Suppose that $A$ is an $m \times n$ matrix. The $(i, j)$-entry of $A$ is denoted as $[A]_{i,j}$.
Suppose that $A$ is an $m \times n$ matrix. Suppose that $i_1$, $i_2$, $\dots$, $i_s$ are distinct integers less than or equal to $m$, and that $j_1$, $j_2$, $\dots$, $j_t$ are distinct integers less than or equal to $n$. The matrix obtained from $A$ by removing rows $i_1$, $i_2$, $\dots$, $i_s$ and columns $j_1$, $j_2$, $\dots$, $j_t$ is denoted as $A({i_1,\dots,i_s}|{j_1,\dots,j_t})$.
If $A$ is an $n \times n$ matrix, we see $A({1,2,\dots,n}|{1,2,\dots,n})$ as a "zero-by-zero matrix" for the sake of convenience.
Suppose that $A$ is an $m \times n$ matrix. Suppose that $i_1$, $i_2$, $\dots$, $i_s$ are distinct integers less than or equal to $m$, and that $j_1$, $j_2$, $\dots$, $j_t$ are distinct integers less than or equal to $n$. The matrix obtained from $A$ by removing all rows other than rows $i_1$, $i_2$, $\dots$, $i_s$ and all columns other than $j_1$, $j_2$, $\dots$, $j_t$ is denoted as $\displaystyle A\binom{i_1,\dots,i_s}{j_1,\dots,j_t}$.
Example. Let
$$ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
\end{bmatrix}.$$
Then
$$A({1,2}|{3}) =
\begin{bmatrix}
11 & 12 & 14 & 15 \\
16 & 17 & 19 & 20 \\
\end{bmatrix},$$
and
$$A\binom{1,2}{3} =
\begin{bmatrix}
3 \\
8 \\
\end{bmatrix}.$$
Suppose that $A$ is an $n \times n$ matrix. The determinant of $A$, $\det {(A)}$ is defined as follows:
$$ \det {(A)}
=
\begin{cases}
[A]_{1,1},
& n = 1; \\
\displaystyle \sum_{i = 1}^{n}
{(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}},
& n \geq 2.
\end{cases}
$$
It should be noted that we can simply write
$$
\det {(A)} = \sum_{i = 1}^{n} {(-1)^{i+1} [A]_{i,1} \det {(A(i|1))}}
$$
if we define the determinant of a "zero-by-zero matrix" to be $1$.
Here are the generalized Laplace expansions, one with respect to columns and the other with respect to rows.
Theorem 1. Suppose that $A$ is an $n \times n$ matrix, that $j_1$, $j_2$, $\dots$, $j_k$ are distinct positive integers less than or equal to $n$, and that $j_1 < j_2 < \dots < j_k$. Then
$$
\begin{aligned}
\det {(A)}
= \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n}
{\det {\left(
A\binom{i_1, i_2, \dots, i_k}
{j_1, j_2, \dots, j_k}
\right)}}
(-1)^{i_1 + i_2 + \dots + i_k
+ j_1 + j_2 + \dots + j_k}
\det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}.
\end{aligned}
$$
Theorem 2. Suppose that $A$ is an $n \times n$ matrix, that $i_1$, $i_2$, $\dots$, $i_k$ are distinct positive integers less than or equal to $n$, and that $i_1 < i_2 < \dots < i_k$. Then
$$
\begin{aligned}
\det {(A)}
= \sum_{1 \leq j_1 < j_2 < \dots < j_k \leq n}
{\det {\left(
A\binom{i_1, i_2, \dots, i_k}
{j_1, j_2, \dots, j_k}
\right)}}
(-1)^{i_1 + i_2 + \dots + i_k
+ j_1 + j_2 + \dots + j_k}
\det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}.
\end{aligned}
$$
How does one prove the two theorems?
Best Answer
It will be easier if it is first proved that the determinant of a square matrix can be computed by expansion about any row or column; see this for the details.
I will use the $\rho$-notation which is defined in the above answer.
I will prove only the first theorem; the proof of the second one is very similar to that of the first one, which I leave as an exercise.
Let the proposition $P(k)$ be as follows:
We will prove that $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.
$P(1)$ is true, because the determinant of a square matrix can be computed by expansion about any column.
Suppose that $P(k-1)$ is true. Our goal is to prove that $P(k)$ is true.
Let $A$ be an $n \times n$ matrix, with $n \geq k$. Let $j_1$, $j_2$, $\dots$, $j_k$ be any $k$ distinct integers less than or equal to $n$, with $j_1 < j_2 < \dots < j_k$. Then $$ \begin{aligned} & \det {(A)} \\ = {} & \sum_{d_1 = 1}^{n} {[A]_{d_1,j_1} (-1)^{d_1 + j_1} \det {(A(d_1|j_1))}} \\ = {} & \sum_{d_1 = 1}^{n} {[A]_{d_1,j_1} (-1)^{d_1 + j_1}} \sum_{\substack{ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} (-1)^{d_2 - \rho(d_2, d_1) + \dots + d_k - \rho(d_k, d_1) + j_2 + \dots + j_k - (k - 1)} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} {[A]_{d_1,j_1} (-1)^{d_1 + j_1}} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} (-1)^{d_2 - \rho(d_2, d_1) + \dots + d_k - \rho(d_k, d_1) + j_2 + \dots + j_k - (k - 1)} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) - 2(k-1) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{\substack{ 1 \leq d_1 \leq n \\ 1 \leq d_2 < \dots < d_k \leq n \\ d_1 \neq d_2,\dots,d_k }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \sum_{\substack{ 1 \leq s \leq k\\ d_1 = i_s \\ d_r = i_{r-1}, \ 2 \leq r \leq s \\ d_r = i_{r}, \ r > s \\ }} (-1)^{ (d_1 + \dots + d_k) + (j_1 + \dots + j_k) + (\rho(d_1, d_2) + \dots + \rho(d_1, d_k)) } [A]_{d_1,j_1} {\det {\left( A\binom{d_2, \dots, d_k}{j_2, \dots, j_k} \right)}} \det {(A({d_1,d_2,\dots,d_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \sum_{1 \leq s \leq k} (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) + (s-1) } [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \left( (-1)^{s-1} \sum_{1 \leq s \leq k} [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \right) (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} \left( (-1)^{s+1} \sum_{1 \leq s \leq k} [A]_{i_s,j_1} {\det {\left( A\binom{i_1,\dots,i_{s-1},i_{s+1},\dots,i_k}{j_2, \dots, j_k} \right)}} \right) (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))} \\ = {} & \sum_{1 \leq i_1 < i_2 < \dots < i_k \leq n} { \det {\left( A\binom{i_1, i_2, \dots, i_k} {j_1, j_2, \dots, j_k} \right)} } (-1)^{ (i_1 + \dots + i_k) + (j_1 + \dots + j_k) } \det {(A({i_1,i_2,\dots,i_k}|{j_1,j_2,\dots,j_k}))}. \end{aligned} $$
We have used the hypothesis in the second equality.
Hence by mathematical induction, $P(k)$ is true for $k = 1$, $2$, $3$, $\dots$.
Note that the proof uses nothing more than the usual Laplace expansion.