Using permutations to find the determinant.

Question

I came across the following problem

Let matrix A be a $n\times n$ square matrix such that $a_{ij}$ = max{ i , j }. Prove that det(A) = $(-1)^{n-1}n$

I have read the post If $a_{ij}=\max(i,j)$, calculate the determinant of $A$ and understood all the answers adressing the same problem.
I just wanted to see an alternative method which does not use row subtractions. As, the determinant fromulla $n(-1)^{n-1}$ is reminiscent of $\frac{d}{dx}(x^{n})$ Is there some way we could use differentiation of determinant technique? How would it be if one goes by the formulla of determinant in terms of permutations?

Answer 1

One alternative approach: $$ A_{n+1}= \begin{pmatrix} 1 & 2 & 3 & \cdots &n & n+1 \\ 2 & 2 & 3 & \cdots & n & n+1 \\ 3&3&3&\cdots &n & n+1\\ \vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\ n & n & n&\cdots& n & n+1 \\ n+1 & n+1 & n+1&\cdots& n+1 & n+1 \end{pmatrix}. $$ Subtract $\frac{n+1}{n}$ times the second to last row from the last row to get $$ \begin{pmatrix} 1 & 2 & 3 & \cdots &n & n+1 \\ 2 & 2 & 3 & \cdots & n & n+1 \\ 3&3&3&\cdots &n & n+1\\ \vdots & \vdots &\vdots &\ddots&\vdots & \vdots\\ n & n & n&\cdots& n & n+1 \\ 0 & 0 & 0 &\cdots & 0 & -\frac{n+1}{n} \end{pmatrix}. $$ Now, because of the block triangular structure, we have $$ \det(A_{n+1}) = - \frac{n+1}{n}\det(A_n). $$

Answer 2

I think if you subtract column $j$ to column $j+1$ for all columns you get an upper triangular matrix with all $1$ on diagonal and can develop determinant easily since on last line all but one term are zeroes.

$$ \begin{pmatrix} 1 & 1 & 1 & \cdots &1 \\ 2 & 0 & 1 & \cdots & 1 \\ 3&0&0&\cdots &1\\ \vdots & \vdots &\vdots &\ddots&\vdots\\ n & 0 & 0&\cdots& 0 \\ \end{pmatrix} =\begin{pmatrix}[i]&T_{n-1}\\ n&0 \end{pmatrix} $$

with $\det(T_{k})=1$.