Laplace Expansion

linear-algebra

Definition

Laplace Expansion

Laplace expansion is a method for computing the determinant of a square matrix by expanding any row or column. The theorem expresses the determinant of a matrix as a sum of products of its entries and their corresponding cofactors.

Let $A=(a_{ij})\in F^{n\times n}$ be a square matrix. The determinant of $A$ can be computed by expanding the $i$-th row as follows:

$$\det A=\sum _{j=1}^n{a}_{ij}\cdot C_{ij}=\sum _{j=1}^n{a}_{ij}\cdot (-1{)}^{i+j}{M}_{ji}$$

or alternatively, by expanding aloing the $j$-th column:

$$\det A=\sum _{i=1}^n{a}_{ij}\cdot C_{ij}=\sum _{i=1}^n{a}_{ij}\cdot (-1{)}^{i+j}{M}_{ij}$$

where $C_{ij}$ is the cofactor and $M_{ij}$ is the minor of $a_{ij}$.

It was named after Pierre-Simon Laplace.

Example

Consider a $3\times 3$ matrix $A$:

$$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$$

To compute the determinant of $A$, the row $i=1$ is chosen.

$$\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|=\sum _{j=1}^3{a}_{1j}\cdot \underset{=C_{1j}}{\underbrace{(-1{)}^{1+j}{M}_{1j}}}$$ $$\begin{array}{rlr}{M}_{11}& =\left|\begin{array}{ccc}\cancel{1}& \cancel{2}& \cancel{3}\\ \cancel{4}& 5& 6\\ \cancel{7}& 8& 9\end{array}\right|& =\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|=5\cdot 9-6\cdot 8=-3\\ C_{11}& =(-1{)}^{1+1}{M}_{11}=-3& \\ M_{12}& =\left|\begin{array}{ccc}\cancel{1}& \cancel{2}& \cancel{3}\\ 4& \cancel{5}& 6\\ 7& \cancel{8}& 9\end{array}\right|& =\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|=4\cdot 9-6\cdot 7=-6\\ C_{12}& =(-1{)}^{1+2}{M}_{12}=6& \\ M_{13}& =\left|\begin{array}{ccc}\cancel{1}& \cancel{2}& \cancel{3}\\ 4& 5& \cancel{6}\\ 7& 8& \cancel{9}\end{array}\right|& =\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|=4\cdot 8-5\cdot 7=-3\\ C_{13}& =(-1{)}^{1+3}{M}_{13}=-3& \end{array}$$

Now, the determinant of $A$ can be computed:

$$\begin{array}{rl}\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|& =\sum _{j=1}^3{a}_{1j}\cdot \underset{=C_{1j}}{\underbrace{(-1{)}^{1+j}{M}_{1j}}}\\ & =1\cdot (-3)+2\cdot 6+3\cdot (-3)\\ & =0\end{array}$$

Thus, the determinant of $A$ is $0$.