Let A be a square matrix of order $n$. Then we have the following well-known result:
$$\textbf{A}\big(\text{cof}(\textbf{A})\big)^T = \big(\text{det} (\textbf{A})\big) \textbf{I}$$
where $\text{cof}(\textbf{A})$ is the cofactor matrix of $\textbf{A}$. Assume that the determinant of $\text{det} (\textbf{A}) \neq 0$. Then,
$$\frac{1}{\text{det} (\textbf{A})}\textbf{A}\big(\text{cof}(\textbf{A})\big)^T = \textbf{I} \Rightarrow \bigg(\big(\text{cof}(\textbf{A})\big)^T\bigg)^{-1} = \frac{1}{\text{det} (\textbf{A})}\textbf{A}$$
Thus, we have shown that the transpose of the cofactor matrix is not singular, so we can also write:
$$\frac{1}{\text{det} (\textbf{A})}\textbf{A}\big(\text{cof}(\textbf{A})\big)^T = \big(\text{cof}(\textbf{A})\big)^T\frac{1}{\text{det} (\textbf{A})}\textbf{A}$$
or
$$\textbf{A}\big(\text{cof}(\textbf{A})\big)^T =\big(\text{cof}(\textbf{A})\big)^T\textbf{A}$$
How do I prove this commutative property if the determinant is zero: $\text{det}(\textbf{A}) = 0$?
We know that $$ A\bigl(\text{cof}(A)\bigr)^T=\det(A)I $$ Our goal is to show $$ \bigl(\text{cof}(A)\bigr)^TA=\det(A)I $$ Identically we have $$\text{cof}(A)=\Bigl(\text{cof}\bigl(A^T\bigr)\Bigr)^T$$ hence \begin{align*} &\bigl(\text{cof}(A)\bigr)^TA=\det(A)I\\[4pt] \iff&\Bigl(\bigl(\text{cof}(A)\bigr)^TA\Bigr)^T\!=\bigl(\det(A)I\bigr)^T\\[4pt] \iff&A^T(\text{cof}(A))=\det(A)I\\[4pt] \iff&A^T\Bigl(\text{cof}\bigl(A^T\bigr)\Bigr)^T\!=\det(A)I\\[4pt] \iff&\det(A^T)I=\det(A)I\\[4pt] \iff&\det(A^T)=\det(A)\\[4pt] \end{align*} which is true.