Theorem: Laplace Expansion Theorem for Determinants

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$:
:$\ds \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$
where:
:the summation $\ds \sum_\lambda$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$
:$\map \sgn \lambda$ is the sign of the permutation $\lambda$.
Let $a_{p q}$ be an element of $\mathbf A$.
Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$.
Then:
:$(1): \quad \ds \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
:$(2): \quad \ds \forall c \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$
Thus the value of a determinant can be found either by:
:multiplying all the elements in a row by their cofactors and adding up the products
or:
:multiplying all the elements in a column by their cofactors and adding up the products.
The identity:
:$\ds D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
is known as the '''expansion of $D$ in terms of row $r$''', while:
:$\ds D = \sum_{k \mathop = 1}^n a_{k c} A_{k c}$
is known as the '''expansion of $D$ in terms of column $c$'''.

Proof:
Because of Determinant of Transpose, it is necessary to prove only one of these identities.
Let:
:<nowiki>$D = \begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix}$</nowiki>
First, note that from Determinant with Row Multiplied by Constant, we have:
:<nowiki>$\begin{vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{r 1} &       0 & \cdots &       0 \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix} = a_{r 1} \begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
      1 &       0 & \cdots &       0 \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix}$</nowiki>
and similarly:
:<nowiki>$\begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
      0 & a_{r 2} & \cdots &       0 \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end {vmatrix} = a_{r 2} \begin {vmatrix}
a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
      0 &       1 & \cdots &       0 \\
 \vdots &  \vdots & \ddots &  \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{vmatrix}$</nowiki>
and so on for the whole of row $r$.
From Determinant as Sum of Determinants:
:<nowiki>$\begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix} = \ds \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix}
a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
      0 & \cdots &       1 & \cdots &       0 \\
 \vdots & \ddots &  \vdots & \ddots &  \vdots \\
a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
\end {vmatrix} }$</nowiki>
Consider the determinant:
:<nowiki>$\begin{vmatrix}
             a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 k} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
              \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} k} & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
                   0 & \cdots &                                  0 &                    1 &                                  0 & \cdots &                    0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} k} & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n k} &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
\end {vmatrix}$</nowiki>
Exchange rows $r$ and $r - 1$, then (the new) row $r - 1$ with row $r - 2$, until finally row $r$ is at the top.
Row $1$ will be in row $2$, row $2$ in row $3$, and so on.
This is permuting the rows by a $k$-cycle of length $r$.
Call that $k$-cycle $\rho$.
Then from Parity of K-Cycle:
:$\map \sgn \rho = \paren {-1}^{r - 1}$
Thus:
:<nowiki>$\begin {vmatrix}
                   0 & \cdots &                    1 & \cdots &                    0 \\
             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n}
\end {vmatrix} = \paren {-1}^{r - 1} \begin {vmatrix}
             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
                   0 & \cdots &                    1 & \cdots &                    0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n}
\end {vmatrix}$</nowiki>
The same argument can be applied to columns.
Thus:
:<nowiki>$\begin {vmatrix}
                   1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\
             a_{1 k} &              a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
             a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } &              \cdots & a_{n n}
\end {vmatrix} = \paren {-1}^{k-1}\begin {vmatrix}
                   0 & \cdots &                    1 & \cdots &                    0 \\
             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n}
\end {vmatrix}$</nowiki>
and so:
:<nowiki>$\begin{vmatrix}
                   1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\
             a_{1 k} &              a_{1 1} & \cdots &               a_{1 \paren {k - 1}} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
             a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
\end {vmatrix} = \paren {-1}^{r + k} \begin {vmatrix}
             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
                   0 & \cdots &                    1 & \cdots &                    0 \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n}
\end {vmatrix}$</nowiki>
Then:
:<nowiki>$\paren {-1}^{r + k} \begin {vmatrix}
             a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
              \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
              \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
             a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
\end{vmatrix}$</nowiki>
is $A_{r k}$, the cofactor of $a_{r k}$ in $D$.
But from Determinant with Unit Element in Otherwise Zero Row, we have:
:<nowiki>$\begin {vmatrix}
      1 &       0 & \cdots &       0 \\
b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
 \vdots &  \vdots & \ddots &  \vdots \\
b_{n 1} & b_{n 2} & \cdots & b_{n n}
\end {vmatrix} = \begin {vmatrix}
b_{2 2} & \cdots & b_{2 n} \\
 \vdots & \ddots &  \vdots \\
b_{n 2} & \cdots & b_{n n}
\end {vmatrix}$</nowiki>
Assembling all the pieces derived above, the result follows.
{{qed}}