Let M = {A = [a,b;c,d] : a,b,c,d ∈ {±3, ±2, ±1, 0}}. Define f: M → Z, as f(A) = det(A), for all A ∈ M, where Z is set of all integers. Then the number of A ∈ M such that f(A) = 15 is equal to………
Let M = {A = [a,b;c,d] : a,b,c,d ∈ {±3, ±2, ±1, 0}}. Define f: M → Z, as f(A) = det(A), for all A ∈ M, where Z is set of all integers. Then the number of A ∈ M such that f(A) = 15 is equal to………
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1 Answer
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M = {A = [a b; c d] | a, b, c, d ∈ {±3, ±2, ±1,0}
f (A) = det (A) = ad-bc=15
Case 1: ad=9 (2 ways: 3×3, -3×-3), bc=-6 (4 ways: 3×-2, -3×2, 2×-3, -2×3). Total = 2×4=8.
Case 2: ad=6 (4 ways), bc=-9 (2 ways). Total = 4×2=8.
Total no. of possible such cases = 8+8=16.
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