1

Use the fact that 11322, 13209, 29189, 56661, and 15096 are all divisible by 17 to show that

$$\begin{vmatrix} 1&1&2&5&1\\ 1&3&9&6&5\\ 3&2&1&6&0\\ 2&0&8&6&9\\ 2&9&9&1&6 \end{vmatrix}$$

is divisible by 17 without directly evaluating the determinant.

kevinbobbkoh
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  • @QiaochuYuan I tried transposing the matrix so that each row will correspond to the numbers provided. And got stuck there. – kevinbobbkoh Sep 19 '17 at 03:50
  • The Answers here are significantly better than the single Answer provided on the earlier Question, so on that account I'm voting to Reopen (and if successful, would recommend marking the earlier Question as a duplicate of this one). – hardmath Sep 20 '17 at 03:35

3 Answers3

5

The determinant of a matrix after adding a multiple of one row to another remains the same.

So you can add $10^4$ of the first row, $10^3$ of the second, $10^2$ of the third and $10^1$ of the fourth row to the fifth row, and the determinant will be preserved. This will give you the last row as

$$(\begin{matrix}11322 & 13209 & 29189 & 56661 & 15096\end{matrix})$$

Afterwards, do cofactor expansion along the fifth row to get

$$\det = 11322A_{5,1} + 13209A_{5,2} + 29189A_{5, 3} + 56661A_{5, 4} + 15096A_{5, 5}$$

which is divisible by $17$ by the divisibility of linear combinations.

Yiyuan Lee
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2

$$ \begin{vmatrix} 1 & 1& 2& 5& 1\\ 1 &3&9&6&5 \\ 3&2&1&6&0\\ 2&0&8&6&9\\ 2&9&9&1&6 \end{vmatrix} = 10^{-10}\begin{vmatrix} 1 \times 10^4 & 1 \times 10^4& 2\times 10^4& 5\times 10^4& 1\times 10^4\\ 1\times 10^3 &3\times 10^3&9 \times 10^3&6 \times 10^3&5 \times 10^3 \\ 3 \times 10^2&2\times 10^2&1\times 10^2&6 \times 10^2&0 \times 10^2\\ 2 \times 10&0 \times 10&8 \times 10&6 \times 10&9 \times 10\\ 2&9&9&1&6 \end{vmatrix} \\ = 10^{-10}\begin{vmatrix} 11322 & 13209& 29189& 56661& 15096\\ 1\times 10^3 &3\times 10^3&9 \times 10^3&6 \times 10^3&5 \times 10^3 \\ 3 \times 10^2&2\times 10^2&1\times 10^2&6 \times 10^2&0 \times 10^2\\ 2 \times 10&0 \times 10&8 \times 10&6 \times 10&9 \times 10\\ 2&9&9&1&6 \end{vmatrix} $$

by adding to the first row, all the other rows. This is a multiple of $17$ as the top row is a multiple of $17$. Furthermore, since the first determinant is an integer, the rest also are integers, despite the fact that there is a division by $10^{-10}$.

As if to confirm, the answer to this determinant is $1666 = 98 \times 17$.

Sarvesh Ravichandran Iyer
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  • Could you explain why is there a division by $10^{10}?$ – kevinbobbkoh Sep 19 '17 at 03:57
  • He multiply each row by a multiple of 10 the total is 10^10 – user577215664 Sep 19 '17 at 03:59
  • When I multiply a row by a constant , in order to keep equality, I also have to divide by that constant, right? In this case, I multiplied the first row by $10^4$, the second one by $10^3$, the third one by $10^2$ and the fourth one by ten. So, we have to divide by $10^4$ for the first row, $10^3$ for the second, $10^2$ for the third and $10$ for the fourth. Putting these divided numbers together means we have to divide by $10^{1+2+3+4} = 10^{10}$. And division by $10^{10}$ is the same as multiplication by $10^{-10}$. – Sarvesh Ravichandran Iyer Sep 19 '17 at 04:01
  • @астонвіллаолофмэллбэрг now that you have a row that is a multiple of 17 you dont need these multiples of 10 inside the determinant ...so it's just 10^-4 – user577215664 Sep 19 '17 at 04:03
  • @Isham Yes, so I can remove the multiples of $10$ which occur in the second third, and fourth rows and therefore reduce the $10^{-10}$ to $10^{-4}$, but that doesn't make a difference, since I've already solved the problem. – Sarvesh Ravichandran Iyer Sep 19 '17 at 04:05
  • @астонвіллаолофмэллбэрг yep your solution is great +1 – user577215664 Sep 19 '17 at 04:06
2

If $M$ is your matrix, considered over the field $\mathbb Z/17 \mathbb Z$, these facts tell you that $$M^T \pmatrix{10^4\cr 10^3\cr 10^2\cr 10\cr 1\cr} = 0 $$ and therefore that $M$ is not invertible over that field.

Robert Israel
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