3
$\begingroup$

I'm trying to show that swapping two columns of a matrix changes the determinant by a factor of $-1$ using the Leibniz formula. Let $A$ be an $n\times n$ matrix and let $B$ be the matrix obtained by swapping column $i$ and column $j$.

Then \begin{align*} \operatorname{det}(B) & = \sum_{\sigma \in S_n}\operatorname{sgn}(\sigma)\, a_{\sigma(1)1} \ldots a_{\sigma(i)j} \ldots a_{\sigma(j)i} \ldots a_{\sigma(n)n} \\ &=\sum_{\pi \in S_n} \operatorname{sgn}(\pi)a_{\pi(1)1} \ldots a_{\pi(i)i} \ldots a_{\pi(j)j} \ldots a_{\pi(n)n} \end{align*} where $\pi=\sigma(i,j).$ I can see that $\operatorname{sgn}(\pi)=-\operatorname{sgn}(\sigma)$ but I don't know where to go now.

$\endgroup$

1 Answer 1

4
$\begingroup$

HINT

Column flip is a multiplication by permutation matrix with determinant -1.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .