Matrix with random entries Let $A$ be a $n\times n$ matrix with independent random entries 
$a_{ij}$ with $E(a_{ij}) ≡ 0$ ($E$ is expectation, $D$ is variance), $D(a_{ij})=\sigma^2$. Calculate $D(\det A)$.
My idea is to associate determinant with a polynomial so that we can use sum and products of its roots, but I cannot think of anything further. Is this a correct way?
 A: Let us consider small matrix to look what happens with variance: for $2\times 2$ matrix $\text{det} A=a_{11}a_{22}-a_{21}a_{12}$ has zero expectation and then its variance equals to the second moment
$$
\text{Var}(\text{det} A) = \mathbb E[(\text{det} A)^2] = \mathbb E[(a_{11}a_{22}-a_{21}a_{12})^2]=\mathbb E(a^2_{11}a^2_{22})+\mathbb E(a^2_{21}a^2_{12})-2\underbrace{\mathbb E(a_{11}a_{22}a_{21}a_{12})}_0 = \mathbb E(a^2_{11})\mathbb E(a^2_{22})+\mathbb E(a^2_{21})\mathbb E(a^2_{12}) =2(\sigma^2)^2. 
$$
For $3\times 3$ matrix 
$$\text{det} A = a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}$$
Again expected value is zero and the second moment is 
$$
\text{Var}(\text{det} A) = \mathbb E[(\text{det} A)^2] = \mathbb E[(a_{11}a_{22}a_{33})^2]+\ldots+ \mathbb E[(a_{12}a_{21}a_{33})^2] = 3!\cdot (\sigma^2)^3
$$
Expectations of products of summands are disappeared since every such product containes few $a_{ij}$ in the first power. Say, 
$$
\mathbb E[a_{11}a_{22}a_{33}\cdot a_{11}a_{23}a_{32}] = \mathbb E[a_{11}^2]\cdot \mathbb E[a_{22}]\cdot \mathbb E[a_{33}]\cdot \mathbb E[a_{23}]\cdot \mathbb E[a_{32}] = 0
$$
In general case, denote by $\alpha=(\alpha_1,\ldots,\alpha_n)$ a permutation of $(1,\ldots,n)$ and by $S_n$ the set of all possible permutations, $|S_n|=n!$ 
Then
$$
\text{det} A = \sum_{\alpha\in S_n} \text{sgn}(\alpha) \cdot a_{1\alpha_1}a_{2\alpha_2} \dots a_{n\alpha_n}.
$$
$$
\mathbb E(\text{det} A) = \sum_{\alpha\in S_n} \text{sgn}(\alpha) \cdot \mathbb E(a_{1\alpha_1})\mathbb E(a_{2\alpha_2}) \dots \mathbb E(a_{n\alpha_n}) =0
$$
and
$$
\text{Var}(\text{det} A)=\mathbb E[(\text{det} A)^2]=\sum_{\alpha\in S_n} \mathbb E[a^2_{1\alpha_1}a^2_{2\alpha_2} \dots a^2_{n\alpha_n}] \pm \underbrace{\sum_{\alpha\in S_n} \sum_{\beta\in S_n \\ \beta\ne\alpha}\mathbb E[a_{1\alpha_1}\dots a_{n\alpha_n}\cdot a_{1\beta_1}\dots a_{n\beta_n}]}_0
$$
$$=\sum_{\alpha\in S_n} \mathbb E[a^2_{1\alpha_1}]\mathbb E[a^2_{2\alpha_2}] \dots \mathbb E[a^2_{n\alpha_n}]=n!\cdot(\sigma^2)^n.
$$
As earlier, the expected value of any product of determinant terms is zero since any different permutations differ at least at one place: 
$\alpha\neq \beta$ implies that there exist $i$: $\alpha_i\neq \beta_i$, so 
$$
\mathbb E[a_{1\alpha_1}\dots a_{n\alpha_n}\cdot a_{1\beta_1}\dots a_{n\beta_n}] = \underbrace{\mathbb E[a_{i\alpha_i}]}_0\mathbb E[a_{i\beta_i}] \cdot \mathbb E\left[\prod_{k\ne i}a_{k\alpha_k}a_{k\beta_k}\right]=0.
$$
