P's question at Yahoo Answers (Determinant of order n)

[Fernando Revilla]

Here is the question:

okay so I need the equation of a determinate matrix that is nxn and has the number "1" from the bottom left corner to the upper right corner for any nxn matrix. to it would be something like [0,0,1] for the top row, [0,1,0] for the second row, and [1,0,0] for the bottom row. any help would be appreciated!

Here is a link to the question:

Matrix with determinates? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello p,

Using a well known determinant's property: $$\Delta_k=\begin{vmatrix} 0 & 0 & \ldots & 0 & 1\\ 0 & 0 & \ldots & 1 & 0 \\ \vdots&&&&\vdots \\ 0 & 1 &\ldots & 0 & 0\\\boxed{1} & 0 &\ldots & 0 & 0\end{vmatrix}=(-1)^{k+1}\begin{vmatrix} 0 & \ldots & 0 & 1\\ 0 & \ldots & 1 & 0 \\ \vdots&&&\vdots \\ 1 &\ldots & 0 & 0\end{vmatrix}=(-1)^{k+1}\Delta_{k-1}$$ Reiterating: $$\begin{aligned}\Delta_n&=(-1)^{n+1}\Delta_{n-1}\\&=(-1)^{n+1}(-1)^{n}\Delta_{n-2}\\&=(-1)^{n+1}(-1)^{n}(-1)^{n-1}\Delta_{n-3}\\&\ldots\\&=(-1)^{n+1}(-1)^{n}(-1)^{n-1}\; \ldots\; (-1)^4\Delta_{2}\\&=(-1)^{n+1}(-1)^{n}(-1)^{n-1}\;\ldots\; (-1)^4\begin{vmatrix}{0}&{1}\\{1}&{0}\end{vmatrix}\\&=(-1)^{n+1}(-1)^{n}(-1)^{n-1}\;\ldots\; (-1)^4(-1)^3\\&=(-1)^{(n+1)+n+(n-1)+\ldots +3}
\end{aligned}$$
We have the sum of the terms of an arithmetic progression $$(n+1)+n+(n-1)+\ldots +3=\dfrac{3+(n+1)}{2}(n-1)=\dfrac{(n+4)(n-1)}{2}
$$ So, $\boxed{\:\Delta_n=(-1)^{(n^2+3n-4)/2}\;}$