How to prove the Cofactor Expansion Theorem for Determinant of a Matrix? I understand that the definition of a determinant of  a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.
 A: Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row/column.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$\newcommand\detname{\,\det}$
$\newcommand\matrixentry[2]{#1_{#2}}$
$\newcommand\submatrix[3]{#1(#2|#3)}$
For the induction, we use the notation $\submatrix{A}{i_1,i_2}{j_1,j_2}$ to denote the $(n-2)\times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$. We assume as inductive hypothesis that for square matrices with $n-1$ rows or less, the determinant can be calculated along any row/column.
For the calculation of the minors there will be a column missing when we express the minor in terms of the entries of the original matrix, so one needs to be careful with the signs. For that we use
$$
\epsilon_{\ell j}=\begin{cases} 0,&\ \ell <j \\ 1,&\ \ell>j\end{cases}
$$
As mentioned above, the idea is that one calculates the minors along the $i^{\rm th}$ row, which is ok by inductive hypothesis. We assume $i>1$, as we are comparing with calculating along the first row. The inductive hypothesis is used below in the second and last equalities.
\begin{align*}
\detname{A}
&=
\sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j}\detname{\submatrix{A}{1}{j}}
\\
&=
\sum_{j=1}^{n}(-1)^{1+j}\matrixentry{A}{1j}
\sum_{\substack{1\leq\ell\leq n\\\ell\neq j}}
(-1)^{i-1+\ell-\epsilon_{\ell j}}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}}
\\
&=
\sum_{j=1}^{n}\sum_{\substack{1\leq\ell\leq n\\\ell\neq j}}
(-1)^{j+i+\ell-\epsilon_{\ell j}}
\matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}}
\\
&=
\sum_{\ell=1}^{n}\sum_{\substack{1\leq j\leq n\\j\neq\ell}}
(-1)^{j+i+\ell-\epsilon_{\ell j}}
\matrixentry{A}{1j}\matrixentry{A}{i\ell}\detname{\submatrix{A}{1,i}{j,\ell}}
\\
&=
\sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell}
\sum_{\substack{1\leq j\leq n\\j\neq\ell}}
(-1)^{j-\epsilon_{\ell j}}
\matrixentry{A}{1j}\detname{\submatrix{A}{1,i}{j,\ell}}
\\
&=
\sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell}
\sum_{\substack{1\leq j\leq n\\j\neq\ell}}
(-1)^{\epsilon_{\ell j}+j}
\matrixentry{A}{1j}\detname{\submatrix{A}{i,1}{\ell,j}}
\\
&=
\sum_{\ell=1}^{n}(-1)^{i+\ell}\matrixentry{A}{i\ell}\detname{\submatrix{A}{i}{\ell}}
\end{align*}
