determinant of an integer matrix

[diogomgf]

I started studying yesterday the determinant of matrices.
There is this exercise where I'm asked to prove by induction that if [MATH]A \in M_{n}[/MATH] has all elements[MATH]\in Z[/MATH], then the [MATH]detA[/MATH] is also [MATH]\in Z[/MATH].

This is very intuitive, because the determinant is obtained only by multiplications and sums of integers in this specific case, so the end result should always be an integer.

To prove it using induction I started by:

[MATH]detA_{1} = a_{1,1}[/MATH], and this is an integer.
[MATH]detA_{p} \in Z \Rightarrow detA_{p+1} \in Z[/MATH].

[MATH] detA_{p} = a_{1,1} \cdot \hat{a}_{1,1} + \cdots + a_{1,p} \cdot \hat{a}_{1,p} [/MATH][MATH] detA_{p+1} = a_{1,1} \cdot \hat{a}_{1,1} + \cdots + a_{1,p} \cdot \hat{a}_{1,p} + a_{1,p+1} \cdot \hat{a}_{1,p+1} = detA + a_{1,p+1} \cdot \hat{a}_{1,p+1} = detA + a_{1,p+1} \cdot \begin{vmatrix}\ a_{1,1} & \cdots & a_{1,p} \\ \cdots & \cdots & \cdots \\ a_{p,1} & \cdots & a_{p,p} \ \end{vmatrix}[/MATH]
And I'm stuck here...

 

[pka]

Here is a most elementary explanation (not a proof) of how that works.
In the calculation of a determinate what are the operations used: multiplication & \(\pm~?\)
Note division (inverse multiplication) is never used. Thus the operation involved in determining the value in the integers always gives an integer,
If you doubt that do at least five examples one of which is a \(4\times 4\).
To evaluate the \(4\times 4\) we must do four \(3\times 3\). That gives an integer
To evaluate the \(3\times 3\) we must do three \(2\times 2\). That gives an integer etc.

 

[diogomgf]

Here is a most elementary explanation (not a proof) of how that works.
In the calculation of a determinate what are the operations used: multiplication & \(\pm~?\)
Note division (inverse multiplication) is never used. Thus the operation involved in determining the value in the integers always gives an integer,
If you doubt that do at least five examples one of which is a \(4\times 4\).
To evaluate the \(4\times 4\) we must do four \(3\times 3\). That gives an integer
To evaluate the \(3\times 3\) we must do three \(2\times 2\). That gives an integer etc.

...
This is very intuitive, because the determinant is obtained only by multiplications and sums of integers in this specific case, so the end result should always be an integer.
...