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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 $19$.

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    Can you pesent what you have tried? – 2017-01-27
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    since the matrix is small it shouldn'T be to difficult to just calculate the determinant – 2017-01-27

3 Answers 3

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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 $19$.

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    Upvote after at most $12$ seconds, chapeau! A very nice idea! (+1) – 2017-01-27
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$$\det\begin{pmatrix}2 & 2 & 8\\ 3& 2 & 3 \\ 4 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}22 & 2 & 8\\ 32& 2 & 3 \\ 45 & 5 & 6\end{pmatrix}=\det\begin{pmatrix}\color{red}{228} & 2 & 8\\ \color{red}{323}& 2 & 3 \\ \color{red}{456} & 5 & 6\end{pmatrix}=\color{red}{19}\cdot\det\begin{pmatrix}12 & 2 & 8\\ 17& 2 & 3 \\ 24 & 5 & 6\end{pmatrix}.$$

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    This is essentially the same as Jorge's prior answer. – 2017-01-27
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    @BillDubuque: I cannot say the opposite, we were just typing at the same time, and he was faster than me :) – 2017-01-27
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    I just didn't use matrices haha – 2017-01-27
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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 $0$.