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 $19$.