Design a positive integral matrix that isn't full rank

Question

Is it possible to design a square $d \times d$ matrix with

1. value $2^n$ on the main diagonal ($n \geq 2$ not given)

2. all off-diagonal elements chosen from the set $\{ 1, 2, 2^2, \dots, 2^{n-1} \}$

3. not have full rank for any choice of $n$ or $d$

I think this is impossible (based on trial and error). But would appreciate a proof or a counterexample.

Answer 1

Some (hilariously inefficient) Matlab code to test this conjecture:

n = 3;
d = 4;
N = n^(d^2 - d);

for i = 0:N-1
    ch = dec2base(i,n,(d^2 - d));
    m = zeros(d^2 - d,1);
    for j = 1:(d^2 - d)
        m(j) = str2num(ch(j));
    end
    P = 2.^reshape(m,[d,d-1]);
    M = 2^n*eye(d);
    for j = 1:d
        col = 1:d;
        col(j) = [];
        M(j,col) = P(j,:);
    end
    if det(M) == 0
        disp('counterexample:')
        disp(M)
    end
end

There is indeed no counterexample for $n = 3,d=4$.