# Determinant of nth order

I want to solve the following determinant:

$D_n= \begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ a_n & a_{n-1} & \cdots & a_2 & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ a_n & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

The idea I had was to 1) get the $a_n$ in front of the determinant, which gets me to:

$D_n= a_n \begin{vmatrix} 1 & a_{n-1} & \cdots & a_2 & x\\ 1 & a_{n-1} & \cdots & x & a_1\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x & \cdots & a_2 & a_1\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

Then I multiplied the nth row with $-1$ and added it to all the other ones, which gives me:

$D_n= a_n \begin{vmatrix} 0 & 0 & \cdots & 0 & x-a_1\\ 0 & 0 & \cdots & x-a_2 & 0\\ 0 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & x-a_{n-1} & \cdots & 0 & 0\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

So, I got a triangular determinant, but with the opposite diagonal, not the main one. How do I transform it into a real triangular determinant? Obviously, my idea to replace every adjacent row (1st and nth, (n-1)th and 2nd...) is wrong? Any ideas?

• Exchanging two adjacent rows multiplies the determinant by $-1$. Now exchange $n$th row with $(n-1)$th, then $(n-1)$th with $(n-2)$th until the last row becomes the first. This needs $n-1$ permutations. Then iterating the procedure we see that the appropriate factor is $(-1)^{ (n-1) +(n -2)+\ldots +1}= (-1)^{n(n-1)/2}$. – Start wearing purple Jan 12 '16 at 23:40
• But is it really $\begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ a_{n} & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$, or rather $\begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ x & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$? – Andreas Caranti Jan 12 '16 at 23:43
• @Startwearingpurple Would the solution then be $(-1)^{n(n-1)/2}(x-a_n!)$? – Quant Jan 12 '16 at 23:44
• @AndreasCaranti If you mean that there is a mistake in the original text of the problem, there isn't. I also double-checked just in case. – Quant Jan 12 '16 at 23:45
• I am not sure about my interpretation of your notation. The answer will be $(-1)^{n(n-1)/2}a_n \prod_{k=1}^{n -1}(x-a_k)$. – Start wearing purple Jan 12 '16 at 23:47

We can simply calculate the determinant of an opposite (lower) triangular matrix:

Let $J_n$ be the $n \times n$ matrix with $1$ on the anti-diagonal and $0$ otherwise (i.e. $J_ne_i = e_{n+1-i}$ for all $1 \leq i \leq n$, where $e_1, \dotsc, e_n$ denotes the standard basis). Given any $m \times n$-matrix $A$ the matrix $AJ_n$ originates from $A$ by vertically mirroring its colums from the middle, i.e. swapping the first column with the last, the second with the second last, etc.

If $A$ is an $n \times n$ square matrix then we get from $J_n^2 = I_n$ that $$\det(A) = \det(J_n) \det(AJ_n).$$ In the case of $D_n$ we get that $D_n J_n$ is the $a_n$-scalar multiple a lower triangular matrix with diagonal entries $x-a_1, x-a_2, \dotsc, x-a_{n-1}, 1$, so $$\det(D_n) = \det(J_n) \det(D_n J_n) = \det(J_n) a_n (x-a_1) \dotsm (x-a_{n-1}).$$

So the only difference is that we need to know $\det(J_n)$. Because $J_n$ is a permutation matrix, corresponding to $\sigma_n \in S_n$ with $\sigma(i) = n+1-i$, we have $\det(J_n) = \mathrm{sgn}(\sigma_n)$. Notice that \begin{align*} \sigma_{2n} &= (1 \;\; 2n) (2 \;\; n-1) \dotsm (n \;\; n+1) \\ \sigma_{2n+1} &= (1 \;\; 2n+1) (2 \;\; n-1) \dotsm (n \;\; n+2). \end{align*} So we can just count the number of transpositions used and get that $$\mathrm{sgn}(\sigma_n) = \begin{cases} \phantom{-}1 & \text{if n \equiv 0,1 \bmod 4}, \\ -1 & \text{if n \equiv 2,3 \bmod 4}, \end{cases} = (-1)^{n(n-1)/2}.$$

So alltogether we have $$\det(D_n) = (-1)^{n(n-1)/2} a_n (x-a_1) \dotsm (x-a_{n-1}).$$

(The nice thing about this is that now that we have calculated $\det(J_n) = (-1)^{n(n-1)/2}$ we can use this to calculate the determinant of opposite triangular and opposite diagonal matrices more ore less in the usual way.)

You have gone almost through most of the way! The determinant then simply is:$$D=(-1)^{n+1}(x-a_1)(-1)^{n+1}(x-a_2)(-1)^{n+1}(x-a_3)\cdots (-1)^{n+1}(x-a_{n-1})(-1)^{n+1}=(x-a_1)(x-a_2)\cdots (x-a_{n-1})$$