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I take this matrix and try put in Smith Normal Form, but the numbers in diagonal not follow disivibility rules. So I make myself a question:

if every matrix has a Smith Normal Form, how to proceed when diagonal matrix not follow disivibility rules?

What is operation we can make to solve this?

\begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 2 & 2 & 13 \end{bmatrix} this matrix belongs to $M(\mathbb{Z})$.

I hope be clear. And so thankfull for any help.

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The smith normal form is $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&22\end{pmatrix}$$

The first step is to use row and column operations to reduce the first row and column, the result of that is $$\begin{pmatrix}1&0&0\\0&-2&-4\\0&-2&11\end{pmatrix}$$ Now we can operate with the $2\times 2$ submatrix reduction giving $$\begin{pmatrix}-2&0\\0&11\end{pmatrix}$$ Next we have the row/column operations, $$\begin{pmatrix}-2&11\\0&11\end{pmatrix}$$

$$\begin{pmatrix}-2&1\\0&11\end{pmatrix}$$

$$\begin{pmatrix}0&1\\22&11\end{pmatrix}$$

$$\begin{pmatrix}0&1\\22&0\end{pmatrix}$$

Thus after some transpositions, the result above.

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  • $\begingroup$ To expand on concern of OP if the 1 in top left of the second matrix had not been a 1 Rene could have used the Euclidean algorithm to ensure that it divided every element of the 2 by 2 submatrix in bottom right corner. This ensures you can always achieve required divisibility property. $\endgroup$ – Patrick Apr 6 '18 at 22:33

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