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Challenge: Create a 3x3 matrix of letters (NO REPEATS) such that the determinant of the matrix forms words from the 'multiplications'. For example, in a simple 2x2 case, take the matrix

$\begin{bmatrix}O & I\\N & R\end{bmatrix}$

the determinant of which is OR-IN. This forms 2 2-letter words. In the case of a 3x3, you will end up with 6 3-letter words if done correctly.

Don't know how to find the determinant of a matrix? Fear not, here's the guide (or read a more detailed explanation here):

$|A| = \begin{vmatrix} a & b & c\\d & e & f\\g & h & i \end{vmatrix} = a\begin{vmatrix} e & f\\h & i \end{vmatrix} - b\begin{vmatrix} d & f\\g & i \end{vmatrix} + c\begin{vmatrix} d & e\\g & h \end{vmatrix} = aei+bfg+cdh-ceg-bdi-afh . $

If that looks too complicated, here's another method that some find easier.

Don't worry about the additions and subtractions, I'm only concerned with the words formed. That said, if your words could be added and subtracted as base 26 numbers according to the determinant and result in a valid word, I would be incredibly impressed (a 100 reputation impressed, even). (This is mathematically impossible, determinant is always 0; sorry to get your hopes up). As usual, words are considered valid iff they are English words found on dictionary.com (read: this does not include (pre/suf)fixes, abbreviations, acronyms, etc.)

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  • $\begingroup$ You may want to add a rule you cannot use the same letters in each column. As the rules are now, pick any three letter word and write it in all columns. Result is 6 times said word, and would make a valid answer. EDIT: never mind, I missed the 'no repeats' part. $\endgroup$
    – Tim Couwelier
    Commented Sep 22, 2015 at 15:17
  • $\begingroup$ @TimCouwelier Good looking out, but I specify no repeated letters in the first sentence of the puzzle. $\endgroup$
    – NeedAName
    Commented Sep 22, 2015 at 15:20
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    $\begingroup$ Also, for anyone attempting this: just repeat the first two columns on the right of the 3x3 block. The 'words' are any diagonal length 3 line in that block, read from top to bottom. $\endgroup$
    – Tim Couwelier
    Commented Sep 22, 2015 at 15:20
  • $\begingroup$ Damnit.. I was about to post an answer, then realised I had ONE repeating letter. (O S B / I R P / A T T make for ort, spa, bit, bra, opt and sit) $\endgroup$
    – Tim Couwelier
    Commented Sep 22, 2015 at 15:27
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    $\begingroup$ If you treat the words as three-digit base 26 numbers instead of products, your "determinant" will always be zero because every letter is added and subtracted equally. $\endgroup$
    – f''
    Commented Sep 22, 2015 at 15:33

5 Answers 5

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Try

$\begin{bmatrix} B & S & L\\I & A & U\\M & D & T \end{bmatrix}$

This produces the following words:

BAT
SUM
LID
LAM
BUD
SIT

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  • $\begingroup$ As the first person to submit a valid answer, I'm accepting yours. Nicely done! $\endgroup$
    – NeedAName
    Commented Sep 22, 2015 at 15:42
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$\begin{bmatrix}F & C & P\\A & O & U\\T & N & B\end{bmatrix}$

making the words FOB, FUN, CAB, CUT, PAN, and POT. Each column on its own is also a word.

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A simple idea to generate solutions

Let the matrix be-

$\begin{bmatrix} a & b & c\\d & e & f\\g & h & i \end{bmatrix}$

Then we need the following to be valid words.

aei afh bdi bfg cdh ceg

This can be treated as a substitution cipher and solved using a brute force dictionary search. This produces lots of solutions. Here are 954 solutions. I stopped there because the engine I was using only shows top 1000 results (of which 954 were unique).

Edit:

I also tried a 4x4 matrix search. It didn't produce any complete result but some got close. The searching key is:

AFKP AFLO AGJP AGLN AHJO AHKN BEKP BELO BGIP BGLM BHIO BHKM CEJP CELN CFIP CFLM CHIN CHJM DEJO DEKN DFIO DFKM DGIN DGJM.

Try here at quipquip. However it might be possible if we allow some repetitions., which I haven't tried yet.

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  • $\begingroup$ That's a really cool application! So I'm curious, what happens if you try on a 4x4 matrix (I've been trying to see if it's possible). The resulting words are: A(FKP GLN HJO HKN GJP FLO) B(EKP GLM HIN HKM GIP ELO) C(EJP FLM HIN HJM FIP ELN) D(EJO FKM GIN GJM FIO EKN). Let me know if you have the time/energy to try it out $\endgroup$
    – NeedAName
    Commented Sep 22, 2015 at 20:02
  • $\begingroup$ quipqiup managed to get through the Cs with CARE CONS CULT CURS COLE CANT MIRE MONK MUDS MURK MODE MINT BILE BANK BUDS BULK BADE BINS, but didn't produce any valid solutions for the full matrix. $\endgroup$
    – f''
    Commented Sep 22, 2015 at 22:10
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To join the gang, here's what I came up with:

$\begin{bmatrix}H & C & N\\A & I & O\\P & G & T\end{bmatrix}$

Makes:

HIT, COP, NAG, NIP, HOG and CAT

Also note that:

The columns in this matrix, just like f'' his answer, form words on their own.

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Sooo, this riddle was stuck in my head for a while, because I really wanted to get a whole sentence in that matrix. Unfortunately it is not perfect yet, hopefuly others have any ideas to fix it.

So here is the matrix.

$|A| = \begin{vmatrix} i & a & o\\r & t & n\\f & g & m \end{vmatrix}$

Leading to

It Man Forgot Farming

Cheers!

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