# Help me with the result of this determinant..

$$D = \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 2 & 1 & 1 & \dots & 1 & 0 \\ 3 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ n-1 & 1 & 0 & \dots & 0 & 0 \\ n & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} =n*1*(-1)^\frac{n(n-1)}{2}$$
I don't quite understand the solution of this determinant. I do understand that if we use Laplace expansion along the last row we get $$D = n* \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix}$$ But how does the remaining determinant euqal: $1*(-1)^\frac{n(n-1)}{2}$?

Edit:

$$\begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{vmatrix} =(-1)^{4} \begin{vmatrix} 0 & 1 \\ 1 & 0 \\ \end{vmatrix} =(-1)^{4+3+2}$$ I thought it should go: $(-1)^{3+2+1}$ or is the power actually the sum of row and column coordinates?

• When you expand about the bottom row you need to take into account the sign which is $(-1)^n$. The full determinant therefore becomes $n(-1)^{n + (n-1) + \ldots +2 + 1}$ Dec 10 '15 at 10:54
• Oh my God, thank you!
– A6EE
Dec 10 '15 at 10:56
• @Winther One more question, I don't understand the sign. My teacher said the sign in this case is $(-1)^{n+1}$. She did an example for a $3x3$ determinant. I'll post it in the edit section. I'm not sure to what power should $-1$ be.
– A6EE
Dec 10 '15 at 11:01
• Yes the power is the sum of the row and column indices. Dec 10 '15 at 11:10
• You are right: it's $(-1)^{n+1}$, not $(-1)^{n}$ as I said above! Dec 10 '15 at 11:11

$$\begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix}$$
You can proceed like that: If we swap the $k$-th row with the $n-k$-th row of the matrix for $k=1,2,\ldots$ then, at some point, we will get a lower triangular matrix with only $1$ on its main diagonal (so its determinant is $1$). Now, each swap corresponds to an elementary operation of the Gauss-Jordan method and thus changes the sign of the determinant (i.e. multiplies it by $-1$). So count the number of swaps to get the result.
First do an elementary transformation that does not change the determinant, namely add to the first column the sum of all the others. You are left with $$\begin{vmatrix} n & 1 & 1 & \dots & 1 & 1 \\ n & 1 & 1 & \dots & 1 & 0 \\ n & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ n & 1 & 0 & \dots & 0 & 0 \\ n & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} = n \cdot \begin{vmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{vmatrix} = n \cdot (-1)^{s}.$$ Here $s$ is the sign of the permutation $\sigma$ on $\{1, 2, \dots, n\}$ that swaps $i$ with $n-i$, since exchanging the $i$-th row with the $(n-i)$-th row of $$\begin{bmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 1 & 1 & \dots & 1 & 0 \\ 1 & 1 & 1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 1 & 1 & 0 & \dots & 0 & 0 \\ 1 & 0 & 0 & \dots & 0 & 0 \\ \end{bmatrix},$$ for $1 \le i \le n/2$, we get a matrix which has visibly determinant $1$.
The permutation $\sigma$ has $s = n/2$ two-cycles if $n$ is even, and $s = (n-1)/2$ two-cycles if $n$ is odd. So indeed $$(-1)^{s} = (-1)^{n (n-1)/2} = \begin{cases} (-1)^{n/2} & \text{if n is even, as (-1)^{n-1} = -1,}\\ (-1)^{(n-1)/2} & \text{if n is odd, as (-1)^{n} = -1.}\\ \end{cases}$$