Question:

**Step 1: Evaluate the determinant** Let $A = \begin{pmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{pmatrix}$. We expand the determinant along the third row: \n$|A| = 0 - 5(ad - bc) + 0 = -5(ad - bc)$. **Step 2: Condition for invertibility** The matrix $A$ is invertible if and only if $|A| \neq 0$, which means $ad - bc \neq 0$, or $ad \neq bc$. **Step 3: Count the total number of matrices and non-invertible matrices** The variables $a, b, c, d$ are chosen from the set $S = \{0, 3, 5, 7, 11, 13, 17, 19\}$. The size of $S$ is 8. The total number of matrices in $R$ is $8^4 = 4096$.\nWe need to find the number of ways to choose $a, b, c, d$ such that $ad = bc$. Let this number be $N$. Then the number of invertible matrices is $4096 - N$. **Step 4: Analyze the equation ad = bc** Notice that the non-zero elements of $S$ are all distinct primes. So $S \setminus \{0\}$ consists of 7 prime numbers.\nCase 1: Both products are 0. $ad = 0$ if at least one of $a$ or $d$ is 0. Number of ways = $8^2 - 7^2 = 15$. Similarly, $bc = 0$ in 15 ways. So there are $15 \times 15 = 225$ ways for $ad = bc = 0$.\nCase 2: Both products are equal and non-zero. The prime factorization of $ad$ must equal that of $bc$. \nSince $a, b, c, d$ are primes, this implies $\{a, d\} = \{b, c\}$.\nSubcase 2a: $a = d$. Then $a^2 = bc \implies b = c = a$. There are 7 choices for this prime, so 7 ways.\nSubcase 2b: $a \neq d$. Then either $(b=a, c=d)$ or $(b=d, c=a)$. Number of ways to choose distinct $a, d$ is $7 \times 6 = 42$. For each pair, there are 2 valid $(b, c)$ pairs. So $42 \times 2 = 84$ ways.\nTotal for Case 2 = $7 + 84 = 91$ ways.\nTotal ways where $ad = bc$ is $N = 225 + 91 = 316$. **Step 5: Calculate the final answer** The number of invertible matrices is $4096 - 316 = 3780$.