Step-by-step Solution

Given Information:

• E and F are 3x3 matrices.
• I is the 3x3 identity matrix.
• The matrix (I - EF) is invertible.
• G = (I - EF)$^{-1}$.

From the definition of G, we have two key relations:

1. (I - EF)G = I => G - EFG = I (Equation 1)
2. G(I - EF) = I => G - GEF = I (Equation 2)

We will now evaluate each statement.

Evaluation of Option C: EFG = GEF

From Equation 1, we can write: $EFG = G - I$ From Equation 2, we can write: $GEF = G - I$ Comparing these two expressions, we get: $EFG = GEF$ Thus, statement C is TRUE.

Evaluation of Option B: (I - FE)(I + FGE) = I

Let's expand the left-hand side (LHS) of the equation: LHS = $(I - FE)(I + FGE) = I(I + FGE) - FE(I + FGE)$ LHS = $I + FGE - FE - FEFGE$

Now, we need to simplify the term $FEFGE$. We can use the result $EFG = G - I$ from Equation 1. $FEFGE = F(EFG)E = F(G - I)E = (FG - F)E = FGE - FE$

Substitute this back into the expression for the LHS: LHS = $I + FGE - FE - (FGE - FE)$ LHS = $I + FGE - FE - FGE + FE$ LHS = $I$

Since LHS = I, the statement is correct. This also implies that the matrix (I - FE) is invertible and its inverse is (I + FGE). Thus, statement B is TRUE.

Evaluation of Option D: (I - FE)(I - FGE) = I

From the evaluation of option B, we found that $(I - FE)^{-1} = I + FGE$. Option D claims that the inverse is $(I - FGE)$. This can only be true if $I + FGE = I - FGE$, which would mean $2FGE = 0$, or $FGE = 0$. This is not true for arbitrary matrices E and F.

Let's expand the LHS of the statement in D: LHS = $(I - FE)(I - FGE) = I - FGE - FE + FEFGE$ Using $FEFGE = FGE - FE$ from the analysis of option B: LHS = $I - FGE - FE + (FGE - FE) = I - 2FE$ For this to be equal to I, we must have $I - 2FE = I$, which implies $2FE = 0$, or $FE=0$. This is not generally true. Thus, statement D is FALSE.

Evaluation of Option A: |FE| = |I - FE| |FGE|

We will use two properties of determinants for square matrices X and Y:

1. $|XY| = |X||Y|$
2. Sylvester's determinant identity: $|I - XY| = |I - YX|$

Applying the second identity with X=E and Y=F, we get: $|I - EF| = |I - FE|$

From the given definition $G = (I - EF)^{-1}$, we can take the determinant of both sides: $|G| = |(I - EF)^{-1}| = \frac{1}{|I - EF|}$

Combining these results, we have: $|I - FE| = |I - EF| = \frac{1}{|G|}$

Now, let's substitute this into the right-hand side (RHS) of the statement in option A: RHS = $|I - FE| |FGE|$ RHS = $(\frac{1}{|G|}) (|F||G||E|)$ Since determinants are scalars, their multiplication is commutative: RHS = $\frac{1}{|G|} |F||E||G| = |F||E|$

Now let's evaluate the left-hand side (LHS) of the statement: LHS = $|FE| = |F||E|$

Since LHS = RHS, the statement is correct. Thus, statement A is TRUE.

Conclusion

The statements A, B, and C are TRUE, while statement D is FALSE.