Prove that $\det(\overline M)=\overline {\det(M)}$ For $M \in M_{n×n}(\mathbb C)$ , let $\overline M$ be the matrix such that $(\overline M)_{ij}=\overline {M_{ij}}$ for all $i,j$, where $\overline {M_{ij}}$ is the complex conjugate of $M_{ij}$.

Prove that $\det(\overline M)=\overline {\det(M)}$.

My attempt: We will use induction on $n$.
Base case $n=1$: Let $M$ be the one-by-one matrix with entry $a+bi$
\begin{align*}
\det(\overline M) &= \det(a-bi)=a-bi \\
\overline {\det(M)} &=\overline {a+bi}=a-bi
\end{align*}
Induction hypothesis: assume holds for $n$, NTS it holds for $n+1$:
\begin{align*}
\det(\overline M) &= \sum_{j=1}^{n+1} {(-1)^{1+j} \overline M_{1j} \cdot \det(\overline {\tilde M_{1j}}})
\\&=(-1)^{1+n+1} \overline {M_{1,n+1}}\cdot \det(\overline {\tilde M_{1,n+1})} + \sum_{j=1}^{n} {(-1)^{1+j} \overline M_{1j} \cdot \det(\overline {\tilde M_{1j}}})
\\&=(-1)^{2+n}\overline {M_{1,n+1}}\cdot \overline{\det(\tilde M_{1,n+1})}+\overline{\det(M')}
\end{align*}
where $M'$ is $n$-by-$n$. I feel like the proof is almost complete but am not sure how to proceed. Any help is greatly appreciated.
 A: The above approach should work but working with such an iterative description of the determinant is just annoying.
Hopefully, you know the following result/definition of the determinant:
Let $A\in \mathbb{C}^{n\times n}$, then $$\det(A)=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}.$$
Indeed, if $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the above formula reduces to the expression $ad-bc$.
Anyhow, once you have this description, note that
$$\det(\overline{A})=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^n\overline{a_{i,\sigma(i)}}=\overline{\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}}=\overline{\det(A)}.$$
A: You seem to be using the recursive definition of the determinant given by
$$ \det(a) = a$$
for $1\times 1$ matrices $(a)$ and
$$ \det(A) = \sum_{j=1}^n (-1)^{1+j} a_{1,j} \, \det\left( A_{1,j}\right), $$
where $A=(a_{ij})$ is an $n\times n$ matrix with $n>1$ and $A_{1,j}$ is the $(n-1)\times (n-1)$ matrix obtained from $A$ by removing the first row and the $j$-th column.
(I use a different notation since I find $\tilde A_{1,j}$ referring to a submatrix and $A_{1,j}$ referring to an entry rather confusing.)
When $B=\overline A$ with entries $b_{ij}=\overline{a_{ij}}$ you already proved that $\det(B)=\det(A)$ in case $n=1$.
For $n>1$ we get: $$\det(B) = \sum_{j=1}^n (-1)^{1+j} b_{1,j} \, \det\left( B_{1,j}\right).$$
Note that $b_{1,j}=\overline{a_{1,j}}$ and $B_{1,j}=\overline{A}_{1,j}=\overline{A_{1,j}}$, since it doesn't matter wether you first conjugate and then remove rows/columns or first remove rows/columns and then conjugate.
Hence, $$\det(B) = \sum_{j=1}^n (-1)^{1+j} \overline{a_{1,j}} \, \det\left( \overline{A_{1,j}}\right).$$
Since $A_{1,j}$ is an $(n-1)\times (n-1)$ matrix, we can assert by induction hypothesis that $\det(\overline{A_{1,j}}) = \overline{\det(A_{1,j})}$ so that
$$\det(B) = \sum_{j=1}^n (-1)^{1+j} \overline{a_{1,j}} \, \overline{\det\left( A_{1,j}\right)}.$$
Since conjugation is compatible with multiplication and addition ($\overline z + \overline w = \overline{z+w}$ and $\overline z \cdot \overline w = \overline{z\cdot w})$, we get
$$\det(B) = \overline{\sum_{j=1}^n (-1)^{1+j} a_{1,j} \, \det\left( A_{1,j}\right)} = \overline{\det(A)}.$$

Note that there is no need to separate the last summand at any time. You might have guessed so since proving summation formulas often involves separating one summand to apply the induction hypothesis, but here we are not proving a summation formula. We are merely using one. The induction is on the size of the matrix which just happens to be the number of summands, but $A_{1,j}$ is already an $(n-1)\times(n-1)$ matrix in all the $n$ summands.
