Number of Invertible Three by Three Matrices with Given Constraints

To find the number of invertible matrices in $R$, we determine the determinant of a general matrix $M \in R$: $\det(M) = \begin{vmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{vmatrix} = -5(ad - bc)$

A matrix $M$ is invertible if and only if $\det(M) \neq 0$, which is equivalent to $ad \neq bc$. The set of available values $S = \{0, 3, 5, 7, 11, 13, 17, 19\}$ contains $8$ elements, where $0$ is zero and the remaining $7$ elements are distinct prime numbers.

The total number of possible matrices is $8^4 = 4096$. We count the number of singular matrices by finding the number of quadruples $(a,b,c,d)$ such that $ad = bc$:

1. If $ad = bc = 0$, there are $15$ choices for $(a,d)$ where at least one variable is $0$, and $15$ choices for $(b,c)$, yielding $15 \times 15 = 225$ quadruples.
2. If $ad = bc \neq 0$, all elements must be chosen from the $7$ distinct primes. This occurs when $(b,c) = (a,d)$ or $(b,c) = (d,a)$, giving $7 + (7 \times 6 \times 2) = 91$ quadruples.

Subtracting the $316$ singular cases from the total $4096$ choices gives: $\text{Number of invertible matrices} = 4096 - (225 + 91) = 3780$