Expansion Theorem for Determinants

Theorem
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $D = \map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Let $a_{p q}$ be an element of $\mathbf A$.

Let $A_{p q}$ be the cofactor of $a_{p q}$ in $D$.

Then:


 * $(1): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
 * $(2): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$

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:
 * $\displaystyle D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$

is known as the expansion of $D$ in terms of row $r$, while:
 * $\displaystyle D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$

is known as the expansion of $D$ in terms of column $r$.

Proof
Because of Determinant of Transpose, it is necessary to prove only one of these identities.

Let:
 * $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}$

First, note that from Determinant with Row Multiplied by Constant, we have:


 * $\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}$

and similarly:


 * $\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}$

and so on for the whole of row $r$.

From Determinant as Sum of Determinants:


 * $\displaystyle \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} = \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} }$

Consider the determinant:
 * $\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}$

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:
 * $\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}$

The same argument can be applied to columns.

Thus:
 * $\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}$

and so:
 * $\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}$

Then:
 * $\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}$ is $A_{r k}$, the cofactor of $a_{r k}$ in $D$.

But from Determinant with Unit Element in Otherwise Zero Row, we have:


 * $\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}$

Assembling all the pieces derived above, the result follows.