Prove there exists a matrix with certain entries and determinant

[Arnold1]

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 $$A \in M(n,n;R)$$, such that $$m_{ij} \in \{-1,0,1\}$$ and $$det A=1995$$ (I think it can be any other number as well, but the book was printed in 1995 :) )

My problem is that I don't know what I should do to prove that there exist a certain matrix. I guess I should show such a matrix, but I don't know how to do that anyway.

 

[Opalg]

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 $$A \in M(n,n;R)$$, such that $$m_{ij} \in \{-1,0,1\}$$ and $$det A=1995$$ (I think it can be any other number as well, but the book was printed in 1995 :) )

My problem is that I don't know what I should do to prove that there exist a certain matrix. I guess I should show such a matrix, but I don't know how to do that anyway.

Let $A_n$ be the $n\times n$ matrix with $+1$ everywhere in the first column and everywhere on the main diagonal, $-1$ everywhere else on the top row, and $0$ in every other position. Its determinant is $$ |A_n| = \begin{vmatrix}1&-1&-1&-1&\ldots&-1 \\ 1&1&0&0&\ldots&0 \\ 1&0&1&0&\ldots&0 \\ 1&0&0&1&\ldots&0 \\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots \\ 1&0&0&0&\ldots&1 \end{vmatrix}$$ Now add each other row to the top row (which does not change the determinant). The top row then becomes $n\ \ 0\ \ 0\ \ 0\ldots$. If you expand along the top row, you see that $|A_n| = n.$ Finally, let $n=1995.$