Finding determinant of nxn matrix with 1s and shifted diagonal 2s

[cdummie]

Homework Statement

I have to solve the following determinant
$D_n=\begin{vmatrix} 1 & 1 & 1 & \cdots & 1 & 1 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 2 & 1 \\ 1 & 1 & 1 & \cdots & 2 & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & 2 & 1 & \cdots & 1 & 1 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 1 & 0 \end{vmatrix}$

Homework Equations

The Attempt at a Solution

So my idea was to multiply the last row by -1 and add it to every other row, that way, i had the following determinant:
$D_n= \begin{vmatrix} 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 1 & 0 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 1 & 0 & \cdots & 0 & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 1 & 0 \end{vmatrix}$

So i have determinant with all zeros above diagonal, so solution to this should be product of all elements of diagonal (i should keep in mind that sign changes since this isn't "regular" diagonal), so i ended up with: $D_n=(-1)^{\frac{n(n+1)}{2}}$. Is this correct?

 

[geoffrey159]

I don't think you are correct. Unless I'm wrong $D_3 = -1$, but with your answer you get 1.
You don't change the determinant by adding to a column a multiple of another column, and you multiply the determinant by $-1$ by exchanging 2 columns. So after your transformation, you exchange columns $(i,n+1-i)$ for $i = 1...\lfloor n/2 \rfloor$ so that you have the determinant of a diagonal (EDIT: TRIANGULAR sorry) matrix that has only 1's on the diagonal, and $\lfloor n/2 \rfloor$ transpositions.

 

[cdummie]

So after your transformation, you exchange columns $(i,n+1-i)$ for $i = 1...\lfloor n/2 \rfloor$ so that you have the determinant of a diagonal matrix that has only 1's on the diagonal, and $\lfloor n/2 \rfloor$ transpositions.

I don't understand what you mean, could you explain me this part?

 

[geoffrey159]

In your attempt to a solution, you wrote

$D_n= \det(C_1,...,C_n) = \begin{vmatrix} 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 1 & 0 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 1 & 0 & \cdots & 0 & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 1 & 0 \end{vmatrix}$

Now each time you exchange two columns, you multiply the determinant by -1. So that by permuting columns $i$ and $n+1 - i$ for all $i = 1... \lfloor n/2 \rfloor$, you get

$D_n = (-1) ^ {\lfloor n/2 \rfloor } \det(C_n,C_{n-1},...,C_1)$

and $\det(C_n,C_{n-1},...,C_1)$ is very easy to compute.