JEE Advanced 2021 — Paper 1 — Question 26

Given $G = (I - EF)^{-1}$, so $(I - EF)G = I$ and $G(I - EF) = I$. From $(I - EF)G = I$, we get $G - EFG = I$, hence $EFG = G - I$. From $G(I - EF) = I$, we get $G - GEF = I$, hence $GEF = G - I$. Therefore $EFG = GEF$, so statement C is true. Using the identity $(I - FE)^{-1} = I + F(I - EF)^{-1}E = I + FGE$, we have $(I - FE)(I + FGE) = I$,

so statement B is true. For determinant: $|FE| = |F||E|$. Also, $|I - FE| = \det(I - FE)$ and $|FGE| = |F||G||E|$.

Since $G = (I - EF)^{-1}$, $|G| = 1/|I - EF|$. Using the identity $\det(I - FE) = \det(I - EF)$, we get $|I - FE| |FGE| = \det(I - FE) \cdot |F||E|/|I - EF| = |F||E| = |FE|$.

Thus statement A is true. For statement D: $(I - FE)(I - FGE) = (I - FE)(I + FGE) - 2(I - FE)FGE = I - 2(I - FE)FGE$.

Since $(I - FE)FGE = FGE - FEFGE$ is not generally zero, the product is not necessarily $I$.

Hence statement D is false.