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If A and B are square matrices of same order such that, AB=BAAB=BA, A−1=A2A^{-1} = A^2 and B−1=B3B^{-1} = B^3, then the value of (AB)2040(AB)^{2040}, is

Solution

✅ Correct Option: 2

From A−1=A2A^{-1}=A^2 we get A3=IA^3=I, and from B−1=B3B^{-1}=B^3 we get B4=IB^4=I. Since AB=BAAB=BA, (AB)2040=A2040B2040(AB)^{2040}=A^{2040}B^{2040}. As 2040 is divisible by both 3 and 4, each factor is II, so the product is the identity matrix.

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