Let $\left\{X_{i}\right\}_{i \in \mathbb{N}}$ be I.I.D. random variables with \begin{align} \mathbb{P}(X_{i} = 1) = \mathbb{P}(X_{i} = -1) = \frac{1}{2}. \end{align}
Now consider the quadratic $n \times n$ matrix $A$ whos entries are the random variables $X_{i}$. The question is now to show that the variance of $\det(A)$ is $n!$. Anyone has an idea, how to proceed? I tried it with induction, but I don't come to a solution. Thanks for help
You can work this out directly from the definition. The determinant is: $\sum_{\pi}b_\pi:=\sum_{\pi}(-1)^{\sigma(\pi)}a_{\pi(1)}\cdots a_{\pi(n)}$. When you calculate the variance, you'll need to account for covariance between terms $b_\pi,b_{\pi'}$. If the pair doesn't share any common $a_i$, the covariance is 0. Otherwise, notice that both terms have mean 0, so the covariance is just $E[b_{\pi}b_{\pi'}]$. By definition if $\pi\neq \pi'$, there must be at least one $a_k$ that belongs to a single $b_\pi$ or $b_{\pi'}$, which will cause the covariance to be 0. So the covariance is always 0.
Finally, $b_\pi$ has the same distribution as $X_1$, as it's just the product of Bernoulli's. So $\mbox{Var}(b_{\pi})=1$, and $\mbox{Var}(\sum_\pi b_\pi)=\sum_\pi \mbox{Var}(b_\pi)=n!$