How to prove that $I\cdot detA = adjA\cdot A $ without assuming $A$ is invertible?

Question

Suppose that $A$ is a square matrix and I is identity matrix with same size. If $A$ is invertible, it is easy to show that $I\cdot detA = adjA\cdot A$, by multiplying each side from the right with $A$ and multiplying the equation with $detA$ for the equation $A^{-1} = \frac{1}{detA} \cdot adjA$. My question is if we do not know matrix A is invertible, how can we prove that the equation above holds?

Answer 1

Let us denote the $k$-th column of $A$ by $A_k$, and let us write the determinant of $A$ as a (multilinear) function of the columns of $A$ as: $$ \det(A) = \det(A_1,\ldots,A_n). $$ Then the $(ik)$-th entry of the adjugate matrix $B=\mathrm{adj}A$ is defined by $$ B_{ik}=\det(A_1,\ldots,A_{i-1},e_k,A_{i+1},\ldots,A_n), $$ where $e_k$ is the $k$-th column of the identity matrix. We can compute the $(ij)$-th entry of the product $BA$ as $$ \begin{split} (BA)_{ij} &=\sum_kB_{ik}A_{kj}\\ &=\sum_k\det(A_1,\ldots,A_{i-1},e_k,A_{i+1},\ldots,A_n)A_{kj}\\ &=\det(A_1,\ldots,A_{i-1},\textstyle\sum_kA_{kj}e_k,A_{i+1},\ldots,A_n)\\ &=\det(A_1,\ldots,A_{i-1},A_j,A_{i+1},\ldots,A_n)\\ &=\delta_{ij}\det(A), \end{split} $$ yielding $$ BA=\det(A)I. $$