# Other proof for the Second Laplace theorem?

In order to use Cramer's rule, we need to prove for a matrix $$A \in M_n$$ , that $$AB= BA = \det(A)I_n$$ where $$B$$ is the adjugate matrix of $$A$$ given by $$B= (b_{ij})\ ,\ b_{i,j}= (−1)^{i+j} \det(A_{j,i})\ \forall i,j\$$
And $$A_{i,j}\in M_{n-1}$$ is the matrix obtained from $$A$$ by removing the $$i^{th}$$ row and $$j^{th}$$ column.

A portion of that result is to prove that $$\forall i\neq j: (AB)_{ij}= \sum\limits_{k=1}^n a_{ik}(−1)^{k+j} \det(A_{j,k})= 0$$
(and I believe this is called the second Laplace theorem)

We can prove this by induction on $$n$$ starting from $$n=2$$, but I am wondering if there is another way to prove this.

Can you prove that $$\forall i\neq j: \sum\limits_{k=1}^n a_{ik}(−1)^{k+j} \det(A_{j,k})= 0$$ directly without induction ?

Let $$i \neq j$$.

Consider the matrix $$M=(m_{ij})_{1 \leq k,l \leq n}$$ defined by $$m_{k,l} = a_{k,l} \quad \text{if } k \neq j \quad \quad \text{and} \quad \quad m_{j,l} = a_{i,l}$$

(i.e. $$M$$ is obtained from $$A$$ by replacing its $$j-$$th row by the original $$i-$$th row of $$A$$).

For every $$k \in \lbrace 1, ..., n \rbrace$$, notice that the submatrix $$M_{j,k}$$ obtained by deleting the $$j-$$th row and the $$k-$$th column of $$M$$ is exactly $$A_{j,k}$$ (since $$M$$ and $$A$$ are the same, except from the $$j-$$th row).

Furthermore, since $$M$$ has two identical rows, then $$\det(M)=0$$.

But expanding $$\det(M)$$ along the $$j-$$th row, one has $$0=\det(M)= \sum_{k=1}^n m_{jk} (-1)^{j+k} \det(M_{j,k})$$

Since $$m_{jk}=a_{ik}$$ and $$M_{j,k}=A_{j,k}$$, you get finally that $$\boxed{ \sum_{k=1}^n a_{ik} (-1)^{j+k} \det(A_{j,k})=0}$$

Your question was already answered by the observation that it's just the expansion of a determinant with two equal rows, which must be zero. This result is sometimes called the theorem on "alien cofactors" because the cofactors (signed minors) in the expansion are alien to the elements -- and maybe also because Laplace was an alien.

I just wanted to note in passing that you don't need to prove this theorem, or the theorem on the adjugate, in order to prove Cramer's rule. If $$Ax=y$$, then breaking $$A$$ into columns $$A_k$$ we have $$\sum_k A_kx_k=y$$, so by alternating multilinearity of the determinant in columns, \begin{align*} \det(A_1,\ldots,A_{i-1},y,A_{i+1},\ldots,A_n)&=\det(A_1,\ldots,A_{i-1},\textstyle\sum_k A_k x_k,A_{i+1},\ldots,A_n)\\ &=\textstyle\sum_k x_k\det(A_1,\ldots,A_{i-1},A_k,A_{i+1},\ldots,A_n)\\ &=x_i\det A \end{align*} This gives Cramer's formula for $$x_i$$ when $$\det A\ne 0$$.