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

Question

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?

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I saw a resolution by the Jacobi method , but could not apply the methodology in this example.

Answer 1

$$ |D| = \begin{vmatrix} 1 & 1 & 9 \\ 1 & 8 & 7 \\ 1 & 5 & 3\end{vmatrix} = \begin{vmatrix} 1 & 1 & 9 \\ 0 & 7 & -2 \\ 0 & 4 & -6\end{vmatrix} = \begin{vmatrix} 1 & 1 & 9 \\ 0 & 7 & -2 \\ 0 & 0 & -\dfrac{34}{7}\end{vmatrix} = 1 \times 7 \times -\dfrac{34}{7} = -34 =-2 \times 17 $$

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.