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 $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be an order-$k$ minor of $D$.

Let $\map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be the cofactor of $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$.

Then:

$\displaystyle D = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k} \, \map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$

A similar result applies for columns.

Proof

Let us define $r_{k + 1}, r_{k + 2}, \ldots, r_n$ such that:

$1 \le r_{k + 1} < r_{k + 2} < \cdots < r_n \le n$
$\rho = \tuple {r_1, r_2, \ldots, r_n}$ is a permutation on $\N^*_n$.

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

Then by Permutation of Determinant Indices we have:

 $\ds D$ $=$ $\ds \sum_\sigma \map \sgn \rho \, \map \sgn \sigma \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}$ $\ds$ $=$ $\ds \sum_\sigma \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + s_i} } \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}$

We can obtain 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 $\tuple {s_1, \ldots, s_k}$ vary over the first set and $\tuple {s_{k + 1}, \ldots, s_n}$ over the second set.

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

$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
$\tuple {s_1, \ldots, s_k} = \map \sigma {u_1, \ldots, u_k}$
$\tuple {s_{k + 1}, \ldots, s_n} = \map \sigma {u_{k + 1}, \ldots, u_n}$

Thus we get:

 $\ds D$ $=$ $\ds \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \sum_{\map \sigma {u_1, \ldots, u_k} } \, \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \prod_{j \mathop = 1}^k a_{\map \rho j \, \map \sigma j}$ $\ds$ $\times$ $\ds \sum_{\map \sigma {u_{k + 1}, \ldots, u_n} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = k + 1}^n a_{\map \rho j \, \map \sigma j}$ $\ds$ $=$ $\ds \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \begin {vmatrix} a_{r_1 u_1} & \cdots & a_{r_1 u_k} \\ \vdots & \ddots & \vdots \\ a_{r_k u_1} & \cdots & a_{r_k u_k} \end {vmatrix} \times \begin {vmatrix} a_{r_{k + 1} u_{k + 1} } & \cdots & a_{r_{k + 1} u_n} \\ \vdots & \ddots & \vdots \\ a_{r_n u_{k + 1} } & \cdots & a_{r_n u_n} \end {vmatrix}$ $\ds$ $=$ $\ds \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map D {r_{k + 1}, \ldots, r_n \mid u_{k + 1}, \ldots, u_n}$ $\ds$ $=$ $\ds \sum_{\map \sigma {u_1, \ldots, u_n} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k}$ $\ds$ $=$ $\ds \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \, \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \sum_{u_{k + 1}, \ldots, u_n} 1$

That last inner sum extends over all integers which satisfy:

$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$

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.

$\blacksquare$

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 $D$.

Also known as

This theorem is also known as the Laplace cofactor expansion.

Examples

Arbitrary $3 \times 3$ Matrix

Let $\mathbf A$ be the matrix defined as:

$\mathbf A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

Then $\map \det {\mathbf A}$ can be calculated using Laplace's Expansion Theorem as follows.

Expanding row $2$:

 $\ds \map \det {\mathbf A}$ $=$ $\ds \paren {-1}^{2 + 1} \times 4 \begin {vmatrix} 2 & 3 \\ 8 & 9 \end {vmatrix}$ $\ds$  $\, \ds + \,$ $\ds \paren {-1}^{2 + 2} \times 5 \begin {vmatrix} 1 & 3 \\ 7 & 9 \end {vmatrix}$ $\ds$  $\, \ds + \,$ $\ds \paren {-1}^{2 + 3} \times 6 \begin {vmatrix} 1 & 2 \\ 7 & 8 \end {vmatrix}$ $\ds$ $=$ $\ds - 4 \paren {2 \times 9 - 3 \times 8}$ $\ds$  $\, \ds + \,$ $\ds 5 \paren {1 \times 9 - 3 \times 7}$ $\ds$  $\, \ds - \,$ $\ds 6 \paren {1 \times 8 - 2 \times 7}$ $\ds$ $=$ $\ds 4 \paren {24 - 18} + 5 \paren {9 - 21} + 6 \paren {14 - 8}$ $\ds$ $=$ $\ds 0$

This shows that $\mathbf A$ is non-invertible.

Source of Name

This entry was named for Pierre-Simon de Laplace.