When does the following type of matrix have full row and column rank?
For each entry, we take the value $v$=row*column. Then for the $b$th bit of $v$, $v_b$ we take the $b$th prime to the $v_b$ power.
EXAMPLES
Say row = 5 and column = 7. This is 5*7=35, which, in binary, is $100011_2$ So we take $$\cdot 13^1 \cdot 11^0 \cdot 7^0 \cdot 5^0 \cdot 3^1 \cdot 2^1$$ $$=2 \cdot 3 \cdot 13$$ $$=78$$ ...So the entry for the 5th row and 7th column would be 78.
Say row = 4 and column = 2. This is 4*2=8, which, in binary, is $001000_2$ So we take $$\cdot 13^0 \cdot 11^0 \cdot 7^1 \cdot 5^0 \cdot 3^0 \cdot 2^0$$ $$=7$$ ...So the entry for the 4th row and 2nd column would be 7.
THE GENERAL MATRIX
$$ \begin{matrix} 2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & 2^0 3^0 5^0 & \dots\\ 2^0 3^0 5^0 & 2^1 3^0 5^0 & 2^0 3^1 5^0 & 2^1 3^1 5^0 & 2^0 3^0 5^1 & \dots\\ 2^0 3^0 5^0 & 2^0 3^1 5^0 & 2^0 3^0 5^1 & 2^0 3^1 5^1 & 2^0 3^0 5^0 7^1 & \dots\\ 2^0 3^0 5^0 & 2^1 3^1 5^0 & 2^1 3^1 5^0 & 2^1 3^0 5^0 7^1 & 2^0 3^0 5^1 7^1 & \dots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\ \\ \end{matrix} $$
We can assume that the matrix is an $N \times N$ matrix.
SOME OBSERVATIONS
Just a general observation: The only counterexample for $N \times N$ matrices with $N$ ranging from 2 to 20 is $N=3$. All other matrices have full rank.
Easy counterexample, take $N = 3$ and you will see that it is not full ranked:
$\left( \begin{array}{ccc} 2^03^05^0=1 & 2^03^05^0=1 & 2^03^05^0=1 \\ 2^03^05^0=1 & 2^13^05^0=2 & 2^03^15^0=3 \\ 2^03^05^0=1 & 2^03^15^0=3 & 2^03^05^1=5 \end{array} \right)$, this matrix is not full rank, so your question's answer would be no, it is not always full rank.