Show that determinant of $\small\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$ is divisible by $19$ Using that the numbers
$228,
323$
 and
$456$
are
divisible
by
$19$.
Show
that
the
determinant of matrix
$\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}$
is
divisible
by
$19$.
 A: $$\det\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}22 & 2 & 8\\ 32& 2 & 3 \\ 45 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}\color{red}{228} & 2 & 8\\ \color{red}{323}& 2 & 3 \\ \color{red}{456} & 5 & 6\end{pmatrix}=\color{red}{19}\cdot\det\begin{pmatrix}12 & 2 & 8\\ 17& 2 & 3 \\ 24 & 5 & 6\end{pmatrix}.$$
A: Let $C_1,C_2,C_3$ be the columns, you want to find the determinant of the matrix $(C_1,C_2,C_3)$. It is the same as the determinant of the matrix $(C_1,C_2,C_3+10C_2+100C_1)$. The elements of the last column of this matrix are all divisible by $19$.
A: There's an alternative and maybe easier way to prove the result, with Gaussian elimination.
Consider the matrix with coefficients over $\mathbb{Z}/19\mathbb{Z}$; the inverse of $2$ is $10$, so
\begin{align}
\begin{bmatrix}
2 & 2 & 8 \\
3 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & 4 \\
3 & 2 & 3 \\
4 & 5 & 6
\end{bmatrix}
&&R_1\gets 10R_1
\\
&\to
\begin{bmatrix}
1 & 1 & 4 \\
0 & -1 & -9 \\
0 & 1 & 9
\end{bmatrix}
&&\begin{aligned}R_2\gets R_2-2R_1\\R_3\gets R_3-4R_1\end{aligned}
\end{align}
which clearly has rank $2$, so its determinant is $0$.
