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Question:

Let d ∈ R, and A = ⎡⎢⎣𕒶4+d(sinθ)𕒶1(sinθ)+2d5(2sinθ)−d(−sinθ)+2+2d⎤⎥⎦, θ ∈ [0,2π]. If the minimum value of det(A) is 8, then a value of d is?

𕒻

𕒹

2(√2+1)

2(√2+2)

Solution:

detA = ∣∣∣∣𕒶4+dsinθ𕒶1sinθ+2d52sinθ−d−sinθ+2+2d∣∣∣∣ (R1 → R1 + R3 𕒶 R2) = ∣∣∣∣1001sinθ+2d52sinθ−d2+2d−sinθ∣∣∣∣ = (2+sinθ)(2+2d−sinθ)−d(2sinθ−d) = 4+4d−sin²θ+2sinθ+2dsinθ−2dsinθ+d² = d²+4d+4−sin²θ = (d+2)²−sin²θ
For a given d, minimum value of det(A) = (d+2)² − 1 = 8
→ (d+2)² = 9
d = 3 − 2 = 1 or d = −3 − 2 = −5
→ d = 1 or −5