Prove that the determinant is a multiple of $17$ without developing it Let, matrix is given as :
$$D=\begin{bmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{bmatrix}$$

Prove that the determinant is a multiple of $17$ without developing it?


I saw a resolution by the Jacobi method , but could not apply the methodology in this example.
 A: Notice that 119, 187 and 153 are all divisible by 17. So multiplying column 2 by 10 and adding to column 3 and multiplying column 1 by 100 and adding to column 3, gives us a column in which each element is divisible by 17:
$D=\left|\begin{matrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{matrix}\right|
=\left|\begin{matrix}
1 & 1 & 19 \\
1 & 8 & 87 \\
1 & 5 & 53\end{matrix}\right|
=\left|\begin{matrix}
1 & 1 & 119 \\
1 & 8 & 187 \\
1 & 5 & 153\end{matrix}\right|
=17\left|\begin{matrix}
1 & 1 & 7 \\
1 & 8 & 11 \\
1 & 5 & 9\end{matrix}\right|$
Thus $D = 17\cdot E$ where $E$ is the determinant of a matrix whose elements are integers which multiplied out using the definition of a determinant will be an integer.
A: $$
|D|
=
\begin{vmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 4 & -6\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 0 & -\dfrac{34}{7}\end{vmatrix}
=
1 \times 7 \times -\dfrac{34}{7}
=
-34
=-2 \times 17
$$
A: Beginning like H.R. but then using the fact that you're only interested in the result mod $17$, you could do without rationals by writing 
$$
|D|
=
\begin{vmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 4 & -6\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 2 & -3 \\
0 & 4 & -6
\end{vmatrix}
=
0\bmod17\;,
$$
where I added $3$ times the last row to the middle row.
A: Jacobi says
$$D=\left|\begin{matrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{matrix}\right|
=\left|\begin{matrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 4 & -6\end{matrix}\right|
=\left|\begin{matrix}
1 & 0 & \frac{65}7 \\
0 & 7 & -2 \\
0 & 0 & -\frac{34}7\end{matrix}\right|
=\left|\begin{matrix}
1 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & -\frac{34}7\end{matrix}\right|
$$
but this takes more operations than Gauss.
A: Note: Adding a multiple of one row to another (and those were the only "row operations" used here) does not change the determinant of a matrix (and those were the only "row operations" used here) but "swap two rows" multiplies the determinant by -1 and "multiply a row by a number" multiplies the determinant by that number.
