We need to count how many matrices in

$$R=\left\{\begin{pmatrix}a&3&b\\ c&2&d\\ 0&5&0\end{pmatrix}: a,b,c,d\in S\right\},$$

where

$$S=\{0,3,5,7,11,13,17,19\},$$

are invertible.

There are $8$ choices each for $a,b,c,d$, so total matrices are

$$8^4=4096.$$

We only need to count those with nonzero determinant.

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1. Compute the determinant

Let

$$A=\begin{pmatrix}a&3&b\\ c&2&d\\ 0&5&0\end{pmatrix}.$$

Expand along the third row:

$$\det(A)=0\cdot C_{31}+5\cdot C_{32}+0\cdot C_{33}.$$

Now

$$C_{32}=(-1)^{3+2}\begin{vmatrix}a&b\\ c&d\end{vmatrix} =-\,(ad-bc).$$

Hence

$$\det(A)=5\big(- (ad-bc)\big)=5(bc-ad).$$

So the matrix is invertible iff

$$bc-ad\ne 0$$

that is,

$$ad\ne bc.$$

Thus we must count quadruples $(a,b,c,d)\in S^4$ such that

$$ad=bc.$$

Then subtract from $4096$.

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2. Count singular matrices: solve $ad=bc$

We count ordered quadruples $(a,b,c,d)\in S^4$ satisfying

$$ad=bc.$$

Let us split into cases.

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Case 1: At least one side is forced to be zero

The only zero in $S$ is $0$.

If $ad=bc=0$, then equality holds. We count all quadruples with both products zero.

• $ad=0$ means $a=0$ or $d=0$.
• $bc=0$ means $b=0$ or $c=0$.

Number of ordered pairs $(a,d)$ with product $0$:

$$8^2-7^2=64-49=15.$$

Similarly, number of ordered pairs $(b,c)$ with product $0$ is also $15$.

Therefore number of quadruples with

$$ad=0,\quad bc=0$$

is

$$15\times 15=225.$$

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Case 2: All products are nonzero

Now $a,b,c,d\ne 0$, so all variables lie in

$$T=\{3,5,7,11,13,17,19\}.$$

We need to count solutions of

$$ad=bc$$

with $a,b,c,d\in T$.

Since all numbers in $T$ are distinct primes, the product of two elements determines the unordered pair of factors uniquely.

So for

$$ad=bc,$$

the multiset $\{a,d\}$ must equal the multiset $\{b,c\}$.

Let us count ordered quadruples.

Subcase 2a: $a=d$

Then product $ad=a^2$. For $bc=a^2$, since elements are prime, we must have

$$b=c=a.$$

So this gives one quadruple for each choice of $a\in T$. Hence number is

$$7.$$

Subcase 2b: $a\ne d$

Choose ordered pair $(a,d)$ from $T$ with distinct entries:

$$7\times 6=42.$$

For fixed distinct $a,d$, the equality $bc=ad$ implies $(b,c)$ must be either

$$(b,c)=(a,d) \quad \text{or} \quad (d,a).$$

So there are $2$ choices.

Thus count is

$$42\times 2=84.$$

Therefore total nonzero solutions are

$$7+84=91.$$

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3. Total singular matrices

So total number of singular matrices is

$$225+91=316.$$

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4. Total invertible matrices

Therefore,

$$4096-316=3780.$$

So the number of invertible matrices is

$$\boxed{3780}.$$

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5. Compare with stored answer

Stored correct answer: $3780$.

Our derived answer matches it.