I stumbled upon this (probably simple) mathematical matrix I just couldn't solve. The problem is I came up with the answer 12 (which isn't one) and now I am so fixated on that that I am having trouble seeing other solutions :P. Can somebody solve it for me? The first image is the question, the second is the possible answers.


Possible answers

  • $\begingroup$ ohhhhhh..... I am a little slow (probably not the greatest idea to take these tests, sort of an ego killer). I got the solution and I will post it, though I doubt any of you intellectual superiors will need such. $\endgroup$ – Jordan Jun 23 '18 at 4:55
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    $\begingroup$ This is not a very good IQ test question since there are many possible answers that are equally valid. They could be tribonacci numbers, lazy caterer sequence, or Fibonacci numbers minus one. Or any one of several sequences containing 1,2,4,7. $\endgroup$ – Riley Jun 23 '18 at 5:00
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    $\begingroup$ Maybe the answer is "secret hidden answer #7: "All of the above." $\endgroup$ – Chowzen Jun 23 '18 at 16:38

The answer is:



Any given diagonal contains the same number. You can get from one diagonal to the next by adding an incrementing number - +1, +2, +3 or +4 (1+1 → 2+2 → 4+3 → 7+4 → 11)

  • $\begingroup$ Plot twist... my solution is the actual one. I just tried to prove it lmao. $\endgroup$ – Jordan Jun 24 '18 at 0:15
  • $\begingroup$ And like Riley said, it doesn't necessarily mean an incrementing number. It could be tribonacci, thus 13 $\endgroup$ – Jordan Jun 24 '18 at 0:16
  • $\begingroup$ I thought of 11 this way: Ignore the matrix and think of the numbers in one row; 1, 2, 4, 7, 11. It's simply +1, +2, +3, +4! And then the matrix is just going through each triplet. Point being, this is a pretty bad IQ question since not only do you have multiple answers, you have multiple ways of getting to the same answer, which isn't useful for a question testing IQ. $\endgroup$ – Aryaman Jun 24 '18 at 10:47

The solution is pretty simple.

All numbers stay the same in diagonals (obvious). To find numbers in the third column, we simply take the sum of the two other columns in a given row and the value of the cell in the row above in the first column. Thus, 2+4+7=13. :)


As Riley informed me... Its better to look at this as tribonacci numbers. In effect, we take the sum of each number in a row and that is equal to that of the third column in the row below...

  • $\begingroup$ Both of your solutions are trying to generalise based on a single observation (f(X) = Y, f(Z) = ?), which may be technically correct, but seems unlikely to be the solution the question creator had in mind. $\endgroup$ – NotThatGuy Jun 23 '18 at 8:24

Well, I guess the answer is



With 1, we get a symmetrical matrix, the symmetry is being formed by the main diagonal from top left to bottom right (1,4,1)

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    $\begingroup$ Very funny, ha-ha. Could you please name a number (any number at all, no need to limit yourself to the given options) that could be substituted for the question mark so that the result would not be a symmetric matrix? $\endgroup$ – Bass Jun 23 '18 at 20:13
  • $\begingroup$ This confirms to Riley's comment. So 1 is just an example. It can be any number for that sake. $\endgroup$ – Mea Culpa Nay Jun 24 '18 at 6:21

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