$a_{ij}=i $ if $i+j=n+1$ and $0$ otherwise; compute det $A$

Question

The entries of the matrix is specified by this rule,

$A=(a_{ij})\in M_n(\mathbb R)$,

$a_{ij}=i$ if $i+j=n+1$

and $0$ otherwise.

Compute det $A$

> I have seen for $n=1,2,3,4,5,6,7......$;det A= $1,-2!,-3!,4!,5!,-6!,-7!,......$ How can i express it in acompact manner?

Answer 1

Note that $$A = \pmatrix{0&&1\\&\cdot&\\n&&0}$$ So $$\det A = (-1)^{\lfloor n/2 \rfloor} \det \pmatrix{n&&0\\&\ddots&\\0&&1} = (-1)^{\lfloor n/2 \rfloor} n!$$ Where the $(-1)^{\lfloor \frac n2 \rfloor}$ is the sign of the inversion permutation $i\mapsto n-i+1$ wich is the product of $\lfloor \frac n2\rfloor$ swapping permutations $\tau_{i, n-i+1}$