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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.

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

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