$3\times3$ matrix of distinct positive primes with determinant $0$ Is there a $3\times3$ matrix of distinct positive primes whose determinant is $0$?
I came across this while attempting a partial answer to a question here (which I forgot to favorite). However, I've made no headway on it.
 A: Note that
$$2\cdot3+5=11$$
$$2\cdot7+17=31$$
$$2\cdot19+29=67$$
Therefore a possible matrix that satisfies your conditions is
$$\begin{bmatrix}3&7&19\\5&17&29\\11&31&67\end{bmatrix}$$
The rows are linearly dependent, twice the first row plus the second equalling the third. The matrix thus has determinant 0.

As mentioned by Mark Dickinson on the other answer, the Green–Tao theorem proves infinitely many matrices of distinct primes with determinant 0 at any size, the smallest such 3×3 example being
$$\begin{bmatrix}3&5&7\\13&17&19\\23&29&31\end{bmatrix}$$
with twice the second row being the sum of the other two rows.
A: The extended Goldbach's conjecture implies that there are infinitely many such matrices. This conjecture implies that not only is every even number greater than 2 the sum of two primes but that large even numbers can all be written as the sum of two primes in many different ways.
Take any 3 distinct sufficiently large primes $p_1, p_2, p_3$. Then by Goldbach's conjecture we can find primes $q_i,r_i$ with $2*p_i = q_i + r_i$ for $i = 1,2,3$. By the extended Goldbach's conjecture, we can choose these primes so that all of the primes are distinct. Clearly the determinant of
$$\begin{bmatrix}p_1&p_2&p_3\\q_1&q_2&q_3\\r_1&r_2&r_3\end{bmatrix}$$
is zero since twice the first row is the sum of the other two.
I don't know if you can prove the existence of infinitely many examples without using any unproven assumption.
On Edit  In the comments @MarkDickinson gave a nice argument which shows that the Green-Tao theorem implies that there are infinitely many solutions, hence no unproven assumptions are needed. Furthermore, Mark's argument implies that there are infinitely many $n \times n$ solutions for any $n \geq 3$ (it is easy to see that no $2 \times 2$ solutions are possible). The Green-Tao theorem states that there are arbitrarily long arithmetic progressions of primes, which immediately implies that for any fixed $k$ there are infinitely many arithmetic progressions of length $k$. 
If $n \geq 3$, pick $n^2$ primes in arithmetic progression. Arrange these column by column into an $n \times n$ matrix $A$. Then the rows are in arithmetic progression. If $n = 3$ the rows satisfy the equation $R_1 - 2R_2 + R_3 = 0$. If $n \geq 4$, the rows satisfy $R_1 - R_2 - R_3 + R_4 = 0$. Thus in all cases the rows are linearly dependent hence the determinant is 0.
This argument can be used to explicitly construct examples for $n \leq 5$, but no explicit arithmetic progression of 36 primes is currently known (according to Wikipedia, the current record is 26). An interesting programming problem would be to find large singular matrices of distinct primes. Note that taking $n^2$ primes in arithemtic progression is overkill. It would suffice to find $n$ disjoint arithmetic progressions of length $n$.
On further edit The full power of Green-Tao is not needed. You get infinitely many such matrices of size $3 \times 3$ or larger as soon as you have 
infinitely many disjoint arithmetic sequences of length 3. Such things can be used to induce a linear dependence among the first 3 rows, after which the remaining rows can be filled in arbitrarily (subject to the constraint that the primes be distinct). Here is a 5x5 example: 
$$ \begin{bmatrix}3&11&13&19&29\\5&17&37&31&41\\7&23&61&43&53\\2&47&59&67&71\\73&79&83&89&97 \end{bmatrix} $$
Note that this $5 \times 5$ matrix consists of the first 25 prime numbers. You can swap out 2 for 101 if you don't like even primes. With very little evidence I conjecture that given any $n \geq 4$, you can arrange the first $n^2$ prime numbers into a singular $n \times n$ matrix. I have verified that it works out to at least $n = 100$.
