# Cute Determinant Question

I stumbled across the following problem and found it cute.

Problem: We are given that $19$ divides $23028$, $31882$, $86469$, $6327$, and $61902$. Show that $19$ divides the following determinant:

$$\left| \begin{matrix} 2 & 3&0&2&8 \\ 3 & 1&8&8&2\\ 8&6&4&6&9\\ 0&6&3&2&7\\ 6&1&9&0&2 \end{matrix}\right|$$

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This is nice! May I ask about the source of the problem - where did you see it? – Martin Sleziak Jun 8 '12 at 12:17
Golan's linear algebra book. – Potato Jun 8 '12 at 16:04

Multiply the first column by $10^4$, the second by $10^3$, third by $10^2$ and fourth by $10$ - this will scale the value of the determinant by $10^{4+3+2+1}=10^{10}$, which is coprime to $19$. Now add the last four columns to the first one - this will not change the value of the determinant. Finally notice the first column now reads $23028, 31882, 86469, 6327$, and $61902$: each is a multiple of $19$ so we can factor a nineteen cleanly out the determinant.

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Perhaps you should add this example to en.wikipedia.org/wiki/Coprime to show an example for a usage for coprimes – mplungjan Jun 8 '12 at 9:14

If the determinant is $0$ it is obvious that $19|0$. Now suppose that the determinant is not $0$.

\begin{align*} 2\cdot10^4+3\cdot10^3+0\cdot10^2+2\cdot10+8\cdot1&=23028\\ 3\cdot10^4+1\cdot10^3+8\cdot10^2+8\cdot10+2\cdot1&=31882\\ 8\cdot10^4+6\cdot10^3+4\cdot10^2+6\cdot10+9\cdot1&=86469\\ 0\cdot10^4+6\cdot10^3+3\cdot10^2+2\cdot10+7\cdot1&=06327\\ 6\cdot10^4+1\cdot10^3+9\cdot10^2+0\cdot10+2\cdot1&=61902 \end{align*}

By Cramer's rule

$$1=\frac{\left|\begin{matrix} 2 & 3 & 0 & 2 & 23028 \\ 3 & 1 & 8 & 8 & 31882 \\ 8 & 6 & 4 & 6 & 86469 \\ 0 & 6 & 3 & 2 & 06327 \\ 6 & 1 & 9 & 0 & 61902 \end{matrix}\right|}{\left|\begin{matrix} 2 & 3 & 0 & 2 & 8 \\ 3 & 1 & 8 & 8 & 2 \\ 8 & 6 & 4 & 6 & 9 \\ 0 & 6 & 3 & 2 & 7 \\ 6 & 1 & 9 & 0 & 2\end{matrix}\right|}$$

Then

$$\left|\begin{matrix} 2 & 3 & 0 & 2 & 8 \\ 3 & 1 & 8 & 8 & 2 \\ 8 & 6 & 4 & 6 & 9 \\ 0 & 6 & 3 & 2 & 7 \\ 6 & 1 & 9 & 0 & 2\end{matrix}\right|=\left|\begin{matrix} 2 & 3 & 0 & 2 & 23028 \\ 3 & 1 & 8 & 8 & 31882 \\ 8 & 6 & 4 & 6 & 86469 \\ 0 & 6 & 3 & 2 & 06327 \\ 6 & 1 & 9 & 0 & 61902 \end{matrix}\right|$$

But last determinant is obviously divisible by $19$.

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I think that I must delete this answer because there is another answer which is accepted and has many up votes. – Gastón Burrull Jun 8 '12 at 5:32
No, yours is more detailed. You are fine. – Potato Jun 8 '12 at 5:43
Your answer is good. Multiple approaches can shed new light on a problem. – Potato Jun 8 '12 at 5:46
@GastónBurrull: Your answer is good. I would not have thought to use Cramer's rule. Please do not delete it. – MJD Jun 8 '12 at 17:55
Just saw this question since it was recently modified. Simpler, at least to me, is to remember that one can always add a multiple of one column to another and not change the determinant. So add $10^4$ times the first column plus $10^3$ times the second column plus $10^2$ times the third column plus $10^1$ times the fourth column to the fifth column. These column operations do not change the determinant and yield the last equation above. – robjohn Jul 12 '15 at 13:04

## Integer proof

Perform the column operation $C_5\leftarrow 10^4C_1+10^3C_2+10^3C_3+10C_4+C_5$: the coefficient of $C_5$ is $1$ so this doesn't change the determinant.

All elements of $C_5$ ($23028$, $31882$, $86469$, $6327$, and $61902$) are now divisible by $19$, so we can factor out $19$: hence the determinant is divisible by $19$.

## Modular proof

In $\mathbb Z/19\mathbb Z$, the columns $10^4C_1+10^3C_2+10^3C_3+10C_4+C_5$ sum to $0$: hence the matrix is not invertible and has determinant $0$. So in $\mathbb Z$, the determinant is a multiple of $19$.

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I like your modular proof! – Potato Jun 8 '12 at 19:15