Let $A$ be an $n \times n$ matrix. Let $C_{ij}$ be the $ij^{th}$ cofactor of $A$, defined by $C_{ij} = (-1)^{i+j} \det A^{ij}$ and let $ C = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & \cdots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{bmatrix} $ be the cofactor matrix. Also let $M$ be the matrix of minors; that is, the matrix that appears in the left hand side of the equation in the question. We will show that
- $\det C = (\det A)^{n-1}$, and
- $\det M = \det C$,
which together establish the claim.
Claim 1. From this wikipedia page, we know that $ C^T A = A C^T = (\det A) I_n. $
If $A$ is invertible, then taking determinants on both sides, we get: $ \det (C^T) \cdot \det A = (\det A)^n \cdot 1. $ Since $\det (C^T) = \det C$ and $\det A \neq 0$, we get $\det C = (\det A)^{n-1}$, and we are done.
On the other hand, if $A$ is noninvertible, then $\det A = 0$ and $ C^T A = AC^T = 0$. In this case, (WLOG assuming $n \geq 1$) it suffices to show that $\det C = 0$, which is equivalent to showing that $C$ is noninvertible as well. Here, we have two subcases:
If $A$ is the zero matrix, then $C$ is also the zero matrix, and hence non-invertible.
Suppose $A$ is nonzero. Then if $C$ is invertible, then $C^T A = 0$ implies that $A = \Big((C^{-1})^T C^T \Big) A = (C^{-1})^T \cdot 0 = 0 ,$ which is a contradiction. Hence $C$ is noninvertible, and we are done.
Claim 2. We now prove that $\det M = \det C$. We have the relation $C_{ij} = (-1)^{i+j} M_{ij} = (-1)^{i+j} \det A^{ij}$. So, if we take the matrix $M$, and multiply the $i^{\rm th}$ row by $(-1)^i$ and $j^{\rm th}$ column by $(-1)^j$, then we end up with the matrix $C$. In this process, the determinant gets multiplied by $ \left(\prod_{i=1}^n (-1)^i \right) \left(\prod_{j=1}^n (-1)^j \right) = \left(\prod_{i=1}^n (-1)^i \right)^2 = \prod_{i=1}^n (-1)^{2i} = 1. $ In other words, $\det C = \det M$.