A result for all invertible matrices?

Question

Consider some invertible $N\times N$ matrix $A$, let $B=A^{-1}$ and let $A_{\left[11,nn\right]}$ be the $\left(N-2\right)\times\left(N-2\right)$ matrix resulting from A after removing row and column 1 and row and column $n \neq 1$. I have seen that in some special case the following equality holds for $n \neq 1$: $$ b_{11}b_{nn}-b_{n1}b_{1n}=\frac{\det A_{\left[11,nn\right]}}{\det A} $$ Is this a general result for all invertible matrices?

Answer 1

Yes, this result is correct in general. Here is a proof.

Without loss of generality, we can take $n=2$. Partition $A$ and $B$ as $$ A = \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}}, \quad B = \pmatrix{B_{11} & B_{12}\\ B_{21} & B_{22}}, $$ where $A_{11}$ and $B_{11}$ have size $2 \times 2$. Because $BA = I$, we can deduce that $$ \pmatrix{B_{11} & B_{12}\\ 0 & I_{N-2}} \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}} = \pmatrix{I_2 & 0\\A_{21} & A_{22}}. $$ Taking the determinant of both sides yields $$ \det\pmatrix{B_{11} & B_{12}\\ 0 & I_{N-2}}\det \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}} =\det \pmatrix{I_2 & 0\\A_{21} & A_{22}} \implies\\ \det(B_{11})\det(A) = \det(A_{22}) \implies\\ \det(B_{11}) = \frac{\det(A_{22})}{\det(A)}, $$ which is what we wanted to show.