Prove that matrix can be divided by 17

Question

We have been given $144228, 532270, 257567, 209270, 289017, 519792$ which are divisible by $17$. Show that: $$ \begin{matrix} 1 & 4& 4& 2& 2& 8 \\ 5 & 3 & 2 & 2 & 7 & 0\\ 2 & 5 &7&5& 6 &7 \\ 2 &0 &9 &2 &7 &0 \\ 2 &8 & 9 &0 &1 & 7 \\ 5 &1& 9 & 7 &9& 2 \\ \end{matrix} $$ is also divided by $17$.

Try to do it without calculating the determinant of the matrix.

Hint: $\Bbb{Z}$ and elementary elimination.

Ok, my idea is using elementary elimination get matrix looks like: $$ \begin{matrix} 1 & 4& 17x4& 2& 2& 8 \\ 5 & 3 & 17a2 & 2 & 7 & 0\\ 2 & 5 &17b7&5& 6 &7 \\ 2 &0 &17c9 &2 &7 &0 \\ 2 &8 & 17d9 &0 &1 & 7 \\ 5 &1& 17f9 & 7 &9& 2 \\ \end{matrix} $$ Then, can I take out $17$ before matrix? That's correct way? When I get $17$ before matrix is prove that is divided by $17$?

Answer 1

Think of the matrix as having coefficients in the field $\mathbb Z / 17 \mathbb Z.$ We have the column vector of integers $$ X = \left( \begin{array}{c} 100000 \\ 10000 \\ 1000 \\ 100 \\ 10 \\ 1 \end{array} \right) $$ which is not the zero vector in this vector space, all entries are nonzero $\pmod{17}.$ If your square matrix is called $A,$ we find $$ AX = 0. $$ That is, the matrix is not full rank, there is an eigenvalue $0.$ Regarded as over the integers, this means this determinant is divisible by $17$

Answer 2

$144228,532270,257567,209270,289017,519792=$

$\begin{bmatrix}10^{30}&10^{24}&10^{18}& 10^{12}&10^{6}&1 \end{bmatrix}\begin{bmatrix} 1 & 4& 4& 2& 2& 8 \\ 5 & 3 & 2 & 2 & 7 & 0\\ 2 & 5 &7&5& 6 &7 \\ 2 &0 &9 &2 &7 &0 \\ 2 &8 & 9 &0 &1 & 7 \\ 5 &1& 9 & 7 &9& 2 \\ \end{bmatrix} \begin{bmatrix}10^5\\10^4\\10^3\\10^2\\10\\1\end{bmatrix}$

Now convert to modulo 17

$\begin{bmatrix} 8&16&15&16&9&1 \end{bmatrix}\begin{bmatrix} 1 & 4& 4& 2& 2& 8 \\ 5 & 3 & 2 & 2 & 7 & 0\\ 2 & 5 &7&5& 6 &7 \\ 2 &0 &9 &2 &7 &0 \\ 2 &8 & 9 &0 &1 & 7 \\ 5 &1& 9 & 7 &9& 2 \\ \end{bmatrix} \begin{bmatrix}6\\4\\14\\15\\10\\1\end{bmatrix}$

And that product is divisible by 17.

However, if you multiply the just the right two matrices, you will get a column that is equivalent the zero vector in $\mathbb Z_{17}$ and there is no need to multiply the $3^{rd}$ matrix.

and if you want to keep the numbers a little smaller you could say:

$\begin{bmatrix} 1 & 4& 4& 2& 2& 8 \\ 5 & 3 & 2 & 2 & 7 & 0\\ 2 & 5 &7&5& 6 &7 \\ 2 &0 &9 &2 &7 &0 \\ 2 &8 & 9 &0 &1 & 7 \\ 5 &1& 9 & 7 &9& 2 \\ \end{bmatrix} \begin{bmatrix}6\\4\\-3\\-2\\10\\1\end{bmatrix}=\begin{bmatrix}34\\102\\68\\51\\34\\85\end{bmatrix}\equiv \mathbf 0 \pmod{17}$

Answer 3

my idea is using elementary elimination get matrix looks like ...

That's the right idea, with the note that adding a multiple of a column to another column does not change the value of the determinant.

Multiply the first column by $10^5$, the second one by $10^4$ and so on until the fifth column gets multiplied by $10$, then add all of them to the last column. This will get you the original $6$ numbers in the last column, each of them divisible by 17, so you can then factor $17$ out of the determinant.