Let $R=\left[\begin{array}{ccc}a& 3& b\\ c& 2& d\\ 0& 5& 0\end{array}\right]$, where $a,b,c,d\in \{0,3,5,7,11,13,17,19\}$. Then the number of invertible matrices in R is:

The correct answer is 3780.

Step 1: Determine the condition for matrix invertibility
A matrix is invertible if and only if its determinant is non-zero.
Given the matrix:

$R=\left[\begin{array}{ccc}a& 3& b\\ c& 2& d\\ 0& 5& 0\end{array}\right]$

Expanding the determinant along the third row (which contains two zeros):

$det\left(R\right)=0-5(ad-bc)+0=-5(ad-bc)$

For $R$ to be an invertible matrix, we require:

$det\left(R\right)\ne 0\Rightarrow -5(ad-bc)\ne 0\Rightarrow ad\ne bc$

Step 2: Total number of possible matrices
The entries $a,b,c,d$ belong to the set:

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

Since $S$ has 8 elements, each of the 4 variables $a,b,c,d$ can be chosen in 8 ways.
Total number of possible matrices $R$ is:

$\text{Total matrices}=8^4=4096$

Step 3: Count the number of non-invertible matrices ($ad=bc$)
We divide the equation $ad=bc$ into two exhaustive cases:

Case 1: $ad=b\mathrm{c<mo>=</mo><mn>0</mn>}$
- For $ad=0$, at least one of $a$ or $d$ must be 0.
- If $a=0$, $d$ can be any of the 8 values (8 choices).
- If $a\ne 0$ (7 choices), then $d=0$ (1 choice).
So, there are $8+7=15$ valid pairs for $(a,d)$.
- Similarly, for $bc=0$, there are 15 valid pairs for $(b,c)$.
Thus, total 4-tuples where $ad=bc=0$ is:

$15\times 15=225$

Case 2: $ad=bc\ne 0$
Here, $a,b,c,d\in \{3,5,7,11,13,17,19\}$ (7 prime numbers).
By the unique prime factorization property of products of prime numbers, $ad=bc$ can occur in two disjoint subcases:

- Subcase 2a: $a=b$ and $d=c$
- $a$ can be any of the 7 prime numbers.
- $d$ can independently be any of the 7 prime numbers.
Number of choices = $7\times 7=49$.

- Subcase 2b: $a=c$ and $d=b$ with $a\ne b$
- $a$ can be any of the 7 prime numbers.
- $b$ can be any of the remaining 6 prime numbers (since $b\ne a$).
Number of choices = $7\times 6=42$.

Total choices for non-zero products = $49+42=91$.

Step 4: Compute total non-invertible and invertible matrices
Total non-invertible matrices:

$\text{Non-invertible matrices}=225+91=316$

Finally, subtracting the non-invertible matrices from the total possible matrices gives the number of invertible matrices:

$\text{Invertible matrices}=4096-316=3780$