For any 3 times 3 matrix M,let |M| denote the | vedclass

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(B) Given $G = (I - EF)^{-1}$,we have $G^{-1} = I - EF$.Since $G G^{-1} = I = G^{-1} G$,we have $G(I - EF) = I = (I - EF)G$.Expanding this,$G - GEF =

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$A, B, D$

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(B) Given $G = (I - EF)^{-1}$,we have $G^{-1} = I - EF$.Since $G G^{-1} = I = G^{-1} G$,we have $G(I - EF) = I = (I - EF)G$.Expanding this,$G - GEF = I = G - EFG$,which implies $GEF = EFG$. Thus,$(C)$ is $TRUE$.Next,consider $(I - FE)(I + FGE) = I + FGE - FE - FEFGE$.Since $G^{-1} = I - EF$,we have $EF = I - G^{-1}$. Substituting this,$FEF = F(I - G^{-1}) = F - FG^{-1}$.Alternatively,$FEFGE = F(EF)GE = F(I - G^{-1})GE = FGE - FG^{-1}GE = FGE - FE$.Substituting back: $I + FGE - FE - (FGE - FE) = I$. Thus,$(B)$ is $TRUE$.From $(I - FE)(I + FGE) = I$,taking the determinant on both sides,$|I - FE| |I + FGE| = |I| = 1$.Also,$FE(I + FGE) = FE + FEFGE = FE + F(EF)GE = FE + F(I - G^{-1})GE = FE + FGE - FE = FGE$.Taking the determinant: $|FE| |I + FGE| = |FGE|$.Since $|I + FGE| = \frac{1}{|I - FE|}$,we get $|FE| \frac{1}{|I - FE|} = |FGE|$,which implies $|FE| = |I - FE| |FGE|$. Thus,$(A)$ is $TRUE$.Therefore,the correct statements are $(A), (B), (C)$.

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