<s>[INST] Rows of a Matrix is divisible by 19, show that its Determinant is also divisible by 19 I came across the following problem while self studying:
Let 
\begin{equation}
A = 
\begin{bmatrix}
2 & 1 & 3 & 7 & 5\\
3 & 8 & 7 & 9 & 8\\
3 & 4 & 1 & 6 & 2\\
4 & 0 & 2 & 2 & 3\\
7 & 9 & 1 & 5 & 4\\
\end{bmatrix}
\end{equation}
Use the fact that 21375, 38798, 34162, 40223, and 79154 are divisible by 19 to show, without evaluating, that $\det[A]$ is divisible by 19.
I noticed that each of these numbers are the entries in the rows of A, but I don't see how that helps me.
 [/INST] Note that $$10^{10} \det(A) = 10^{4+3+2+1+0} \det(A) =  \det \begin{pmatrix} 2 \cdot 10^{4} & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\
3\cdot 10^{4} & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\
3\cdot 10^{4} & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\
4\cdot 10^{4} & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\
7 \cdot 10^{4} & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4  \end{pmatrix}$$  $$ =  \det \begin{pmatrix} 21375 & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\
38798 & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\
34162 & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\
40223 & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\
79154 & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4  \end{pmatrix}$$ which is evidently divisibly by $19$ when calculated via expansion by minors along the first column. 
</s>