Let R = left{ begin{bmatrix} a & 3 & b c & 2 | vedclass

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(B) matrix $M \in R$ is invertible if and only if its determinant $|M| 
eq 0$. The determinant of the matrix is given by: $|M| = a(2 	imes 0 - 5 	i

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$3780$

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(B) matrix $M \in R$ is invertible if and only if its determinant $|M| 
eq 0$. The determinant of the matrix is given by: $|M| = a(2 	imes 0 - 5 	imes d) - 3(c 	imes 0 - 0 	imes d) + b(c 	imes 5 - 2 	imes 0)$ $|M| = a(-5d) - 3(0) + b(5c) = 5bc - 5ad = 5(bc - ad)$. For the matrix to be invertible,$|M| 
eq 0$,which implies $bc - ad 
eq 0$,or $bc 
eq ad$. The set of values for $a, b, c, d$ is $S = \{0, 3, 5, 7, 11, 13, 17, 19\}$,which contains $8$ elements. Total number of matrices in $R$ is $8^4 = 4096$. We need to find the number of cases where $bc = ad$. Let $X = bc$ and $Y = ad$. If $0 
otin S$,there are $7^4$ ways. Since $0 \in S$,we consider: $1$. If $ad = 0$,then either $a=0$ or $d=0$. Number of ways = $8^2 - 7^2 = 15 - 0 = 15$. $2$. If $bc = 0$,then either $b=0$ or $c=0$. Number of ways = $8^2 - 7^2 = 15$. Total ways where $bc = ad = 0$ is $15 	imes 15 = 225$. Now consider $bc = ad = k$,where $k 
eq 0$. The possible values for $k$ are products of two non-zero elements from $S$. For each $k$,let $N(k)$ be the number of pairs $(x, y)$ such that $xy = k$. Since all non-zero elements are prime,$bc = ad$ implies ${b, c} = {a, d}$ or specific combinations. Total cases where $bc = ad 
eq 0$ is $91$. Thus,$|M| = 0$ in $225 + 91 = 316$ cases. Number of invertible matrices = $4096 - 316 = 3780$.

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

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