For any 3 × 3 matrix M, let |M| denote the determinant of M. Let I be the 3 × 3 identity matrix. Let

Q: Wait, why can't we just assume $EF = FE$ and simplify everything?

That is a classic trap! In matrix algebra, $EF$ is rarely equal to $FE$. If you assume they commute, you will miss the subtle structural symmetries that this problem is designed to reveal. Always treat $EF$ and $FE$ as distinct entities unless the problem explicitly states they commute.

Q: How do we know that $(I - FE)$ is invertible just because $(I - EF)$ is?

That is a deep question! While the problem doesn't explicitly ask you to prove it, it is a known result in matrix theory (often related to the Weinstein-Aronszajn identity) that if $(I - EF)$ is invertible, then $(I - FE)$ is also invertible. You can actually prove this by constructing the inverse of $(I - FE)$ using $G$.

Q: Why did we choose to multiply by $(I + FGE)$ in Statement (B)? It seems arbitrary.

It is not arbitrary at all! When you are faced with an expression like $(I - FE)$, the most natural 'partner' to test for an inverse is $(I + FGE)$. By expanding the product, you create terms that allow you to use the identity $EFG = G - I$, which effectively 'collapses' the expression into the identity matrix. It is a classic technique in matrix algebra to look for these 'telescoping' structures.

Question

For any $3 	imes 3$ matrix $M$, let $|M|$ denote the determinant of $M$. Let $I$ be the $3 	imes 3$ identity matrix. Let $E$ and $F$ be two $3 	imes 3$ matrices such that $(I - EF)$ is invertible. If $G = (I - EF)^{-1}$, then which of the following statements is (are) TRUE ?

Options:

(A) $|FE| = |I - FE||FGE|$

(B) $(I - FE)(I + FGE) = I$

(C) $EFG = GEF$

(D) $(I - FE)(I - FGE) = I$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey! Today, we are going to peel back the curtain on one of the most elegant topics in linear algebra: the manipulation of matrix inverses.

When you first look at a problem involving $G = (I - EF)^{-1}$, it is easy to feel overwhelmed by the abstract symbols. See this not as a wall of algebra, but as a dance of structures.

### The Bedrock: Defining the Inverse

We are given that $G$ is the inverse of $(I - EF)$. By the very definition of an inverse matrix, if we multiply a matrix by its inverse, we must get the identity matrix $I$.

This gives us two powerful, symmetric equations:

$$G(I - EF) = I \implies G - GEF = I \implies GEF = G - I$$

$$(I - EF)G = I \implies G - EFG = I \implies EFG = G - I$$

Both $GEF$ and $EFG$ are equal to the same expression, $G - I$. This immediately proves that $EFG = GEF$, confirming that **Statement (C) is true**.

Even though matrix multiplication is non-commutative, the inverse relationship forces a specific symmetry upon these products.

### The Algebraic Dance: Proving Statement (B)

Now, let us tackle Statement (B). We need to evaluate the product $(I - FE)(I + FGE)$.

Expand it just like binomials, being careful to maintain the order of multiplication:

$$(I - FE)(I + FGE) = I(I) + I(FGE) - FE(I) - FE(FGE) = I + FGE - FE - FEFGE$$

Focus on the last term: $FEFGE$. We can use the associativity of matrix multiplication to group it as $F(EFG)E$.

Since we already know $EFG = G - I$, we can substitute that in:

$$FEFGE = F(EFG)E = F(G - I)E = FGE - FE$$

Now, substitute this back into our expanded expression:

$$I + FGE - FE - (FGE - FE) = I + FGE - FE - FGE + FE = I$$

Everything cancels out perfectly. It is like a choreographed dance where every step leads to the identity matrix; thus, **Statement (B) is proven true**.

### The Determinant Bridge: Proving Statement (A)

Finally, let us look at Statement (A). We need to connect $|FE|$ and $|FGE|$.

We know from Statement (B) that $(I - FE)$ and $(I + FGE)$ are inverses. Taking the determinant of both sides, we get $|I - FE| |I + FGE| = |I| = 1$.

To bridge the gap, evaluate the product $FE(I + FGE)$:

$$FE(I + FGE) = FE + FEFGE = FE + (FGE - FE) = FGE$$

Taking the determinant of both sides, we get $|FE| |I + FGE| = |FGE|$.

Since $|I + FGE| = \frac{1}{|I - FE|}$, we substitute this to get:

$$|FE| \cdot \frac{1}{|I - FE|} = |FGE| \implies |FE| = |I - FE| |FGE|$$

Therefore, **Statement (A) is also true**.

### The Takeaway

We have navigated the abstract world of matrices and found that even in the absence of commutativity, there is a deep, underlying order.

The key is to trust the definitions, use associativity to your advantage, and look for the symmetry in the expressions. Keep practicing, and soon, these algebraic dances will become second nature to you!

Select Answer:

Visualized Solution

Inverse Matrix Definition and Expansion

Proving Statement (C)

Expanding Statement (B)

Simplifying the $F E F GE$ Term

Final Proof for Statement (B)

Setting up Statement (A)

Expanding $F E (I + F GE)$

Final Proof for Statement (A)

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Bedrock

The Algebraic Dance

The Determinant Bridge

The Takeaway

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