Determinant formed by cofactors

Question

We are given $$\Delta_0 = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \neq 0.$$

And let $\Delta_1$ denote the determinant formed by the cofactors of elements of $\Delta_0$, and $\Delta_2$ denote the determinant formed by cofactors of $\Delta_1$. Then we have to find value of $\Delta_n$ in terms of $\Delta_0$.

I tried a lot, but did not get any start.

Answer 1

For this, we first need to know the concept of Reciprocal Determinant. It is defined as: When in a given determinant, each element is replaced by it’s cofactor, then the determinant so formed is called reciprocal determinant of the given determinant. If the original determinant is $D$, then the reciprocal determinant is given by $D'$.

Theorem: If $D$ is a determinant of order $n$, and $D'$ be its reciprocal determinant then $D' =D^{n-1}$.
Proof: Let
$$ D= \det{\begin{vmatrix} a_1 & b_1 & \cdots & n_1\\ a_2 & b_2& \cdots & n_2\\ \vdots & \vdots & \ddots & \vdots \\ a_n & b_n & \cdots & n_n\\\ \end{vmatrix}} $$ and $$ D'= \det{\begin{vmatrix} A_1 & B_1 & \cdots & N_1\\ A_2 & B_2& \cdots & N_2\\ \vdots & \vdots & \ddots & \vdots \\ A_n & B_n & \cdots & N_n\\\ \end{vmatrix}} $$ where uppercase letters denote the cofactors of the lowercase letters. Also, we know that $a_iA_j+b_iB_j+\ldots+n_iN_j=D$ where $i=j$ and $a_iA_j+b_iB_j+\ldots+n_iN_j=0$ when $i\neq j$. Thus on multiplying $D$ and $D'$, we have $$ DD'= \det{\begin{vmatrix} D & 0 & \cdots & 0\\ 0 & D & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & D\\\ \end{vmatrix}} =D^{n}. $$ Thus $D'=D^{n-1}$.

Hope you can then take it from here on successive iterations of this theorem.