Let and be two invertible matrices of order . If and , then is equal to
- A.1
- B.
- C.
- D.16
Show answer and solution
Correct option: C
Solution
From det(ABAT) = 8: det(A)·det(B)·det(A) = det(A)²·det(B) = 8.
From det(AB⁻¹) = 8: det(A)/det(B) = 8, so det(A) = 8·det(B).
Substituting: (8·det(B))²·det(B) = 64·det(B)³ = 8, giving det(B)³ = 1/8, so det(B) = 1/2 and det(A) = 4.
Now det(BA⁻¹BT) = det(B)·(1/det(A))·det(B) = det(B)²/det(A) = (1/4)/4 = 1/16.
From det(AB⁻¹) = 8: det(A)/det(B) = 8, so det(A) = 8·det(B).
Substituting: (8·det(B))²·det(B) = 64·det(B)³ = 8, giving det(B)³ = 1/8, so det(B) = 1/2 and det(A) = 4.
Now det(BA⁻¹BT) = det(B)·(1/det(A))·det(B) = det(B)²/det(A) = (1/4)/4 = 1/16.
[JEE Main 2019 · 11 Jan · Shift II]