the determinant of the adjugate matrix removing the last line and the last column $A=(a_{ij})$ is an $n\times n$ matrix and $\det(A)=d\neq 0$. Removing the last line and the last column of $\text{adj}(A),$ we get the $(n-1) \times (n-1)$ matrix $B$.
Find $\det(B)$.
Thanks.
 A: $\det(B)$ is just the right-bottom entry of $adj(adj(A))$. And $adj(adj(A))=d^{n-2}A$.So $\det(B)=d^{n-2}A_{n,n}$.
I learned it from somebody just now.
And in another hand,we can use Jacobi's theorem.(see comments)
A: Here is the algebraic derivation. First write $$\bf{A} = \left(\matrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right) \,\,\,\text{and}\,\,\,\, \bf{A}^{-1}=\pmatrix{\bf{E} & \bf{f} \\ \bf{g}^\top & h}$$
and define determinant $\Delta = \det\bf{A}$. The dimension of $\bf{A}$ is $n$. The question then is the determinant of the matrix $\left(\Delta\bf{I}\cdot\bf{E}\right)$ which has dimension $n-1$. Let $k$ be non-zero (otherwise the result is trivial; multiplying the terms of $A^{-1}A$ shows $E$ to be singular).
Rewriting the determinant:
$$\Delta = \det\left[\overbrace{\pmatrix{\bf{I} & -\mathbf{b}\frac{1}{k} \\ \mathbf{0}^\top & 1\\}}^{\det = 1}\pmatrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right]$$
or
$$\Delta = \det\left(\mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top\right)\cdot k$$
Since the $\bf{b}$ column was zeroed out, the determinant is the multiplication of the blocks' determinants. Also the unimodular row operation matrix does not change the determinant (standard algebra). 
Now with that form of the determinant handy, we focus on the $\bf{E}$ matrix. Using a previously derived formula (also worked out algebraically) we write:
$$\mathbf{E}^{-1} = \mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top$$
Hmm, that looks familiar, it is the matrix that determined $\Delta$. So  now we have
$$\Delta = \det\left(\mathbf{E}^{-1}\right)\cdot k$$
or
$$\frac{1}{\det\left(\mathbf{E}^{-1}\right)}= \det(\mathbf{E}) = \frac{k}{\Delta}$$
And we finally arrive at the formula of interest (so long as we remember that the dimension of $\mathbf{E}$ is $n-1$)
$$\det\left(\Delta\bf{I}\cdot\bf{E}\right) = \Delta^{n-1}\frac{k}{\Delta} = \Delta^{n-2}k$$
