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

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

Then $$D = \sum_{1 \le u_1 < \cdots < u_k \le n} D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right) \tilde D \left({r_1, r_2, \ldots, r_k | u_1, u_2, \ldots, u_k}\right)$$.

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 $\mathbb{N}^*_n$.

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

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.