Prove there exists a matrix

Hi. Here is a problem I found in my algebra book and I don't know how to solve it. Could you please help me?

Show that there exists a matrix $\displaystyle A \in M(n,n;R)$, such that $\displaystyle m_{ij} \in \{-1,0,1\}$ and $\displaystyle det M=1995$

My problem is that I don't know what I should do to __prove__ that there __exist__ a certain matrix.

Re: Prove there exists a matrix

I may be oversimplifying things, but I think that all you need to do is show a matrix for which this is the case.

You need to find a square matrix (obviously), in which all entries that aren't on the diagonal are either -1, 0 or 1.

This means you could just give a diagonal or triangular matrix, with values on the diagonal that when multiplied together, amount to 1995.

Re: Prove there exists a matrix

Quote:

Originally Posted by

**Zangeki** You need to find a square matrix (obviously), in which all entries **that aren't on the diagonal** are either -1, 0 or 1.

Why did you decide that the diagonal elements can be something other than -1, 0 or 1?

Quote:

Originally Posted by

**Zangeki** This means you could just give a diagonal or triangular matrix, with values on the diagonal that when multiplied together, amount to 1995.

Well, if you multiply -1, 0 and 1, you won't get 1995...

Edit: Added the first remark.

Re: Prove there exists a matrix

Yes, you are over simplifying. The determinant of a diagonal or triangular matrix is just the product of the numbers on the diagonal.

If the numbers in a **diagonal** or **triangular** matrix are only -1, 0, or 1, the only possible determinants are -1, 0, and 1 so that is not possible.

Re: Prove there exists a matrix

Ah, I assumed **i** couldn't be equal to **j**. My bad.

Re: Prove there exists a matrix

I'm sorry. I made a mistake above but I cannot edit the post. There should be detA=1995, not detM

Re: Prove there exists a matrix

I have replied to this problem in another forum.