# determinant of anti-diagonal matrix

Let $A=\operatorname{adiag}(a_{1},\ldots,a_{n})$ be an anti-diagonal matrix. Using the sum over all permutations formula for the determinant of a matrix and since all but possibly the anti-diagonal elements are null we get directly at the result

 $\operatorname{det}A=\operatorname{sgn}(n,n-1,\ldots,1)\prod_{i=1}^{n}a_{i}$

so all that remains is to calculate the sign of the permutation. This can be done directly.

To bring the last element to the beginning $n-1$ permutations are needed so

 $\operatorname{sgn}(n,n-1,\ldots,1)=(-1)^{n-1}\operatorname{sgn}(1,n,n-1,\cdots% ,2)$

Now bring the last element to the second position. To do this $n-2$ permutations are needed. Repeat this procedure $n-1$ times to get the permutation $(1,\ldots,n)$ which has positive sign.

Summing every permutation, it takes

 $\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$

permutations to get to the desired permutation.

So we get the final result that

 $\operatorname{det}\operatorname{adiag}(a_{1},\ldots,a_{n})=(-1)^{\frac{n(n-1)}% {2}}\prod_{i=1}^{n}a_{i}$

Notice that the sign is positive if either $n$ or $n-1$ is a multiple of $4$ and negative otherwise.

Title determinant of anti-diagonal matrix DeterminantOfAntidiagonalMatrix 2013-03-22 15:50:25 2013-03-22 15:50:25 cvalente (11260) cvalente (11260) 6 cvalente (11260) Result msc 15-00