Determinant in terms of certain $2\times 2$ minors

Question

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?

Answer 1

A simple proof is given in the Art of Problem Solving (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 $-a_{in}/a_{nn}$: $$\det{A}= \left| \begin{array}{cccccc} a_{1,1}-a_{n,1}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,j}-a_{n,j}\dfrac{a_{1,n}}{a_{n,n}} & \dots & a_{1,n-1}-a_{n,n-1}\dfrac{a_{1,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{i,1}-a_{n,1}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,j}-a_{n,j}\dfrac{a_{i,n}}{a_{n,n}} & \dots & a_{i,n-1}-a_{n,n-1}\dfrac{a_{i,n}}{a_{n,n}} & 0 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{n-1,1}-a_{n,1}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,j}-a_{n,j}\dfrac{a_{n-1,n}}{a_{n,n}} & \dots & a_{n-1,n-1}-a_{n,n-1}\dfrac{a_{n-1,n}}{a_{n,n}} & 0 \\ a_{n,1} & \dots & a_{n,j} & \dots & a_{n,n-1} & a_{n,n} \end{array} \right|$$ $$=a_{n,n} \cdot \left| \begin{array}{ccccc} \dfrac{b_{1,1}}{a_{n,n}} & \dots & \dfrac{b_{1,j}}{a_{n,n}} & \dots & \dfrac{b_{1,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{i,1}}{a_{n,n}} & \dots & \dfrac{b_{i,j}}{a_{n,n}} & \dots & \dfrac{b_{i,n-1}}{a_{n,n}} \\ \dots & \dots & \dots & \dots & \dots \\ \dfrac{b_{n-1,1}}{a_{n,n}} & \dots & \dfrac{b_{n-1,j}}{a_{n,n}} & \dots & \dfrac{b_{n-1,n-1}}{a_{n,n}} \end{array} \right|= \dfrac{1}{{a_{n,n}}^{n-2}} \cdot \det{B} $$

Answer 2

This can also be done in terms of the Schur complement determinant formula.

The matrix $B$ is by definition $a_{1,1}A/{1}$, where $A/{1}$ is the Schur complement of $A$ with respect to the $1 \times 1$ submatrix with that one diagonal entry. The determinant formula for Schur complements says $$ \det(A)=a_{1,1}\det(A/{1})=\frac{\det(B)}{a_{1,1}^{n-2}}, $$ which is what we wanted.

Answer 3

Thank you to @carlo-beenakker, @erick-wong, @quizzical, and @sam-hopkins for helpful answers and comments.

Investigating further, I found that the statement in the question appears to be a well-known result of Chio from his 1853 manuscript Mémoire sur les fonctions connues sous le nom de résultantes ou de déterminans. I was unable to obtain Chio's manuscript but found a summary of it in Muir's 1911 work The Theory of Determinants in the Historical Order of Development, Volume II. Footnote 2 of this paper discusses the rendering of Chio's name. Sections 2 and 3 of this paper discuss the relationship between Chio's result and a more general result stated without proof by Sylvester in his 1851 article On the relation between the minor determinants of linearly equivalent quadratic functions.

I found an appropriate reference to a proof to be Eves's 1966 textbook Elementary Matrix Theory, which includes the statement in this question as Theorem 3.6.1.