Why is the determinant of an integer matrix always an integer? Why is the determinant of an integer matrix always an integer? I am using the definition of the determinant as being the product of the diagonal entries of a matrix in upper triangular form.
 A: The explicit definition of determinant only uses multiplication and addition (and subtraction); it does not use division. When you multiply/add integers, you get, again, integers.
A: You can avoid fractions by something similar to Euclid's algorithm.   Rather than make the numbers below a pivot be zero, subtract a multiple of the pivot row so that all numbers below the pivot are smaller than the pivot.  Then choose the smallest nonzero number in that column as the new pivot.  The procedure ends of course, with just the greatest common divisor of the original values in the pivot place.  Then you go onto the next column.
A: When we try to reduce the matrix to upper triangular form, we encounter the problem that the element we are trying to eliminate may not be a multiple of the pivot.  We can deal with this by multiplying the row of that element by the pivot.  This multiplies the determinant by the pivot, but it also multiplies the diagonal element by the pivot.  Once we have reduced the matrix to triangular form, we can cancel the product of all the pivots from both sides.
I know I haven't expressed this very well, so let me do an example.
$$\begin{align}
D&=\begin{vmatrix}2&3&6\\4&1&2\\5&1&7\end{vmatrix}\\
D&=\begin{vmatrix}2&3&6\\0&-5&-10\\5&1&7\end{vmatrix}\tag{1}\\
2D&=\begin{vmatrix}2&3&6\\0&-5&-10\\10&2&14\end{vmatrix}\tag{2}\\
2D&=\begin{vmatrix}2&3&6\\0&-5&-10\\0&-13&-16\end{vmatrix}\\
10D&=\begin{vmatrix}2&3&6\\0&-5&-10\\0&-65&-80\end{vmatrix}\\
10D&=\begin{vmatrix}2&3&6\\0&-5&-10\\0&0&50\end{vmatrix}\\
10D&=2\cdot-5\cdot50=-500\\
D&=-50
\end{align}$$
In $(1)$, we can just subtract twice row $1$ from row $2$.  In $(2)$, we have the problem that $2\nmid5$, so we multiply row $3$ by $2$. The remainder is similar.
A: If every entry in the diagonal is an integer, let's say we have a $5$x$5$ upper-triangular matrix where the diagonal entries are $a,b,c, d,e $. The determinant of this matrix is $abcde$. A property of integers which you can probably use without proof is that two integers multiplied together is an integer. Because any product can be subdivided into smaller products, this shows that any product of integers will have to be an integer. Hence the determinant will be an integer.

If the matrix that we start with is not upper triangular, it is possible to make it upper triangular via row operations. It is possible for the result to have non-integer entries.
However, we can make a matrix upper triangular without getting $1$'s in the diagonal. i.e., we do not divide any rows by integers. Then the resuling marix will surely be an integer matrix. And the determinant will not be multiplied by anything other than integers ( multiplying a row by a number affect the value of the determinant), so the determinant will also be an integer.
