# 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 $\mathbf A$:

$\displaystyle \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$

where:

the summation $\displaystyle \sum_\lambda$ 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:

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

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 $r$.

### Corollary

Let $\delta_{rs}$ be the Kronecker delta.

Then:

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

That is, if you multiply each element of a row or column by the cofactor of another row or column, the sum of those products is zero.

## 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 $r$.

$\displaystyle \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} = \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ 0 & \cdots & 1 & \cdots & 0 \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} }$

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.

$\blacksquare$