cof($AB$) = cof($A$)$\cdot$cof($B$) Let $A$ and $B$ be two $n\times n$ matrices and let cof$(A)$ and cof$(B)$ denote the cofactor matrix of A and B respectively. Then show that cof($AB$) = cof($A$)$\cdot$cof($B$)? I am trying to show it using the definition of cofactor i.e. $[\text{cof}(A)]_{ij} := \sum_{\sigma \in S_{n-1}} sgn(\sigma) \prod_{t = 1}^{n-1} a_{f_i(t)f_j(\sigma(t))} $ where $, f_k(t):= \begin{cases} t &t < k \\\ t + 1 &t \geq k\end{cases} \text{for } k \in [n-1]$ and using the definition of matrix multiplication. However, it is not clear as to how to do so. If there is a simpler proof, it would be appreciated. I looked at a similar question asked before: Proof of adjoint(ab) = adjoint(b)adjoint(a), however no rigorous answer was given, so I ask this question again. P.S: I am using this property to show a similar property for adjunct of a matrix, so I wouldn't want a solution which takes that property as a starting point.

**Prove it for invertible matrices:**

If $A$ and $B$ are invertible then the statement holds because: $$ A^{-1} = \frac{1}{\det A}(\text{ cof } A)^T $$ Can be turned into: $$ \text{cof }A = \left[(\det A)A^{-1}\right]^T $$ Then $\text{cof } A \text{ cof } B = \text{ cof } (AB)$ follows.

**Generalization:**

$\text{cof } A \text{ cof } B = \text{ cof } (AB)$ can be seen as a polynomial equation with $2n^2$ unknowns with integer coefficients. In a polynomial ring with coefficients in $\mathbb{Z}$, we may use the fact that $A$ and $B$ will be invertible in the field of algebraic fractions with coefficients in $\mathbb{Q}$. Then because of our first step, the equation holds in general because we are allowed to replace the unknowns with any elements from a commutative ring.