Myujth

I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?

This is the matrix :

$$beginbmatrixn&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaendbmatrix$$

Let $M_n$ be your matrix.

Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you

1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.
   
   This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left




2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).
   
   This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.

After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
beginbmatrix
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
endbmatrix$$

From this, you can deduce

$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^n-1(lambda-1) + n^n-2lambda$$

This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_ij = a_ij - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.

Then $det(A) = det(B)(1+ e^TB^-1e)$. Since $B$ is triangular, $B^-1e$ is not hard to find.

This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.