Intuition of finding the formula of Laplace's expansions of the Determinant?

How did Laplace find the formula $$\left |A \right |=\sum_{i=1}^{n}(-1)^{i+j}(A)_{ij}M_{ij}$$? What is the intuition of the evalution of this formula?

Note: I'm not asking for proof that the formula is indeed a formula for the determinant, I'm asking about how does one evaluate and discovers a formula like this.

Another note: In the formula, $$1\leq j\leq n$$ is fixed. The minor $$M_{ij}$$ is the determinant of the $$(n-1)*(n-1)$$ matrix that results from $$A$$ by removing the i-th row and the j-th column.

• Just one precision: what is $M$? And you should add the summation on $j$, too, please! – Milloupe May 9 at 14:21
• @Milloupe $M_{ij}$ is to be understood as a minor, thus no summation on $j$ ! – Jean Marie May 9 at 14:26
• @JeanMarie, sorry, English not being my native language, it wasn't obvious to me that M stood for minor, thank you! – Milloupe May 9 at 14:30
• Look at math.stackexchange.com/q/845196 with a non-classical answer mentionning the important "Clifford algebra" vision. – Jean Marie May 9 at 14:32
• @JeanMarie Well, that's one intuition, but I'm pretty sure Laplace did not know about Clifford algebras. – Robert Israel May 9 at 15:01

I don't know Laplace's actual steps, but the expansion follows from Cramer's Rule, which was known before Laplace. Namely, if $$A$$ is a nonsingular $$n \times n$$ matrix the solution of the equation $$A x = e_i$$, where $$e_i$$ is the vector with $$1$$ in position $$i$$ and $$0$$ otherwise, is $$x_j = (-1)^{i+j} M_{ij}/\det(A)$$. Then the $$i$$'th entry of $$A x = e_i$$ gives you $$\sum_{j} a_{ij} x_j = \sum_{j} \frac{(-1)^{i+j} a_{ij} M_{ij}}{\det(A)} = 1$$ Now solve for $$\det(A)$$.