Calculate determinant of $n$-th order

Question

I have problem with creating a upper triangular matrix (in order to calculate the determinant) of the following matrix:

$$\begin{pmatrix} 1& 2& 3& ...& n-2 &n-1 &1 \\ 1& 2& 3& ...& n-2& 1&n \\ 1& 2& 3& ...& 1&n-1 &n \\ .& .& .& .& .& .& \\ .& .& .& .& .& .& \\ 1& 2& 1& ...& n-2 &n-1 &n \\ 1& 1& 3& ...& n-2& n-1 &n \\ 1& 2& 3& ...& n-2& n-1 &n \end{pmatrix}$$

One of my attempts included subtracting the first row from the others which got me to this point:

$$\begin{pmatrix} 1& 2& 3& ...& n-2 &n-1 &1 \\ 0& 0& 0& ...& 0& 2-n&n-1 \\ 0& 0& 0& ...& 3-n&0 &n-1 \\ .& .& .& .& .& .& \\ .& .& .& .& .& .& \\ 0& 0& -2& ...& 0 &0 &n-1 \\ 0& -1& 0& ...& 0&0 &n-1 \\ 0& 0& 0& ...& 0&0 &n-1 \end{pmatrix}$$

Any ideas on how to move on from here ?

Answer 1

In a first step, let us switch to the matrix with columns' order completely reversed :

$$M_n=\begin{pmatrix} \color{red}{1}& (n-1)& (n-2)& ...& 3 &2 &1 \\ n& \color{red}{1}& (n-2) & ... &3& 2& 1 \\ n& (n-1)& \color{red}{1}& ...& 3&2 &1 \\ .& .& .& .& .& .& \\ .& .& .& .& .& .& \\ n& (n-1)& (n-2)& ...& \color{red}{1} &2 &1 \\ n& (n-1)& (n-2)& ...& 3& \color{red}{1} &1 \\ n& (n-1)& (n-2)& ...& 3& 2 &\color{red}{1} \end{pmatrix}$$

in this way the disrupting $\color{red}{"1"}$ entries are now situated on the main diagonal.

Remark : Changing the order of the columns of a matrix only changes the sign of its determinant (it is multiplied by the signature of the permutation).

The key fact comes from the following factorization :

$$M_n=P_nQ_n\tag{1}$$

where

$$P_n=\left(\begin{array}{c|cccccc} 1&-(n-1)&&&&&&\\ 1&&-(n-2)&&&&&\\ 1&&&-(n-3)&&&&\\ .& .& .& .& .& .& .&\\ 1&&&&&&&-1\\ \hline 1&&&&&&&\end{array}\right)$$

and

$$Q_n=\left(\begin{array}{cccccc|c} n&(n-1)&&.&3&2&1\\ \hline 1&0&&&&0&0\\ 0&1&&&&0&0\\ &&1&&&&\\ .& .& .& .& .& .& \\ 0&&&&&1&0\end{array}\right)$$

I have generaly replaced $0$ by void places, but in some cases where their presence clarify the situation.

(relationship (1) can be verified by block computations.)

Now, we can conclude because

$$\det(P_n)=(-1)^{(n-1)}.(-1)^{(n-1)} . (n-1)!=(n-1)!$$

(Laplace expansion along the first column) and

$$\det(Q_n)=(-1)^{(n-1)}$$

(expansion along the last column).

Don't forget, in a final step, to take into account the signature of the initial permutation which is rather peculiar (see results in this answer here).

Numerical results :

for $n=1,2,3...$ we get resp. $1,1,−2,−6,24,120,−720,−5040,40320...$

If you want to verify, here is the little Matlab program I have used :

 for n=1:10;
    M=ones(n,1)*(n:-1:1)-diag((n-1):-1:0);% initial
    P=[ones(n-1,1),-diag(n-1:-1:1);1,zeros(1,n-1)];
    Q=[(n:-1:1);eye(n-1),zeros(n-1,1)];
    M-P*Q, % always zero
    T(n)=det(M);
 end
 T

Answer 2

HINT:

We could use the Cauchy determinant formula

$$\det(A + B C) = \det A + C \operatorname{adj}(A) B$$

where $A$ is $n\times n$, $B$ is $n\times 1$, and $C$ is $1 \times n$, $\operatorname{adj}(A)$ the adjugate of $A$.

Now we have $B= (1,2,\ldots, n)^t$, $C= (1,1,\ldots,1)$, and $A = \operatorname{diag}(0,-1, \ldots, -n)$. Now $A + B C$ is the matrix obtained by permuting the columns, like @Jean Marie: did. Note that $\operatorname{adj}(A)$ has only one non-zero element $(-1)^{n-1} (n-1)!$ at position $(1,1)$. This simplifies the problem a lot. Will skip the details.