Prove the following determinant identities. Definition of determinant:
$$\det{A} := \sum_{i}^{n} (-1)^{i+j}A_{ji}\det{M_{ji}}$$
where $M_{ij}$ is the minor with respect to $A_{ji}$ for some row $j$ in $A$.
Show
$$\det{A} = \det{A^T}$$
$$(\det{A})^* = \det{A^*},$$
where $$A^* = \overline{{A^T}}$$ is the complex conjugate of the matrix.
 A: Let $A$ be an $n\times n$ complex matrix. I will use the following as the definition of the determinant, 
$$ \det(A) = \frac{1}{n!} \sum \epsilon_{i_1 \dots i_n} \epsilon_{j_1\dots j_n} A_{i_1 j_1}A_{i_2 j_2} \cdots A_{i_n j_n}, $$
where the $i_k$ and $j_k$ are each summed from $1$ to $n$. 

Now we consider the first property $\det(A^T) = \det(A)$. 
$$ \det(A^T) = \frac{1}{n!} \sum \epsilon_{i_1 \dots i_n} \epsilon_{j_1\dots j_n} A^T_{i_1 j_1}A^T_{i_2 j_2} \cdots A^T_{i_n j_n}, $$
We first use the fact that $A^T_{ij} = A_{ji}$. 
$$ = \frac{1}{n!} \sum \epsilon_{i_1 \dots i_n} \epsilon_{j_1\dots j_n} A_{j_1 i_1}A_{j_2 i_2} \cdots A_{j_n i_n}, $$
The $i_k$ are dummy indices and so we can change their label without changing the value of the summation. Replace each $i_k$ with the label $s_k$.
$$ = \frac{1}{n!} \sum \epsilon_{s_1 \dots s_n} \epsilon_{j_1\dots j_n} A_{j_1 s_1}A_{j_2 s_2} \cdots A_{j_n s_n}, $$
Now change each label $j_k$ to have the label $i_k$. 
$$ = \frac{1}{n!} \sum \epsilon_{s_1 \dots s_n} \epsilon_{i_1\dots i_n} A_{i_1 s_1}A_{i_2 s_2} \cdots A_{i_n s_n}, $$
Now change each label $s_k$ to have the label $j_k$. 
$$ = \frac{1}{n!} \sum \epsilon_{j_1 \dots j_n} \epsilon_{i_1\dots i_n} A_{i_1 j_1}A_{i_2 j_2} \cdots A_{i_n j_n}, $$
Now use the commutative property of real numbers to exchange the order of 
the levi-cevita symbols.
$$ = \frac{1}{n!} \sum  \epsilon_{i_1\dots i_n} \epsilon_{j_1 \dots j_n} A_{i_1 j_1}A_{i_2 j_2} \cdots A_{i_n j_n} = \det(A), $$

