Let $A$ be an $n\times n$ matrix with entries $a\_{i,j}$. Define an $(n-1)\times(n-1)$ matrix $B$ with entries $b\_{i,j}=a\_{1,1}a\_{i+1,j+1}-a\_{1,j+1}a\_{i+1,1}$. Then $\det(B)=a\_{1,1}^{n-2}\det(A)$.


I can prove this by direct computation, but it seems like something that may be well known or follow from other properties of determinants. Do you either know a reference for this result or know how to give a simple proof of it?



Determinant in terms of certain $2\times 2$ minors
A simple proof is given in the [Art of Problem Solving](https://artofproblemsolving.com/community/c7h1328834p7152628) (it is entered as an "olympiade problem").


I reproduce the two-line proof for the record, with the change that $a\_{11}\mapsto a\_{nn}$: 
Add the $n$-th row of $A$ to the $i$-th row, multiplied by $-a\_{in}/a\_{nn}$:
$$\det{A}= \left| \begin{array}{cccccc}
a\_{1,1}-a\_{n,1}\dfrac{a\_{1,n}}{a\_{n,n}} & \dots & a\_{1,j}-a\_{n,j}\dfrac{a\_{1,n}}{a\_{n,n}} & \dots & a\_{1,n-1}-a\_{n,n-1}\dfrac{a\_{1,n}}{a\_{n,n}} & 0 \\ 
\dots & \dots & \dots & \dots & \dots & \dots \\ 
a\_{i,1}-a\_{n,1}\dfrac{a\_{i,n}}{a\_{n,n}} & \dots & a\_{i,j}-a\_{n,j}\dfrac{a\_{i,n}}{a\_{n,n}} & \dots & a\_{i,n-1}-a\_{n,n-1}\dfrac{a\_{i,n}}{a\_{n,n}} & 0 \\ 
\dots & \dots & \dots & \dots & \dots & \dots \\ 
a\_{n-1,1}-a\_{n,1}\dfrac{a\_{n-1,n}}{a\_{n,n}} & \dots & a\_{n-1,j}-a\_{n,j}\dfrac{a\_{n-1,n}}{a\_{n,n}} & \dots & a\_{n-1,n-1}-a\_{n,n-1}\dfrac{a\_{n-1,n}}{a\_{n,n}} & 0 \\ 
a\_{n,1} & \dots & a\_{n,j} & \dots & a\_{n,n-1} & a\_{n,n}
\end{array} \right|$$
$$=a\_{n,n} \cdot \left| \begin{array}{ccccc}
\dfrac{b\_{1,1}}{a\_{n,n}} & \dots & \dfrac{b\_{1,j}}{a\_{n,n}} & \dots & \dfrac{b\_{1,n-1}}{a\_{n,n}} \\ 
\dots & \dots & \dots & \dots & \dots \\ 
\dfrac{b\_{i,1}}{a\_{n,n}} & \dots & \dfrac{b\_{i,j}}{a\_{n,n}} & \dots & \dfrac{b\_{i,n-1}}{a\_{n,n}} \\ 
\dots & \dots & \dots & \dots & \dots \\ 
\dfrac{b\_{n-1,1}}{a\_{n,n}} & \dots & \dfrac{b\_{n-1,j}}{a\_{n,n}} & \dots & \dfrac{b\_{n-1,n-1}}{a\_{n,n}}
\end{array} \right|= \dfrac{1}{{a\_{n,n}}^{n-2}} \cdot \det{B} $$
