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

The given matrices are of the form
$$A=\begin{pmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{pmatrix},$$
where $$a,b,c,d$$ come from the set $$S=\{0,3,5,7,11,13,17,19\}$$ (8 elements).

First find the determinant of $$A$$. Expanding along the third row,

$$\det(A)=0\;C_{31}+5\,C_{32}+0\,C_{33}=5\,(-1)^{3+2}\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$

$$\Rightarrow\quad \det(A)=-5\,(ad-bc).$$

The matrix is invertible $$\iff$$ $$\det(A)\neq0 \iff ad-bc\neq0.$$
Hence we must count the quadruples $$(a,b,c,d)\in S^4$$ for which $$ad\neq bc$$ and subtract this from the total number of quadruples.

Total quadruples: $$|S|^4=8^4=4096.$$

Let $$N$$ be the number of “bad’’ quadruples with $$ad=bc$$. We will find $$N$$ and then compute $$4096-N$$.

For $$ad$$ to be zero, at least one of $$a,d$$ is $$0$$. For $$bc$$ to be zero as well, at least one of $$b,c$$ is $$0$$.

• $$a=0$$ (1 choice) and $$d$$ arbitrary (8 choices):
  Pairs $$(b,c)$$ satisfying $$bc=0$$ are obtained by “at least one zero’’:
  $$(b=0, c\in S)$$ (8) and $$(b\in S, c=0)$$ (8) minus the double-count $$(0,0).$$
  Thus $$15$$ such pairs. So the count here is $$1\times8\times15=120.$$

• $$a\neq0$$ (7 choices) and $$d=0$$ (1 choice): the same 15 pairs for $$(b,c)$$.
  Count $$7\times1\times15=105.$$

Total for Case I: $$120+105=225.$$

Restrict to $$T=S\setminus\{0\}=\{3,5,7,11,13,17,19\}$$ (7 elements).
Put $$p=ad=bc\neq0$$. For every fixed product value $$p$$, let $$n_p$$ be the number of ordered pairs $$(x,y)\in T^2$$ with $$xy=p$$. Then

number of ordered quadruples with product $$p$$ is $$n_p^2$$ (choose $$(a,d)$$ and $$(b,c)$$ independently).

Because all numbers in $$T$$ are distinct primes, the only way two products coincide is by swapping the order of a pair or taking a square:

• Squares: $$(3,3),(5,5),\dots,(19,19)$$ - 7 products, each with $$n_p=1.$$

• Products of two distinct primes: there are $${7\choose2}=21$$ such products; for each, the ordered pairs $$(x,y),(y,x)$$ give $$n_p=2.$$

Hence

$$\sum n_p^2 \;=\;7\cdot1^2 + 21\cdot2^2 = 7 + 84 = 91.$$

Total “bad’’ quadruples: $$N = 225 + 91 = 316.$$

Therefore the number of invertible matrices is

$$4096 - 316 = 3780.$$

Answer: 3780