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Calculate the determinant of the matrix $$ \begin{pmatrix} 10^{10} & 10^{10^{10}} & 11^{11^{11}} & 1 & 0 \\ 2^{2^2} & 3^{3^3} & 7^{7^7} & 0 & 1 \\ 11 & 17 & 12 & 2 & 7 \\ 2 & 3 & 5 & 1 & 1 \\ 9 & 14 & 7 & 1 & 6 \\ \end{pmatrix} $$

enter image description here

My wonderful Russian Professor put this one up on the board. Obviously it's not something one can put in Wolfram Alpha. I can't see any obvious linear dependencies between rows or columns, I've tried assigning variables to the big values (just for ease of notation) and doing row operations, transposing to get it into upper triangular it all still ends in a mess.

Now, I know our Prof HATES computation and is terrible at it, so he wouldn't write this up unless there is some structural simplicity I can't see...?

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    $\begingroup$ just going to take a wild guess and say it's 0. $\endgroup$
    – djechlin
    May 10, 2016 at 23:52
  • $\begingroup$ @djechlin I wasn't seeing the 2nd element of the 5th row as a 14... the answer is dependent on it. $\endgroup$
    – Benjamin R
    May 11, 2016 at 1:04
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – Martin Sleziak
    May 11, 2016 at 2:32
  • $\begingroup$ Is there a reason why some of the ones in your determinant are underlined? $\endgroup$
    – Martin Sleziak
    May 11, 2016 at 2:32
  • $\begingroup$ @MartinSleziak Sorry, I normally write everything in TeX. I didn't write this, it's literally a photo I took of the whiteboard from class. I can delete the question if you want? $\endgroup$
    – Benjamin R
    May 11, 2016 at 2:36

1 Answer 1

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Hint: look at the sum of the last two rows.

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    $\begingroup$ Is the fact that your professor is Russian ... so determinant ? $\endgroup$
    – Jean Marie
    May 10, 2016 at 23:25
  • $\begingroup$ @JeanMarie 😎😎😎😎😎😎😎😎 $\endgroup$
    – Benjamin R
    May 10, 2016 at 23:28
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    $\begingroup$ Ah, R3 = R4 + R5! $\endgroup$
    – Benjamin R
    May 10, 2016 at 23:41
  • $\begingroup$ This is assuming, of course, that the element at $\alpha_{52}$ = 14. My problem was I thought it was an 11 or 17... $\endgroup$
    – Benjamin R
    May 11, 2016 at 1:00
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    $\begingroup$ It does not look at all like how your prof writes 1's and 7's (though in fairness, the prof's 1's are not all consistent: probably the result of being away from Russia too long!). $\endgroup$
    – Robert Israel
    May 11, 2016 at 1:30

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