Key Concept: Determinant of a $2 \times 2$ Matrix For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its determinant, denoted as $\det(A)$ or $|A|$, is calculated using the formula: $\det(A) = ad - bc$ This formula represents the difference between the product of the main diagonal elements ($a$ and $d$) and the product of the anti-diagonal elements ($b$ and $c$).

Understanding the Problem We are given a set $M$ of $2 \times 2$ matrices where each entry $a, b, c, d$ must belong to the specific finite set $S = \{ \pm 3, \pm 2, \pm 1, 0 \}$. We need to find the number of matrices $A \in M$ such that $f(A) = \det(A) = 15$. This means we are looking for the total count of distinct ordered quadruplets $(a, b, c, d)$ such that $a, b, c, d \in S$ and their determinant $ad - bc = 15$.

Step 1: Determine the Possible Values for Products $ad$ and $bc$

The elements $a, b, c, d$ are chosen from the set $S = \{-3, -2, -1, 0, 1, 2, 3\}$. Let $X = ad$ and $Y = bc$. Our goal is to find the number of ways to choose $a,b,c,d$ such that $X - Y = 15$.

First, let's find all possible product values for any two elements from $S$:

• If either element is $0$, the product is $0$.
• If both elements are non-zero, the possible products are:
  • $1 \times 1 = 1$, $1 \times (-1) = -1$
  • $1 \times 2 = 2$, $1 \times (-2) = -2$
  • $1 \times 3 = 3$, $1 \times (-3) = -3$
  • $2 \times 2 = 4$, $2 \times (-2) = -4$
  • $2 \times 3 = 6$, $2 \times (-3) = -6$
  • $3 \times 3 = 9$, $3 \times (-3) = -9$ (Products of two negative numbers yield positive results, e.g., $(-1) \times (-2) = 2$, which are already covered by positive products.) So, the set of all possible product values for $ad$ or $bc$ is $P = \{0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9\}$.

Important Constraint Check: Consider the equation $ad - bc = 15$:

• If $ad = 0$, then $0 - bc = 15 \implies bc = -15$. However, $-15$ is not in the set of possible product values $P$ (the minimum product is $-9$). Therefore, $ad$ cannot be $0$. This implies that $a \ne 0$ and $d \ne 0$.
• If $bc = 0$, then $ad - 0 = 15 \implies ad = 15$. However, $15$ is not in $P$ (the maximum product is $9$). Therefore, $bc$ cannot be $0$. This implies that $b \ne 0$ and $c \ne 0$.

This critical observation simplifies our problem: all four entries $a, b, c, d$ must be non-zero. They must be chosen from the set $S' = \{\pm 1, \pm 2, \pm 3\}$.

Step 2: Enumerate Ways to Form Products from $S'$ Now, let's count the number of distinct ordered pairs $(x,y)$ from $S'$ such that their product $xy$ equals a specific value. This is crucial for correctly counting the matrices.

• $xy = 1$: $(1,1), (-1,-1)$ (2 ways)
• $xy = -1$: $(1,-1), (-1,1)$ (2 ways)
• $xy = 2$: $(1,2), (2,1), (-1,-2), (-2,-1)$ (4 ways)
• $xy = -2$: $(1,-2), (-2,1), (-1,2), (2,-1)$ (4 ways)
• $xy = 3$: $(1,3), (3,1), (-1,-3), (-3,-1)$ (4 ways)
• $xy = -3$: $(1,-3), (-3,1), (-1,3), (3,-1)$ (4 ways)
• $xy = 4$: $(2,2), (-2,-2)$ (2 ways)
• $xy = -4$: $(2,-2), (-2,2)$ (2 ways)
• $xy = 6$: $(2,3), (3,2), (-2,-3), (-3,-2)$ (4 ways)
• $xy = -6$: $(2,-3), (-3,2), (-2,3), (3,-2)$ (4 ways)
• $xy = 9$: $(3,3), (-3,-3)$ (2 ways)
• $xy = -9$: $(3,-3), (-3,3)$ (2 ways)

Step 3: Find Pairs $(ad, bc)$ such that $ad - bc = 15$ We need to find pairs of product values $(X, Y)$ from $P \setminus \{0\}$ (as $0$ is already excluded) such that $X - Y = 15$. The maximum possible value for $X=ad$ is $3 \times 3 = 9$. The minimum possible value for $Y=bc$ is $3 \times (-3) = -9$. Therefore, the maximum possible value for $X-Y$ is $9 - (-9) = 18$. The minimum possible value for $X-Y$ is $-9 - 9 = -18$. Since $15$ is within this range $[-18, 18]$, solutions are possible.

Let's systematically check values for $X=ad$, starting from the maximum possible product:

• Case 1: $X = ad = 9$ If $ad=9$, then $9 - Y = 15 \implies Y = -6$. Both $X=9$ and $Y=-6$ are valid product values in $P$.

  • From Step 2, the number of ways to obtain $ad=9$ is 2.
  • From Step 2, the number of ways to obtain $bc=-6$ is 4. Since the choices for $(a,d)$ and $(b,c)$ are independent, the number of matrices for this case is the product of the ways: $2 \times 4 = 8$.

• Case 2: $X = ad = 6$ If $ad=6$, then $6 - Y = 15 \implies Y = -9$. Both $X=6$ and $Y=-9$ are valid product values in $P$.

  • From Step 2, the number of ways to obtain $ad=6$ is 4.
  • From Step 2, the number of ways to obtain $bc=-9$ is 2. The number of matrices for this case is $4 \times 2 = 8$.

• Case 3: $X = ad = 4$ If $ad=4$, then $4 - Y = 15 \implies Y = -11$. However, $-11$ is not a possible product value in $P$. Thus, there are no matrices for this case.

• Cases with $X \le 3$ (e.g., $ad=3, 2, 1$) If $ad=3$, then $3 - Y = 15 \implies Y = -12$. Not in $P$. If $ad=2$, then $2 - Y = 15 \implies Y = -13$. Not in $P$. If $ad=1$, then $1 - Y = 15 \implies Y = -14$. Not in $P$. Any smaller positive value for $ad$ will also result in $bc$ being an impossible product.

• Cases with $X < 0$ (e.g., $ad=-1, -2, \dots, -9$) If $ad$ is negative, $bc$ would need to be an even smaller negative number (e.g., if $ad=-1, bc=-16$; if $ad=-9, bc=-24$). None of these very negative values for $bc$ are possible products in $P$. Thus, there are no matrices for these cases.

Step 4: Total Number of Matrices The total number of matrices $A \in M$ such that $f(A) = 15$ is the sum of the counts from all valid cases. Total matrices = (Matrices from Case 1) + (Matrices from Case 2) Total matrices = $8 + 8 = 16$.

Tips and Key Takeaways

• Systematic Decomposition: Break down complex counting problems into simpler parts. First, identify the range of possible values for intermediate products ($ad, bc$). Second, find pairs of these intermediate products that satisfy the main condition ($ad-bc=15$). Third, count the ways to form each product independently.
• Ordered Pairs Matter: Remember that for matrix elements, $(a,d)$ is distinct from $(d,a)$ unless $a=d$. Similarly, $(b,c)$ is distinct from $(c,b)$. Our counting in Step 2 accounts for these ordered pairs.
• Implicit Constraints: Always check for implicit constraints. Here, the condition $ad-bc=15$ implicitly meant that none of $a,b,c,d$ could be zero, which significantly narrowed down the choices.
• Double-Check Range: Ensure that the target value (15 in this case) falls within the possible range of the expression ($ad-bc$) to avoid unnecessary calculations.

The final answer is $\boxed{\text{16}}$.