If A=[2−112],B=[1101],C=ABAT and X=ATC2A, then detX is equal to :(A)243(B)729✓ Correct(C)27(D)891Answer: (B)SolutionA=(2−112), detA=2+1=3. B=(1101), detB=1.C=ABAT⟹detC=detA⋅detB⋅detAT=(detA)2detB=9×1=9.X=ATC2A⟹detX=det(AT)⋅(detC)2⋅detA=(detA)2(detC)2=9×81=729JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option