Matrices and Determinants JEE 2021: Identities and Solution

Which identities are true in the JEE Advanced 2021 matrix question?

The matrices and determinants jee 2021 question is solved through one chain of inverse identities, not four separate checks. This is a hard JEE Advanced 2021 Paper 1 Mathematics question. Based on the length of the identity-based solution, its expected solving time is about 180 seconds. It is multiple-correct, so each statement needs proof or rejection.

Let the two matrices have order: $3\times 3$

Assume that the following matrix is invertible: $I-EF$

Define: $G=(I-EF{)}^{-1}$

Test these four statements:

• A $|FE|=|I-FE||FGE|$
• B $(I-FE)(I+FGE)=I$
• C $EFG=GEF$
• D $(I-FE)(I-FGE)=I$

Do not decide from the signs. The order of every matrix factor matters.

How do you begin the full solution from the defining inverse relation?

Start from the defining inverse relation. Expanding it gives the identity that controls the rest of the solution. The same relation proves statement C because an invertible matrix commutes with its inverse. This does not require the two original matrices to commute.

Start with: $(I-EF)G=I$

Expand: $G-EFG=I$

Rearrange: $G=I+EFG$

Since an invertible matrix commutes with its inverse: $G(I-EF)=(I-EF)G$

Expand both sides: $G-GEF=G-EFG$

Cancel the common term: $GEF=EFG$

Therefore: $EFG=GEF$

Thus, statement C is true.

No step assumes: $EF=FE$

The equality follows only from the inverse relation.

How does the inverse identity prove B and reject D?

Direct multiplication shows that the correct inverse contains a plus sign. It proves statement B and distinguishes it from statement D without changing the order of any factors. Products such as the one below cannot be rearranged as though matrices were ordinary numbers.

Expand: $(I-FE)(I+FGE)$

This gives: $I+FGE-FE-FEFGE$

Regroup: $FGE-FEFGE=F(G-EFG)E$

Hence:

From the defining inverse relation: $G-EFG=I$

Substitute: $I-FE+FIE$

Therefore: $I-FE+FE=I$

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

Since all the matrices are square, the displayed right inverse is the inverse: $(I-FE{)}^{-1}=I+FGE$

Therefore, statement B is true.

The inverse contains the positive term, not the negative term. Statement D is not generally true.

For a concrete three by three counterexample, take:

Then: $EF=\frac{1}{4}I$ and:

Now:

Therefore:

This is not the identity matrix. Hence, statement D is false.

How does Sylvester’s determinant theorem prove statement A?

Statement A follows from Sylvester’s determinant theorem and determinant multiplicativity. The required cancellation is valid because the original condition makes the determinant involving the first product nonzero. Sylvester’s theorem gives the same nonzero determinant when the product order is reversed.

In this setting, Sylvester’s determinant theorem gives: $|I-EF|=|I-FE|$

Since: $G=(I-EF{)}^{-1}$ we have:

Evaluate the right side of statement A: $|I-FE||FGE|$

By determinant multiplicativity:

Substitute the determinant of the inverse:

Now use: $|I-FE|=|I-EF|$

This determinant is nonzero, so it can be cancelled: $|F||E|$

By determinant multiplicativity: $|F||E|=|FE|$

Therefore: $|I-FE||FGE|=|FE|$

Thus, statement A is true.

The verified answer is:

Why is the minus-sign inverse identity wrong?

The error is carrying the minus sign from the original matrix directly into its supposed inverse. Inversion does not distribute over subtraction. The sign and factor order in an inverse identity must be established through multiplication, not copied from the original expression.

In general:

It is also not generally true that: $(I-X{)}^{-1}=I-X$

The correct derived identity is: $(I-FE{)}^{-1}=I+FGE$

The earlier counterexample fails even though both matrices are scalar multiples of the identity:

It gives:

Matrix order must also remain fixed. The product can be factored as: $FEFGE=F(EFG)E$

Its factors cannot be swapped. Treating the original matrices as commuting would invalidate the proof.

Which related practice questions test the same identities?

These are related practice questions, not additional JEE PYQs. The first requires the same inverse proof with general square matrices. The second uses determinant multiplicativity with given values. Solve both by reproducing the identities, not by recalling the verified options from the matrices and determinants jee 2021 question.

Related practice question 1

How do you prove the inverse identity for general square matrices?

Use the same product check as statement B: define the inverse, expand the proposed product and preserve factor order. No commutation assumption is needed. Let the two square matrices have order shown below, then define the inverse. $n\times n$

Let: $H=(I-AB{)}^{-1}$

Prove: $(I-BA{)}^{-1}=I+BHA$

From the definition: $(I-AB)H=I$

Therefore: $H-ABH=I$

Check the required product: $(I-BA)(I+BHA)$

Expand: $I+BHA-BA-BABHA$

Regroup: $I-BA+B(H-ABH)A$

Use: $H-ABH=I$

Thus: $I-BA+BIA=I-BA+BA=I$

Hence: $(I-BA{)}^{-1}=I+BHA$

Related practice question 2

How do you calculate the required determinant from the given values?

Use determinant multiplicativity and the determinant of an inverse. The matrices have order three by three, and the determinant of the inverse is obtained directly from the given value. Keep the multiplication order unchanged, even though the final determinant values are ordinary numbers.

Suppose:

Let: $G=(I-EF{)}^{-1}$

Then:

By determinant multiplicativity:

Therefore:

When practising, write the product check before memorising the inverse identity. This prevents both the wrong sign and the wrong matrix order.

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