(Solved): a) In this exercise, we introduce you to Hill cipher. In this cipher, we convert our message to num ...

a) In this exercise, we introduce you to Hill cipher. In this cipher, we convert our message to numbers, just as in affine cipher. However, instead of encrypting character by character, we encrypt pairs of characters by multiplying them with an invertible matrix with co-efficients in ${\mathbf{Z}}_{26}$. Here is an example: Suppose we want to ENCRYPT "ALLISWELL". Since we require the plaintext to have even number of characters, we pad the message with the character ' $X$ '. We break up the message into pairs of characters AL, LI, SW, EL and LX. We convert each pair of characters into a pair elements in ${\mathbf{Z}}_{26}$ as follows: Next, we choose an inveritble $2\times 2$ matrix with coefficients in ${\mathbf{Z}}_{26}$, for example, $A=\left[\begin{array}{ll}\overline{3}& 1\\ \overline{7}& \overline{4}\end{array}\right]$. This matrix has determinant $\overline{3}?\overline{4}?\overline{7}?\overline{1}=\overline{5}$ and $\overline{5}$ is a unit in $Z_{26}$ with inverse $\overline{21}$. We write each pair of elements in $Z_{26}$ as a column vector and multiply it by $A$ : $A\left[\frac{\overline{0}}{11}\right]=\left[\frac{11}{18}\right],A\left[\begin{array}{c}11\\ \overline{8}\end{array}\right]=\left[\begin{array}{c}15\\ \overline{5}\end{array}\right],\dots$ We then convert each pair of numbers to a pair of characters and write them down. In this example, we get the cipher text "LSPFYGXUEN" corresponding to the plain text "ALLWELL". To decrypt, we convert pairs of characters to pairs of numbers and multiply by $A^{?1}={\overline{5}}^{?1}\left[\begin{array}{cc}\overline{4}& \overline{?1}\\ \overline{?7}& \overline{3}\end{array}\right]=\overline{21}\left[\begin{array}{cc}\overline{4}& \overline{?1}\\ \overline{?7}& \overline{3}\end{array}\right]=\left[\begin{array}{cc}\overline{6}& \overline{5}\\ \overline{9}& \overline{11}\end{array}\right]$ and we have $\left[\begin{array}{ll}\overline{6}& \overline{5}\\ \overline{9}& \frac{\overline{11}}{11}\end{array}\right]\left[\begin{array}{l}\frac{11}{18}\end{array}\right]=\left[\frac{\overline{0}}{18}\right],\dots$ Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill's cipher with the matrix $\left[\begin{array}{cc}\overline{3}& \overline{1}\\ \overline{0}& \overline{9}\end{array}\right]$ as the encryption matrix.