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To encode the message I 'Return at noon' in a matrix first make a 1 to 1 mapping from the letters of the alphabet to the set of integers. One possible mapping is
\[a \rightarrow 1, \; b \rightarrow 2,..., \; z \rightarrow 26\]

The message becomes 18, 5, 20, 21, 18, 14, 1, 20, 14, 15, 15, 14. We can write this as a matrix with 4 rows and 3 columns.
\[ \left( \begin{array}{cccc} 18 & 5 & 20 \\ 21 & 18 & 14 \\ 1 & 20 & 14 \\ 15 & 15 & 14 \end{array} \right) =A\]

Now take any 3 by 3 matrix  
\[B\]
  subject to the conditions
1. The entries of  
\[B\]
  are integers.
2.  
\[A\]
  is invertible and the entries of  
\[A^{-1}\]
  are integers.
We can take  
\[B= \left( \begin{array}{ccc} 1 & 2 & -1 \\ 2 & 5 & 2 \\ -1 & -2 & 2 \end{array} \right)\]
.
Then  
\[B^{-1}= \left( \begin{array}{ccc} 14 & -2 & 9 \\ -6 & 1 & -4 \\ -1 & 0 & 1 \end{array} \right) \]
.
\[AB=\left( \begin{array}{cccc} 18 & 5 & 20 \\ 21 & 18 & 14 \\ 1 & 20 & 14 \\ 15 & 15 & 14 \end{array} \right) \left( \begin{array}{ccc} 1 & 2 & -1 \\ 2 & 5 & 2 \\ -1 & -2 & 2 \end{array} \right)= \left( \begin{array}{cccc} 18 & 12 & 32 \\ 43 & 104 & 43 \\ 27 & 74 & 67 \\ 31 & 77 & 43 \end{array} \right)=C\]

To decode the message find  
\[A^{-1}C\]
  and use the inverse mapping, which must be sent separately.