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Matrices & Determinants: Square Matrices Order Adj Adj Equal

JEE Maths question with a full step-by-step solution.

Question
If AA and BB are square matrices of order 3 such that det⁡(A)=−2\det(A) = -2 and det⁡(B)=1\det(B) = 1, then det⁡(A−1 adj(B−1) adj(2A−1))\det\big(A^{-1}\,\text{adj}(B^{-1})\,\text{adj}(2A^{-1})\big) is equal to:
A88
B−8-8correct
C11
D−1-1
Solution
Step 1: Apply the product rule and det⁡(adj M)=(det⁡M)n−1\det(\text{adj}\,M) = (\det M)^{n-1} with n=3n = 3
det⁡(A−1 adj(B−1) adj(2A−1))=det⁡(A−1)⋅det⁡(adj(B−1))⋅det⁡(adj(2A−1))\det\big(A^{-1}\,\text{adj}(B^{-1})\,\text{adj}(2A^{-1})\big) = \det(A^{-1})\cdot\det(\text{adj}(B^{-1}))\cdot\det(\text{adj}(2A^{-1}))
 Step 2: Evaluate each factor
det⁡(A−1)=1det⁡A=−12\det(A^{-1}) = \frac{1}{\det A} = -\frac{1}{2}
det⁡(adj(B−1))=(det⁡B−1)2=12=1\det(\text{adj}(B^{-1})) = (\det B^{-1})^2 = 1^2 = 1
det⁡(2A−1)=23det⁡(A−1)=8(−12)=−4  ⟹  det⁡(adj(2A−1))=(−4)2=16\det(2A^{-1}) = 2^3\det(A^{-1}) = 8\left(-\frac{1}{2}\right) = -4 \implies \det(\text{adj}(2A^{-1})) = (-4)^2 = 16
 Step 3: Multiply
(−12)(1)(16)=−8\left(-\frac{1}{2}\right)(1)(16) = -8
 Answer: (2)
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