Laplace's Expansion Theorem

Theorem
Let $D$ be the determinant of order $n$.

Let $r_1, r_2, \ldots, r_k$ be integers such that:
 * $1 \le k < n$


 * $1 \le r_1 < r_2 < \cdots < r_k \le n$

Let $D \left({r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}\right)$ be an order-$k$ minor of $D$.

Let $\tilde D \left({r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}\right)$ be the cofactor of $D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$.

Then:
 * $\displaystyle D = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} D \left({r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}\right) \tilde D \left({r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}\right)$

A similar result applies for columns.

Proof
Let us define $r_{k+1}, r_{k+2}, \ldots, r_n$ such that
 * $1 \le r_{k+1} < r_{k+2} < \cdots < r_n \le n$
 * $\rho = \left({r_1, r_2, \ldots, r_n}\right)$ is a permutation on $\N^*_n$.

Let $\sigma = \left({s_1, s_2, \ldots, s_n}\right)$ be a permutation on $\N^*_n$.

Then by Permutation of Determinant Indices we have:

We can get all the permutations $\sigma$ exactly once by separating the numbers $1, \ldots, n$ in all possible ways into a set of $k$ and $n-k$ numbers.

We let $\left({s_1, \ldots, s_k}\right)$ vary over the first set and $\left({s_{k+1}, \ldots, s_n}\right)$ over the second set.

So the summation over all $\sigma$ can be replaced by:


 * $\left({u_1, \ldots, u_n}\right) = \sigma \left({1, \ldots, n}\right)$
 * $u_1 < u_2 < \cdots < u_k, u_{k+1} < u_{k+2} < \cdots < u_n$
 * $\left({s_1,\ldots, s_k}\right) = \sigma \left({u_1, \ldots, u_k}\right)$
 * $\left({s_{k+1},\ldots, s_n}\right) = \sigma \left({u_{k+1}, \ldots, u_n}\right)$

Thus we get:

That last inner sum extends over all integers which satisfy:
 * $\left({u_1, \ldots, u_n}\right) = \sigma \left({1, \ldots, n}\right)$
 * $u_1 < u_2 < \cdots < u_k, u_{k+1} < u_{k+2} < \cdots < u_n$

But for each set of $u_1, \ldots, u_k$, then the integers $u_{k+1}, \ldots, u_n$ are clearly uniquely determined.

So that last inner sum equals 1 and the theorem is proved.

The result for columns follows from Determinant of Transpose.

Comment
This gives us an expansion of the determinant $D$ in terms of $k$ specified rows.

We form all possible order-$k$ minors of $D$ which involve all of these rows, and multiply each of them by their cofactors.

The sum of these products is equal to $D$.

We note that when $k = 1$ this becomes the Expansion Theorem for Determinants.

Also known as
This theorem is also known as the Laplace cofactor expansion.