The $n-1$ order principle minor of the adjoint matrix of $A$ has determinant $|A|^{n-2}a_{nn}$. Let $A$ be a $n\times n$ matrix, $A^*$ is its adjoint, that is, $A=\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1n}&\cdots&A_{nn}\end{pmatrix}$, where $A_{ij}$ is the cofactor of $a_{ij}$ in $A$. Show that $det\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1,n-1}&\cdots&A_{n-1,n-1}\end{pmatrix}=|A|^{n-2}a_{nn}$. How to show? We do not know whether or not $A$ or the $n-1$ matrix is invertible?
 A: To prove an identity of this sort, the usual technique is to prove it for a matrix $A$ containing $n^2$ indeterminates, so that $A$ is always invertible. Then, by "substituting" some concrete numbers into the indeterminates, we automatically get an identity for this new $A$ with number entries, even if it is singular.
(There is also a similar but subtly different perspective from Zariski topology, which loosely says that the set of all invertible matrices are "dense" in the set of all matrices, so that in most cases, if a matrix identity holds for all invertible matrices, we can pass the identity to the limit to make it valid for all square matrices. If the field in question is $\mathbb F=\mathbb R$ or $\mathbb C$, you can also ignore Zariski topology and consider the usual topology of $M_n(\mathbb F)$.)
Anyway, suppose $A$ is invertible. The identity then follows from a simple block-matrix argument that was portrayed in Prasolov, Problems and Theorems in Linear Algebra, theorem 2.5.1. Partition $A$ as $\pmatrix{X&y\\ z&a_{nn}}$ and partition $\operatorname{adj}(A)$ in a conforming manner as $\pmatrix{U&v\\ w&m_{nn}}$, where $a_{nn}$ is the bottom right entry of $A$ and $m_{nn}$ is the corresponding $(n-1)$-rowed minor. Since $A\operatorname{adj}(A)=\det(A)I_n$, we have
$$
\pmatrix{X&y\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)I_n.
$$
Hence
$$
\pmatrix{z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)\pmatrix{0&1}=\pmatrix{0&\det(A)}
$$
and
$$
\pmatrix{I&0\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\pmatrix{U&v\\ 0&\det(A)}.
$$
Take determinants on both sides, we obtain $a_{nn}\det(\operatorname{adj}(A))=\det(A)\det(U)$. However, from $A\operatorname{adj}(A)=\det(A)I_n$ we also obtain $\det(\operatorname{adj}(A))=\det(A)^{n-1}$. Therefore $a_{nn}\det(A)^{n-2}=\det(U)$.
