Let $$M = A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in \{\pm 3, \pm 2, \pm 1, 0\}$$. Define $$f: M \to Z$$, as $$f(A) = \det A$$, for all $$A \in M$$ where $$Z$$ is set of all integers. Then the number of $$A \in M$$ such that $$f(A) = 15$$ is equal to ___.

We have a general matrix from the set

$$A=\begin{pmatrix}a & b\\c & d\end{pmatrix},\qquad a,b,c,d\in\{-3,-2,-1,0,1,2,3\}.$$

For any such matrix the determinant is defined by the well-known formula

$$\det A = ad-bc.$$

In this question we must count how many ordered quadruples $$(a,b,c,d)$$ satisfy the condition

$$ad-bc = 15.$$

First concentrate on the term $$ad$$. With the given entries the largest possible product is $$3\cdot 3 = 9$$ and the smallest is $$(-3)\cdot 3=-9$$. Hence

$$-9\le ad\le 9.$$

Rewrite the determinant condition as

$$bc = ad-15.$$ So the required product $$bc$$ is $$bc = t,\qquad\text{where }t = ad-15.$$

Because $$ad\in[-9,9]$$, the possible values of $$t$$ lie between

$$-9-15=-24\quad\text{and}\quad 9-15=-6,$$ that is, $$-24\le t\le -6.$$

But $$bc$$ itself is the product of two elements from the same set $$\{-3,-2,-1,0,1,2,3\}$$, and such a product can only take values in the interval $$[-9,9]$$. Therefore the overlap of the two ranges is

$$-9\le t\le -6.$$

Thus $$t$$ can be $$-9,-8,-7,-6$$. We next decide which of these actually occur.

Because $$t=ad-15$$, we solve for $$ad$$

$$ad = t+15.$$ Substituting each possible $$t$$ gives $$$ \begin{aligned} t=-9 &\;\Longrightarrow\; ad = 6,\\ t=-8 &\;\Longrightarrow\; ad = 7,\\ t=-7 &\;\Longrightarrow\; ad = 8,\\ t=-6 &\;\Longrightarrow\; ad = 9. \end{aligned} $$$

Remember that $$ad$$ itself must be attainable with $$a,d\in\{-3,-2,-1,0,1,2,3\}$$. Let us list all achievable products and their corresponding ordered pairs:

$$\bullet\;ad=6$$: possible through $$2\cdot3$$, $$3\cdot2$$, $$(-2)\cdot(-3)$$, $$(-3)\cdot(-2)$$. That gives the four ordered pairs $$(a,d)=(2,3),\;(3,2),\;(-2,-3),\;(-3,-2).$$

$$\bullet\;ad=7$$: impossible because neither $$7$$ nor $$-7$$ can be written as a product of two numbers from the allowed set.

$$\bullet\;ad=8$$: impossible for the same reason; $$8$$ has factors $$2$$ and $$4$$ or $$-2$$ and $$-4$$, but $$\pm4\notin\{-3,-2,-1,0,1,2,3\}$$.

$$\bullet\;ad=9$$: possible through $$3\cdot3$$ and $$(-3)\cdot(-3)$$, giving the two ordered pairs $$(a,d)=(3,3),\;(-3,-3).$$

Hence only $$ad=6$$ or $$ad=9$$ are feasible. Their corresponding required values of $$bc$$ are

$$$ \begin{aligned} ad=6 &\;\Longrightarrow\; bc = 6-15 = -9,\\ ad=9 &\;\Longrightarrow\; bc = 9-15 = -6. \end{aligned} $$$

Now we count the ordered pairs $$(b,c)$$ that give each product.

$$\bullet\;bc=-9$$: because $$3\cdot3=9$$ we need opposite signs. The pairs are $$(b,c)=(3,-3),\;(-3,3).$$ So there are $$2$$ possibilities.

$$\bullet\;bc=-6$$: here $$2\cdot3=6$$. Again the signs must be opposite, yielding $$(b,c)=(2,-3),\;(-2,3),\;(3,-2),\;(-3,2).$$ This gives $$4$$ possibilities.

Finally form the matrices by combining the independent choices for $$(a,d)$$ and $$(b,c)$$.

$$$ \begin{aligned} ad=6 &\text{ (4 choices)} \quad\&\quad bc=-9 &\text{ (2 choices)} &\;\Longrightarrow\; 4\times2 = 8\text{ matrices},\\ ad=9 &\text{ (2 choices)} \quad\&\quad bc=-6 &\text{ (4 choices)} &\;\Longrightarrow\; 2\times4 = 8\text{ matrices}. \end{aligned} $$$

Adding the two disjoint cases we obtain the total

$$8+8 = 16.$$

So, the answer is $$16$$.