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.

  • $\begingroup$ Hi, perhaps try using the {pmatrix} environment. What is $M$? What is $C$? $\begin{pmatrix} 2 & -1 \\ 11 & ?\end{pmatrix}$. $\endgroup$
    – snar
    Apr 1, 2014 at 22:14
  • $\begingroup$ Thank you it was very helpfull $\endgroup$
    – Jane Doe
    Apr 1, 2014 at 22:20
  • $\begingroup$ Do you understand the concept of a matrix inverse? $\endgroup$ Apr 1, 2014 at 22:29
  • $\begingroup$ Hint: Cramer's Rule $\endgroup$ Apr 1, 2014 at 22:29

2 Answers 2


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


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.

  • $\begingroup$ For a $2\times 2$ matrix, you can just write down the inverse. $\endgroup$
    – D Wiggles
    Apr 2, 2014 at 6:01
  • $\begingroup$ I know, but it still involves (1) remembering the formula (2) some minor calculations including a matrix multiplication with fractions and convert it into letters (which takes some time, unless you know off the top of your head what the xth letter in the alphabet is). The method above takes almost no time if you don't bother to verify your answer (which is absolutely unnecessary). The key is that the method works even if you don't remember the formula for inverting a 2x2. $\endgroup$ Apr 2, 2014 at 6:48

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