Crack the Matrix Code!

Algebra Level 3

Your friend gives you a code that reads: $(116, 85, 64, 43, 27, 18, 76, 57, 83, 60, 71, 52, 57, 38, 131, 93, 58, 39, 32, 23)$ He tells you that the original message is a math question written with words that he encoded using a $2 \times 2$ encoder matrix (see method outlined below). He won't tell you the numbers of the encoder matrix, but he does tell you that the first word of the original message is "WHAT".

Crack the code and give the numerical answer to the coded math question.

To encode a message using a square encoder matrix, first convert each letter to its corresponding number, then partition those numbers into groups that are the same length as the dimensions of the encoder matrix, and then multiply each partition by the encoder matrix. So to encode the message "BRILLIANT" with the square encoder matrix $\begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix}$ , first convert each of letters to its corresponding number (using $A = 1$ , $B = 2$ , $C = 3$ , and so on), so "BRILLIANT" converts to $(2, 18, 9, 12, 12, 9, 1, 14, 20)$ , because $B = 2$ , $R = 18$ , $I = 9$ , and so on. Then since the encoder matrix is a $2 \times 2$ matrix, partition these numbers into groups of $2$ (if the length of message is not the correct multiple, add $0$ s to the message until it is) and multiply each partition by the encoder matrix. Therefore, since

$\begin{pmatrix}2 & 18\end{pmatrix} \cdot \begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix} = \begin{pmatrix}74 & 130\end{pmatrix}$

$\begin{pmatrix}9 & 12\end{pmatrix} \cdot \begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix} = \begin{pmatrix}57 & 102\end{pmatrix}$

$\begin{pmatrix}12 & 9\end{pmatrix} \cdot \begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix} = \begin{pmatrix}48 & 87\end{pmatrix}$

$\begin{pmatrix}1 & 14\end{pmatrix} \cdot \begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix} = \begin{pmatrix}57 & 100\end{pmatrix}$

$\begin{pmatrix}20 & 0\end{pmatrix} \cdot \begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix} = \begin{pmatrix}20 & 40\end{pmatrix}$

the message "BRILLIANT" encodes to $(74, 130, 57, 102, 48, 87, 57, 100, 20, 40)$ .

To decode a message , partition the coded message into groups that are the same length as the dimensions of the encoder matrix, multiply each group by the inverse of the encoder matrix, and convert each number to its corresponding letter. So to decode $(74, 130, 57, 102, 48, 87, 57, 100, 20, 40)$ with the square encoder matrix $\begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix}$ , multiply each partitioned group of $2$ by the inverse of the encoder matrix $\begin{pmatrix}1 & 2 \\ 4 & 7\end{pmatrix}$ , which is $\begin{pmatrix}\frac{7}{1 \cdot 7 - 2 \cdot 4} & \frac{-2}{1 \cdot 7 - 2 \cdot 4} \\ \frac{-4}{1 \cdot 7 - 2 \cdot 4} & \frac{1}{1 \cdot 7 - 2 \cdot 4}\end{pmatrix} = \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix}$ . Since

$\begin{pmatrix}74 & 130\end{pmatrix} \cdot \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix} = \begin{pmatrix}2 & 18\end{pmatrix}$

$\begin{pmatrix}57 & 102\end{pmatrix} \cdot \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix} = \begin{pmatrix}9 & 12\end{pmatrix}$

$\begin{pmatrix}48 & 87\end{pmatrix} \cdot \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix} = \begin{pmatrix}12 & 9\end{pmatrix}$

$\begin{pmatrix}57 & 100\end{pmatrix} \cdot \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix} = \begin{pmatrix}1 & 14\end{pmatrix}$

$\begin{pmatrix}20 & 40\end{pmatrix} \cdot \begin{pmatrix}-7 & 2 \\ 4 & -1\end{pmatrix} = \begin{pmatrix}20 & 0\end{pmatrix}$

the coded message $(74, 130, 57, 102, 48, 87, 57, 100, 20, 40)$ decodes to $(2, 18, 9, 12, 12, 9, 1, 14, 20, 0)$ , and using $A = 1$ , $B = 2$ , $C = 3$ , and so on, $(2, 18, 9, 12, 12, 9, 1, 14, 20, 0)$ decodes to "BRILLIANT".

The answer is 81.

1 solution

David Vreken
Apr 24, 2018

Let the missing $2 \times 2$ decoder matrix be $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ . Since the first two letters of the decoded message are $W$ and $H$ , and $W = 23$ and $H = 8$ , and the first two numbers of the coded message are $116$ and $85$ , we have $\begin{pmatrix}116 & 85\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}23 & 8\end{pmatrix}$ which means $116a + 85c = 23$ and $116b + 85d = 8$ .

Since the second two letters of the decoded message are $A$ and $T$ , and $A = 1$ and $T = 20$ , and the second two numbers of the coded message are $64$ and $43$ , we have $\begin{pmatrix}64 & 43\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}1 & 20\end{pmatrix}$ which means $64a + 43c = 1$ and $64b + 43d = 20$ .

Solving the system of equations $116a + 85c = 23$ and $64a + 43c = 1$ gives $a = -2$ and $c = 3$ , and solving the system of equations $116b + 85d = 8$ and $64b + 43d = 20$ gives $b = 3$ and $d = -4$ . Therefore, the decoder matrix is $\begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix}$ .

Multiplying the rest of the message by the decoder matrix gives:

$\begin{pmatrix}27 & 18\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}0 & 9\end{pmatrix}$

$\begin{pmatrix}76 & 57\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}19 & 0\end{pmatrix}$

$\begin{pmatrix}83 & 60\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}14 & 9\end{pmatrix}$

$\begin{pmatrix}71 & 52\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}14 & 5\end{pmatrix}$

$\begin{pmatrix}57 & 38\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}0 & 19\end{pmatrix}$

$\begin{pmatrix}131 & 93\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}17 & 21\end{pmatrix}$

$\begin{pmatrix}58 & 39\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}1 & 18\end{pmatrix}$

$\begin{pmatrix}32 & 23\end{pmatrix} \cdot \begin{pmatrix}-2 & 3 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}5 & 4\end{pmatrix}$

for $(0, 9, 19, 0, 14, 9, 14, 5, 0, 19, 17, 21, 1, 18, 5, 4)$ .

Using $A = 1$ , $B = 2$ , $C = 3$ , and so on, and using the given first word of "WHAT", this message decodes to "WHAT IS NINE SQUARED" and the answer to that is $\boxed{81}$ .

Crack the Matrix Code!

The answer is 81.

1 solution

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