# 0.2 Matrices  (Page 9/11)

 Page 9 / 11

Once again, we remind the reader that not every system of equations can be solved by the matrix inverse method. Although the Gauss-Jordan method works for every situation, the matrix inverse method works only in cases where the inverse of the square matrix exists. In such cases the system has a unique solution.

We summarize our discussion in the following table.

## The method for finding the inverse of a matrix

1. Write the augmented matrix $\left[\begin{array}{ccc}A& \mid & {I}_{n}\end{array}\right]$ .
2. Write the augmented matrix in step 1 in reduced row echelon form.
3. If the reduced row echelon form in 2 is $\left[\begin{array}{ccc}{I}_{n}& \mid & B\end{array}\right]$ , then $B$ is the inverse of $A$ .
4. If the left side of the row reduced echelon is not an identity matrix, the inverse does not exist.

## The method for solving a system of equations when a unique solution exists

1. Express the system in the matrix equation $\text{AX}=B$ .
2. To solve the equation $\text{AX}=B$ , we multiply on both sides by ${A}^{-1}$ .

## Application of matrices in cryptography

In this section, we see a use of matrices in encoding and decoding secret messages. There are many techniques used, but we will use a method that first converts the secret message into a string of numbers by arbitrarily assigning a number to each letter of the message. Next we convert this string of numbers into a new set of numbers by multiplying the string by a square matrix of our choice that has an inverse. This new set of numbers represents the coded message. To decode the message, we take the string of coded numbers and multiply it by the inverse of the matrix to get the original string of numbers. Finally, by associating the numbers with their corresponding letters, we obtain the original message.

In this section, we will use the correspondence where the letters A to Z correspond to the numbers 1 to 26, as shown below, and a space is represented by the number 27, and all punctuation is ignored.

 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

Use the matrix $A=\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]$ to encode the message: ATTACK NOW!

We divide the letters of the message into groups of two.

AT TA CK –N OW

We assign the numbers to these letters from the above table, and convert each pair of numbers into $2×1$ matrices. In the case where a single letter is left over on the end, a space is added to make it into a pair.

$\left[\begin{array}{c}A\\ T\end{array}\right]=\left[\begin{array}{c}1\\ \text{20}\end{array}\right]\left[\begin{array}{c}T\\ A\end{array}\right]=\left[\begin{array}{c}\text{20}\\ 1\end{array}\right]\left[\begin{array}{c}C\\ K\end{array}\right]=\left[\begin{array}{c}3\\ \text{11}\end{array}\right]\left[\begin{array}{c}_\\ N\end{array}\right]=\left[\begin{array}{c}\text{27}\\ \text{14}\end{array}\right]\left[\begin{array}{c}O\\ W\end{array}\right]=\left[\begin{array}{c}\text{15}\\ \text{23}\end{array}\right]$

So at this stage, our message expressed as $2×1$ matrices is as follows.

$\left[\begin{array}{c}1\\ \text{20}\end{array}\right]\left[\begin{array}{c}\text{20}\\ 1\end{array}\right]\left[\begin{array}{c}3\\ \text{11}\end{array}\right]\left[\begin{array}{c}\text{27}\\ \text{14}\end{array}\right]\left[\begin{array}{c}\text{15}\\ \text{23}\end{array}\right]$

Now to encode, we multiply, on the left, each matrix of our message by the matrix A. For example, the product of A with our first matrix is

$\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{c}1\\ \text{20}\end{array}\right]=\left[\begin{array}{c}\text{41}\\ \text{61}\end{array}\right]$

By multiplying each of the matrices in ( I ) by the matrix A, we get the desired coded message given below.

$\left[\begin{array}{c}\text{41}\\ \text{61}\end{array}\right]\left[\begin{array}{c}\text{22}\\ \text{23}\end{array}\right]\left[\begin{array}{c}\text{25}\\ \text{36}\end{array}\right]\left[\begin{array}{c}\text{55}\\ \text{69}\end{array}\right]\left[\begin{array}{c}\text{61}\\ \text{84}\end{array}\right]$

Decode the following message that was encoded using matrix $A=\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]$ .

$\left[\begin{array}{c}\text{21}\\ \text{26}\end{array}\right]\left[\begin{array}{c}\text{37}\\ \text{53}\end{array}\right]\left[\begin{array}{c}\text{45}\\ \text{54}\end{array}\right]\left[\begin{array}{c}\text{74}\\ \text{101}\end{array}\right]\left[\begin{array}{c}\text{53}\\ \text{69}\end{array}\right]$

Since this message was encoded by multiplying by the matrix A in [link] , we decode this message by first multiplying each matrix, on the left, by the inverse of matrix A given below.

${A}^{-1}=\left[\begin{array}{cc}3& -2\\ -1& 1\end{array}\right]$

For example,

$\left[\begin{array}{cc}3& -2\\ -1& 1\end{array}\right]\left[\begin{array}{c}\text{21}\\ \text{26}\end{array}\right]=\left[\begin{array}{c}\text{11}\\ 5\end{array}\right]$

By multiplying each of the matrices in ( II ) by the matrix ${A}^{-1}$ , we get the following.

$\left[\begin{array}{c}\text{11}\\ 5\end{array}\right]\left[\begin{array}{c}5\\ \text{16}\end{array}\right]\left[\begin{array}{c}\text{27}\\ 9\end{array}\right]\left[\begin{array}{c}\text{20}\\ \text{27}\end{array}\right]\left[\begin{array}{c}\text{21}\\ \text{16}\end{array}\right]$

Finally, by associating the numbers with their corresponding letters, we obtain the following.

$\left[\begin{array}{c}K\\ E\end{array}\right]\left[\begin{array}{c}E\\ P\end{array}\right]\left[\begin{array}{c}_\\ I\end{array}\right]\left[\begin{array}{c}T\\ _\end{array}\right]\left[\begin{array}{c}U\\ P\end{array}\right]$

And the message reads: KEEP IT UP.

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