ( $E$ ). a. Verify that the pair (6; 3) is a solution of $( E )$. b. Show that the pair of integers ( $a$; $b$ ) is a solution of ( $E$ ) if and only if $3 ( a - 6 ) = 5 ( b - 3 )$. Deduce the set of solutions of equation ( $E$ ).
Part B
We set $Q = \left( \begin{array} { l l } 6 & 3 \\ 5 & 3 \end{array} \right)$. Using Part A, determine the inverse matrix of $Q$.
Encoding with matrix $Q$ To encode a two-letter word using the matrix $Q = \left( \begin{array} { l l } 6 & 3 \\ 5 & 3 \end{array} \right)$ we use the following procedure: Step 1: We associate with the word the matrix $X = \binom { x _ { 1 } } { x _ { 2 } }$ where $x _ { 1 }$ is the integer corresponding to the first letter of the word and $x _ { 2 }$ the integer corresponding to the second letter of the word according to the correspondence table below:
A
B
C
D
E
F
G
H
I
J
K
L
M
0
1
2
3
4
5
6
7
8
9
10
11
12
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
13
14
15
16
17
18
19
20
21
22
23
24
25
Step 2: The matrix $X$ is transformed into the matrix $Y = \binom { y _ { 1 } } { y _ { 2 } }$ such that $Y = Q X$. Step 3: The matrix $Y$ is transformed into the matrix $R = \binom { r _ { 1 } } { r _ { 2 } }$ such that $r _ { 1 }$ is the remainder of the Euclidean division of $y _ { 1 }$ by 26 and $r _ { 2 }$ is the remainder of the Euclidean division of $y _ { 2 }$ by 26.
\section*{Part A}
We consider matrices $M$ of the form $M = \left( \begin{array} { l l } a & b \\ 5 & 3 \end{array} \right)$ where $a$ and $b$ are integers.\\
The number $3 a - 5 b$ is called the determinant of $M$. We denote it $\operatorname { det } ( M )$.\\
Thus $\operatorname { det } ( M ) = 3 a - 5 b$.
\begin{enumerate}
\item In this question we assume that $\operatorname { det } ( M ) \neq 0$ and we set $N = \frac { 1 } { \operatorname { det } ( M ) } \left( \begin{array} { c c } 3 & - b \\ - 5 & a \end{array} \right)$. Justify that $N$ is the inverse of $M$.
\item We consider the equation $( E ) : \quad \operatorname { det } ( M ) = 3$.
We wish to determine all pairs of integers ( $a ; b$ ) that are solutions of equation ( $E$ ).\\
a. Verify that the pair (6; 3) is a solution of $( E )$.\\
b. Show that the pair of integers ( $a$; $b$ ) is a solution of ( $E$ ) if and only if $3 ( a - 6 ) = 5 ( b - 3 )$.\\
Deduce the set of solutions of equation ( $E$ ).
\end{enumerate}
\section*{Part B}
\begin{enumerate}
\item We set $Q = \left( \begin{array} { l l } 6 & 3 \\ 5 & 3 \end{array} \right)$.
Using Part A, determine the inverse matrix of $Q$.
\item Encoding with matrix $Q$
To encode a two-letter word using the matrix $Q = \left( \begin{array} { l l } 6 & 3 \\ 5 & 3 \end{array} \right)$ we use the following procedure:\\
Step 1: We associate with the word the matrix $X = \binom { x _ { 1 } } { x _ { 2 } }$ where $x _ { 1 }$ is the integer corresponding to the first letter of the word and $x _ { 2 }$ the integer corresponding to the second letter of the word according to the correspondence table below:
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
A & B & C & D & E & F & G & H & I & J & K & L & M \\
\hline
0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline\hline
N & O & P & Q & R & S & T & U & V & W & X & Y & Z \\
\hline
13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\
\hline
\end{tabular}
\end{center}
Step 2: The matrix $X$ is transformed into the matrix $Y = \binom { y _ { 1 } } { y _ { 2 } }$ such that $Y = Q X$.\\
Step 3: The matrix $Y$ is transformed into the matrix $R = \binom { r _ { 1 } } { r _ { 2 } }$ such that $r _ { 1 }$ is the remainder of the Euclidean division of $y _ { 1 }$ by 26 and $r _ { 2 }$ is the remainder of the Euclidean division of $y _ { 2 }$ by 26.
\end{enumerate}