determinant divisible by 13

Question

Question: Given: $195,403$ and $247$ are divisible by 13.

Prove (without actually calculating the determinant) that

$$\det \begin{bmatrix} 1 & 9 & 5 \\ 2 & 4 & 7 \\ 4 & 0 & 3 \end{bmatrix}$$

is divisible by 13.

What I did:

Apart from calculating the determinant and seeing that it's true, I couldn't really think of anything else...

Thanks

See also: http://math.stackexchange.com/questions/155439/cute-determinant-question — Martin Sleziak, Dec 08 '14 at 18:25

score 7 · Answer 1 · answered May 08 '13 at 17:42

7

Hint: If you multiply the first column by $100$, how does that change the determinant? Then if you add the third column to the first, how does that change the determinant?

answered May 08 '13 at 17:42

Ross Millikan

383,099

Did what you said, and still don't get it... I multiply the 1st column so the det is multiplied by 100. The adding the 3rd column doesn't change it. If I also add the 2nd to the first multiplied by 10 I can get the first column to be 195,247,403) and from that I can extract 13, but on the other side of the equation I still have 100detA - So it's not clearly divisible by 13 yet... – jreing May 08 '13 at 18:06
1

@user1685224: $100|A|$ is divisible by $13$ if and only if $|A|$ is. Having mulitplied the second column by $10$ the determinant is now multiplied by $1000$, but that doesn't change the argument. Now think about expanding the determinant by using the first column. What happens? – Ross Millikan May 08 '13 at 18:14
I think I got you, but I would love a complete answer, just to make sure. I don't think I've ever heard of that theorem you mention (if and only if...), I only knew it was true one way (x -> 100x) – jreing May 08 '13 at 18:25
1

@user1685224: because $1000$ does not have $13$ as a factor, $1000n$ can only have $13$ as a factor if $n$ does. It goes both ways for any number that doesn't have $2$ or $5$ as a factor. Since your matrix has integer entries, the determinant must be an integer. – Ross Millikan May 08 '13 at 18:33
Okay, Thanks alot! – jreing May 08 '13 at 18:39
1

Note that it is not necessary to use $\gcd(1000,13)=1$ since $\vec v_3 \leftarrow \vec v_3 + 10\vec v_2 + 100\vec v_1$ is a column operation that does not affect the determinant. – Erick Wong May 08 '13 at 19:03

score 1 · Accepted Answer · answered May 08 '13 at 18:56

We start by proving that $x$ is divisible by 13 if and only if $100x$ is also divisible by 13:
$\Rightarrow $ If x is divisible by 13 then $x=13n \Rightarrow 100x=100(13n)=13(100n)$ So it's divisible as well.
$\Leftarrow$ If 100x is divisible by 13 then $100x=13n$ but since 100 is not divisible by 13, $x$ must have 13 as a factor. Therefore $x$ is divisible by 13.
Now we prove what was asked: $100 detA=100\cdot det \begin{bmatrix} 1 & 9 & 5 \\ 2 & 4 & 7 \\ 4 & 0 & 3 \end{bmatrix}$= $det \begin{bmatrix} 100 & 9 & 5 \\ 200 & 4 & 7 \\ 400 & 0 & 3 \end{bmatrix}$= (elementary column operations that don't change the determinant) =$det \begin{bmatrix} 195 & 9 & 5 \\ 247 & 4 & 7 \\ 403 & 0 & 3 \end{bmatrix}$= (extracting 13 from the 1st column) $13 det\hat A$ ($det \hat A$ is the determinant after the extraction of 13 from the 1st column)
Now since $100detA$ is divisible by 13, $detA$ is also divisible by 13.

determinant divisible by 13

2 Answers2