Let \(M = \left\{ A = \begin{pmatrix} a & b \\
c & d \end{pmatrix} : a, b, c, d \in \{ \pm 3, \pm 2, \pm 1, 0 \} \right\}\). Define \(f: M \to \mathbb{Z}\) as \(f(A) = \det(A)\), for all \(A \in M\), where \(\mathbb{Z}\) is the set of all integers. Then the number of \(A \in M\) such that \(f(A) = 15\) is equal to __________.
Show Hint
First list all possible products of the given set. Then find pairs whose difference is the target value. This systematic approach ensures no combination is missed.
Step 1: Understanding the Concept:
The determinant of a \(2 \times 2\) matrix is given by \(ad - bc\).
We need to find the number of integer quadruplets \((a, b, c, d)\) from the given set such that their combination results in 15. Step 3: Detailed Explanation:
The allowed set of values for \(a, b, c, d\) is \(X = \{0, \pm 1, \pm 2, \pm 3\}\).
The possible products of two elements from this set are:
\(P = \{ 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9 \}\).
We need \(ad - bc = 15\).
Let \(x = ad\) and \(y = bc\). We need \(x - y = 15\) where \(x, y \in P\).
Looking at the differences between elements of \(P\):
Case 1: \(x = 9\) and \(y = -6\).
Case 2: \(x = 6\) and \(y = -9\).
No other combination works (e.g., \(9 - (-9) = 18\); \(4 - (-9) = 13\)).
Count for Case 1 (\(ad=9, bc=-6\)):
- Products for \(ad=9\): \((3,3)\) and \((-3,-3)\). (2 ways)
- Products for \(bc=-6\): \((2,-3), (-3,2), (-2,3), (3,-2)\). (4 ways)
Total Case 1 \(= 2 \times 4 = 8\).
Count for Case 2 (\(ad=6, bc=-9\)):
- Products for \(ad=6\): \((2,3), (3,2), (-2,-3), (-3,-2)\). (4 ways)
- Products for \(bc=-9\): \((3,-3), (-3,3)\). (2 ways)
Total Case 2 \(= 4 \times 2 = 8\).
Total number of matrices \(= 8 + 8 = 16\). Step 4: Final Answer:
The total number of matrices is 16.
