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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.

Bryden C
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1 Answers1

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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.

Sameer Kailasa
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    Wow, a superb solution (and very fast indeed!) – Peter Aug 03 '18 at 19:19
  • Amazing! (+1) $ $ – Nigel Overmars Aug 03 '18 at 19:20
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    You don't need the $10^{10}$; you can just say $$ \det\begin{pmatrix} 2 & 1 & 3 & 7 & 5 \ 3 & 8 & 7 & 9 & 8 \ 3 & 4 & 1 & 6 & 2 \ 4 & 0 & 2 & 2 & 3 \ 7 & 9 & 1 & 5 & 4 \end{pmatrix} = \det\begin{pmatrix} 2 & 1 & 3 & 7 & 21375 \ 3 & 8 & 7 & 9 & 38798 \ 3 & 4 & 1 & 6 & 34162 \ 4 & 0 & 2 & 2 & 40223 \ 7 & 9 & 1 & 5 & 79154 \end{pmatrix} = 19 \det\begin{pmatrix} 2 & 1 & 3 & 7 & 1125 \ 3 & 8 & 7 & 9 & 2042 \ 3 & 4 & 1 & 6 & 1798 \ 4 & 0 & 2 & 2 & 2117 \ 7 & 9 & 1 & 5 & 4166 \end{pmatrix}$$ – hmakholm left over Monica Aug 03 '18 at 19:25
  • Can you justify what's going on between the second to last and last steps? How does that first column change like that? – Bryden C Aug 03 '18 at 19:28
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    @BrydenC: Basic determinant rule: You can add any scalar multiple of a column to any other column without changing the determinant. – hmakholm left over Monica Aug 03 '18 at 19:29
  • Yes, that makes sense to me now. Thank you. – Bryden C Aug 03 '18 at 19:30
  • @HenningMakholm You are right, but the answer appears clearer to me. – Peter Aug 03 '18 at 19:36