Expansion Theorem for Determinants

Theorem
Let $\mathbf A = \left[{a}\right]_n$ be a square matrix of order $n$.

Let $D = \det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

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

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

Then:


 * $(1): \quad \displaystyle \forall r \in \left[{1 \,.\,.\, n}\right]: D = \sum_{k \mathop = 1}^n a_{rk} A_{rk}$
 * $(2): \quad \displaystyle \forall r \in \left[{1 \,.\,.\, n}\right]: D = \sum_{k \mathop = 1}^n a_{kr} A_{kr}$

What this theorem states is that 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:
 * $\displaystyle D = \sum_{k \mathop = 1}^n a_{rk} A_{rk}$

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

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_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rk} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix}$

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


 * $\begin{vmatrix}

a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r1} & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix} = a_{r1} \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}$

... and similarly:


 * $\begin{vmatrix}

a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & a_{r2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix} = a_{r2} \begin{vmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{vmatrix}$

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

From Determinant as Sum of Determinants:


 * $\displaystyle \begin{vmatrix}

a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1} & \cdots & a_{rk} & \cdots & a_{rn} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix} = \sum_{k \mathop = 1}^n \left({a_{rk} \begin{vmatrix} a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\  0 & \cdots & 1 & \cdots & 0 \\ \vdots & \ddots & \vdots & \ddots & \vdots \\  a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix}}\right)$

Consider the determinant:
 * $\begin{vmatrix}

a_{11} & \cdots & a_{1\left({k-1}\right)} & a_{1k} & a_{1\left({k+1}\right)} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)\left({k-1}\right)} & a_{\left({r-1}\right)k} & a_{\left({r-1}\right)\left({k+1}\right)} & \cdots & a_{\left({r-1}\right)n} \\ 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)\left({k-1}\right)} & a_{\left({r+1}\right)k} & a_{\left({r+1}\right)\left({k+1}\right)} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{n\left({k-1}\right)} & a_{nk} & a_{n\left({k+1}\right)} & \cdots & a_{nn} \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:
 * $\operatorname{sgn} \left({\rho}\right) = \left({-1}\right)^{r-1}$

Thus:
 * $\begin{vmatrix}

0 & \cdots & 1 & \cdots & 0 \\ a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)k} & \cdots & a_{\left({r-1}\right)n} \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)k} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix} = \left({-1}\right)^{r-1} \begin{vmatrix} a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)k} & \cdots & a_{\left({r-1}\right)n} \\ 0 & \cdots & 1 & \cdots & 0 \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)k} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix}$

The same argument can be applied to columns.

Thus:
 * $\begin{vmatrix}

1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{1k} & a_{11} & \cdots & a_{1\left({k-1}\right)} & a_{1\left({k+1}\right)} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)k} & a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)\left({k-1}\right)} & a_{\left({r-1}\right)\left({k+1}\right)} & \cdots & a_{\left({r-1}\right)n} \\ a_{\left({r+1}\right)k} & a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)\left({k-1}\right)} & a_{\left({r+1}\right)\left({k+1}\right)} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{nk} & a_{n1} & \cdots & a_{n\left({k-1}\right)} & a_{n\left({k+1}\right)} & \cdots & a_{nn} \end{vmatrix} = \left({-1}\right)^{k-1}\begin{vmatrix} 0 & \cdots & 1 & \cdots & 0 \\ a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)k} & \cdots & a_{\left({r-1}\right)n} \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)k} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix}$

and so:
 * $\begin{vmatrix}

1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ a_{1k} & a_{11} & \cdots & a_{1\left({k-1}\right)} & a_{1\left({k+1}\right)} & \cdots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)k} & a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)\left({k-1}\right)} & a_{\left({r-1}\right)\left({k+1}\right)} & \cdots & a_{\left({r-1}\right)n} \\ a_{\left({r+1}\right)k} & a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)\left({k-1}\right)} & a_{\left({r+1}\right)\left({k+1}\right)} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{nk} & a_{n1} & \cdots & a_{n\left({k-1}\right)} & a_{n\left({k+1}\right)} & \cdots & a_{nn} \end{vmatrix} = \left({-1}\right)^{r+k} \begin{vmatrix} a_{11} & \cdots & a_{1k} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)k} & \cdots & a_{\left({r-1}\right)n} \\ 0 & \cdots & 1 & \cdots & 0 \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)k} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nk} & \cdots & a_{nn} \end{vmatrix}$

Then:
 * $\left({-1}\right)^{r+k} \begin{vmatrix}

a_{11} & \cdots & a_{1\left({k-1}\right)} & a_{1\left({k+1}\right)} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{\left({r-1}\right)1} & \cdots & a_{\left({r-1}\right)\left({k-1}\right)} & a_{\left({r-1}\right)\left({k+1}\right)} & \cdots & a_{\left({r-1}\right)n} \\ a_{\left({r+1}\right)1} & \cdots & a_{\left({r+1}\right)\left({k-1}\right)} & a_{\left({r+1}\right)\left({k+1}\right)} & \cdots & a_{\left({r+1}\right)n} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{n\left({k-1}\right)} & a_{n\left({k+1}\right)} & \cdots & a_{nn} \end{vmatrix}$ is $A_{rk}$, the cofactor of $a_{rk}$ in $D$.

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


 * $\begin{vmatrix}

1 & 0 & \cdots & 0 \\ b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nn} \end{vmatrix} = \begin{vmatrix} b_{22} & \cdots & b_{2n} \\ \vdots & \ddots & \vdots \\ b_{n2} & \cdots & b_{nn} \end{vmatrix}$

Assembling all the pieces derived above, the result follows.

Comment
This result is significant, because it provides a way of calculating the value of a determinant from the values of determinants of lower order.

This is a far more convenient and programmable process than working out all the permutations of the order of the determinant you're calculating, working out which elements of the rows and columns each permutation requires, multiplying them all out, adding them together ...

However, from the derivation of this result, it can be seen that there's another possible technique to explore. We can try reducing one row or column of a matrix (by adding and/or subtracting other rows or columns, etc.) to one with zeroes everywhere except in one place. Then you only have one determinant to calculate.

This technique is taken to its logical conclusion in the exploitation of the Determinant of Triangular Matrix.