In other words, your question is whether $A=[a_{i,j}]_{i,j=1}^n$ and $d=[d_1,\dots,d_n]^T$ are independent. This is of course so for $n=1$.
However, $A$ and $d$ are not independent already for $n=2$. Indeed, let $$R:=\frac{d_1^2}{d_1^2+d_2^2} =\frac{\left(r a_{2,2}-a_{1,2}\right)^2}{\left(r a_{2,1}-a_{1,1}\right)^2+\left(r a_{2,2}-a_{1,2}\right)^2},$$ where $r:=b_1/b_2$ and $b=[b_1,b_2]^T$, so that $r$ has the standard Cauchy distribution. So, the conditional expectation $$f(A):=E(R|A)$$ is the integral of a rational expression, so that $f$ is a certain elementary function, which is continuous where it is defined, and it is defined on an open set of full measure wrt the distribution of the random matrix $A$. Moreover, $f(A)=1/3$ if $A=\begin{bmatrix}2&1\\ 1&1\end{bmatrix}$ and $f(A)=1/4$ if $A=\begin{bmatrix}3&1\\ 1&1\end{bmatrix}$. Details of the calculation of $f(A)$ and its two particular values $1/3$ and $1/4$ just mentioned are given in this Mathematica notebook.
So, the distribution of the random variable $f(A)$ is nondegenerate. So, $$Ef(A)\frac{d_1^2}{d_1^2+d_2^2}=Ef(A)R=Ef(A)E(R|A)=Ef(A)^2 \\ > (Ef(A))^2 =Ef(A)\,ER=Ef(A)\,E\frac{d_1^2}{d_1^2+d_2^2},$$ which proves that $A$ and $d$ are not independent. $\quad\Box$