#10 on GRE Form 8767 In order to send an undetected message to an agent in the field, each letter in the message is replaced by the 
number of its position in the alphabet and that number is entered in a matrix M. Thus, for example, "DEAD" becomes the matrix $$M = \begin{pmatrix} 4 & 5 \\ 1 & 4\end{pmatrix} .$$ In order to further avoid detection, each message with four letters is sent to the agent encoded as $MC$, where $$C = \begin{pmatrix} 2 & -1 \\ 1 & 1\end{pmatrix} .$$ If the agent receives the matrix $$\begin{pmatrix} 51 & -3 \\ 31 & -8\end{pmatrix}$$ then the message is 
(A) RUSH (B) COME (C) ROME (D) CALL 
(E) not uniquely determined by the information given 
I know that the answer is C.  I am not sure what the negative number are suppose to represent. Sorry I do not know how to format correctly if you can fix it please do! The first two numbers of the matrix are the top and the second two numbers are the bottom.
 A: Inverse of $C$ is $$C^{-1} = \begin{pmatrix} 1/3 & 1/3 \\ -1/3 & 2/3\end{pmatrix} .$$ So $$MCC^{-1}=\begin{pmatrix} 51 & -3 \\ 31 & -8\end{pmatrix}\begin{pmatrix} 1/3 & 1/3 \\ -1/3 & 2/3\end{pmatrix}=\begin{pmatrix} 18& 15 \\13 & 5\end{pmatrix}, $$ which is ROME
A: Considering the GRE is done in pen and paper, computing inverses by hand, while easy if you know the formula, is perhaps slower than this process of elimination logic. Let
$$M = \begin{pmatrix} a & b \\ c & d\end{pmatrix} .$$
So in order for $MC$ to be equal to your desired matrix (i.e. have the last column negative), we must have
$a>b$ and $c>d$ 
This follows from the rules of matrix multiplication via the (-1,1) column in $C$. (C) and (D) satisfy the first inequality. However (D) does not satisfy the second. (C) ROME, is the only one that satisfies both. Given that $C$ is clearly invertable (you can tell $det(C) \neq0$ without actually calculating it explicitly) (E) cannot be the answer. This leaves only (C).
If you want to check your answer though, write down the M for ROME 
$$M = \begin{pmatrix} 18 & 15 \\ 13 & 5\end{pmatrix} .$$
and verify that MC does in fact give you the desired matrix. Explicitly computing the inverse as done in Francis's solution, might have been more time consuming depending on how fast you are at various parts of linear algebra.
