Given two matrices, prove that their product is identity We are given two matrices: $A=[a_{ij}]$ and cofactor $B_{ji} = (-1)^{i+j} \frac{\det A_{ij}}{\det A} $. Prove that then $A \cdot B = I$.
I know that $(AB)_{ki} = \sum _j a_{kj}b_j =  \sum _j a_{kj} (-1)^{i+j} \frac{\det A_{ij}}{\det A}$.
When $k=i$, we have $ \sum _j a_{kj} (-1)^{i+j} \frac{\det A_{ij}}{\det A}= \frac{\det A_{ij}}{\det A} =1$.
But I don't know what to do for $k \neq i$.
Could you help me?
 A: Let $A_{rj}$ denote the matrix obtained from the $(n\times n)$-matrix $A$ by deleting  ${\rm row}_r(A)$ and ${\rm col}_j(A)$, and pushing the remains together. Then "expanding $\det(A)$ with respect to ${\rm row}_r(A)$" means the formula
$$\det(A)=\sum_{j=1}^n (-1)^{r+j} a_{rj}\det(A_{r,j})\ .\tag{1}$$
Assume now that $s\ne r$ and that $A'$ denotes the matrix obtained from $A$ by replacing ${\rm row}_r(A)$ bei ${\rm row}_s(A)$. Then $A'_{rj}=A_{rj}$ for all $j$. Since $A'$ has two equal rows we get
$$0=\det(A')=\sum_{j=1}^n (-1)^{r+j} a_{sj}\det(A_{r,j})\ .\tag{2}$$
We now introduce the matrix $B$ with $$b_{jr}:=(-1)^{r+j}{\det(A_{r,j})\over\det(A)}\ .$$
Then
$$\sum_{j=1}^n  a_{sj}b_{jr}={1\over\det(A)}\sum_{j=1}^n (-1)^{r+j} a_{sj}\det(A_{r,j})\ ,$$
and here the right side is $=1$ if $s=r$ by $(1)$, and $=0$ if $s\ne r$ by $(2)$.
A: Notice that it is the same as the laplace's formula just in the different direction.
This is called cramers rule und gives you the inverse of a Matrix.
A: The same argument for $k \ne i$ just gives you a new matrix $A'$ on the numerator instead of $A$.  If you think about it, the $i$th row of $A'$ is equal to the $k$th row of $A$, so...
