Laplace Theorem

Let $|A|$ be the determinant of order $n$. Fix $m$ rows, $1\leq m\leq n$, then we have

$$|A|=\sum_{1\leq k_1

By the definition of the determinant, we have

$$|A|=\sum_{(k_1,\ldots,k_n)\in S_n}\mathrm{sgn}(k_1,\ldots,k_n)a_{1k_1}a_{2k_2}\cdots a_{nk_n}.$$

Recall the result, expanding the determinant along the first row, we have

$$|A|=\sum_{i=1}^n a_{1i}\hat A\left(\begin{matrix}1\\i\end{matrix}\right).$$

If we fix $m$ rows, namely $1\leq i_1 $$|A|=\sum_{(k_1,\ldots,k_n)\in S_n}\mathrm{sgn}(k_1,\ldots,k_n)a_{i_1k_1}a_{i_2k_2}\cdots a_{i_mk_m}a_{i_{m+1}k_{m+1}}\cdots a_{i_nk_n}(-1)^{\sum_{j=1}^m(i_j-j)}.$$

where $i_{m+1}\leq \cdots\leq i_n$ are the remaining rows. Then we split the sum into two parts by first $m$ columns and last $n-m$ columns, we have the summation

$$\sum_{(k_1,\ldots,k_n)\in S_n}\mathrm{sgn}(k_1,\ldots,k_n)=\sum_{1\leq j_1Note that

$$\sum_{(k_1,\ldots,k_m)\in S(j_1,\ldots,j_m)}\mathrm{sgn}(k_1,\ldots,k_m)a_{i_1k_1}a_{i_2k_2}\cdots a_{i_mk_m}=A\left(\begin{matrix}i_1&i_2&\cdots&i_m\\j_1&j_2&\cdots&j_m\end{matrix}\right),$$

and with the factor, we have the remaining part

$$\sum_{(k_{m+1},\ldots,k_n)\in S([n]\setminus\{j_1,\ldots,j_m\})}(-1)^{\sum_{c=1}^m(i_c+j_c)}\mathrm{sgn}(k_{m+1},\ldots,k_n)a_{i_{m+1}k_{m+1}}\cdots a_{i_nk_n}=\hat A\left(\begin{matrix}i_1&i_2&\cdots&i_m\\j_1&j_2&\cdots&j_m\end{matrix}\right).$$

Finally we prove the theorem by combining the above two equations.