[MathWorld](https://mathworld.wolfram.com/SylvestersDeterminantIdentity.html) presents the following two versions of Sylvester's determinant identity, relating to an $n\times n$ matrix $\mathbb{A}$:


First:
$$
|\mathbb{A}||A\_{r\,s,p\,q}| = |A\_{r,p}||A\_{s,q}| - |A\_{r,q}| |A\_{s,p}|
$$


where $r$ and $s$ ($p$ and $q$) are sets that indicate which rows (columns) of $\mathbb{A}$ are to be **deleted** (correcting MathWorld's typo).


Second:  

$$
|\mathbb{A}|\left[ a\_{k\,k}^{(k-1)}\right]^{n-k-1}
=
\left|
\begin{matrix}
a\_{k+1\, k+1}^{(k)} & \cdots & a\_{k+1\, n}^{(k)} \\
\vdots & \ddots & \vdots \\
a\_{n\, k+1}^{(k)} & \cdots & a\_{n\, n}^{(k)} \\
\end{matrix}\right|
$$


where


$$ 
a\_{i\, j}^{(k)}
=
\left|
\begin{matrix}
a\_{11} & \cdots & a\_{1\,k} & a\_{1 \, j} \\
\vdots & \ddots & \vdots& \vdots \\
a\_{k\ 1} & \cdots & a\_{k\,k}& a\_{k\, j} \\
a\_{i\ 1} & \cdots & a\_{i\, k} & a\_{i\, j} \\
\end{matrix}
\right|
$$


for $k<i$, $j \leq n$.


Would anyone help me prove that these two versions are indeed equivalent.


Note: As has been pointed our below, MathWorld's claim is obviously (using a counterexample) incorrect; the second version implies only a special case (when r,s,p,q are single numbers) of the first 'version'.



Two versions of Sylvester identity
For the special case that $r,s,p,q$ are single elements, it is shown in [these notes](http://www.dm.unibo.it/~morigi/prova1_file/Talks_Web/MRRusso.pdf) (page 7) how the first identity (known as the [Desnanot-Jacobi identity](https://en.wikipedia.org/wiki/Dodgson_condensation#Desnanot%E2%80%93Jacobi_identity_and_proof_of_correctness_of_the_condensation_algorithm)) follows from the second identity.


Apply the second identity to the matrix  

![](https://ilorentz.org/beenakker/MO/sylvester.png)  

we thus arrive at the first identity, illustrated graphically as  

![](https://i.stack.imgur.com/dQINA.png)  

[source](http://poisson.ms.u-tokyo.ac.jp/~mi/posters/poster1_fpsac13.pdf)




---


**Update**: Since I could not find the first identity in the literature, for the more general case when $r,s,p,q$ each contain more than a single element, I tried to check it for an example. I took $n=6$, $r=1,2$, $s=5,6$, $p=1,2$, $q=5,6$. For the $6\times 6$ matrix $A$ I took
$$A=\left(
\begin{array}{cccccc}
 0 & 1 & 0 & 0 & 1 & 1 \\
 0 & 0 & 1 & 0 & 1 & 0 \\
 2 & 0 & 2 & 2 & 0 & 0 \\
 2 & 1 & 0 & 0 & 2 & 1 \\
 2 & 0 & 2 & 1 & 0 & 2 \\
 1 & 0 & 0 & 0 & 2 & 2 \\
\end{array}
\right)$$
The left-hand-side of the first identity is 0,  

$\det A \det \left(
\begin{array}{cc}
 2 & 2 \\
 0 & 0 \\
\end{array}
\right) = 24\cdot 0 = 0$,  

but for the right-hand-side I find a nonzero answer:


$$ \det \left(
\begin{array}{cccc}
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 2 & 0 & 2 & 2 \\
 2 & 1 & 0 & 0 \\
\end{array}
\right) \det \left(
\begin{array}{cccc}
 2 & 2 & 0 & 0 \\
 0 & 0 & 2 & 1 \\
 2 & 1 & 0 & 2 \\
 0 & 0 & 2 & 2 \\
\end{array}
\right) - \det \left(
\begin{array}{cccc}
 2 & 0 & 2 & 2 \\
 2 & 1 & 0 & 0 \\
 2 & 0 & 2 & 1 \\
 1 & 0 & 0 & 0 \\
\end{array}
\right) \det \left(
\begin{array}{cccc}
 0 & 0 & 1 & 1 \\
 1 & 0 & 1 & 0 \\
 2 & 2 & 0 & 0 \\
 0 & 0 & 2 & 1 \\
\end{array}
\right)=$$
$$\qquad\qquad= (-4) \cdot 4 - (-2)\cdot (-2)=-20\neq 0$$
Incidentally, I did find a determinantal identity of a somewhat similar form in [Tao's blog](https://terrytao.wordpress.com/2017/08/28/dodgson-condensation-from-schur-complementation/) (last equation, Karlin's identity). But it is not quite of the form of the first identity in the OP.


So unless I have made a mistake, my conclusion is that the first identity in the OP only holds when $r,s,p,q$ are single elements, but not more generally.


