If A is a 3 x 3 non-singular matrix such that AA' = A'A and B = A⁻¹A', then BB' equals: (a) B⁻¹ (b)

Q: Wait, why does $(A^{-1})'$ become $(A')^{-1}$? Is that always true?

Yes, it is a fundamental property! Since $AA^{-1} = I$, taking the transpose of both sides gives $(AA^{-1})' = I' = I$. By the reversal law, $(A^{-1})'A' = I$. Multiplying both sides by $(A')^{-1}$ on the right, we get $(A^{-1})' = (A')^{-1}$. It is a very useful tool in your JEE toolkit!

Q: Can I just assume $A$ is a symmetric matrix since $AA' = A'A$?

Not necessarily! The condition $AA' = A'A$ defines a 'Normal Matrix'. While all symmetric matrices are normal, not all normal matrices are symmetric. Do not assume $A = A'$ unless the problem explicitly states it. Stick to the property $AA' = A'A$ provided.

Q: Why did we regroup the terms in $BB'$? Is matrix multiplication commutative?

Great catch! Matrix multiplication is generally NOT commutative ($AB \neq BA$). However, it IS associative, meaning $(XY)Z = X(YZ)$. We regrouped the terms to isolate the product $A'A$ so we could apply the given condition $A'A = AA'$. We didn't change the order of the matrices, just the order of operations!

Question

If $A$ is a $3 	imes 3$ non-singular matrix such that $AA' = A'A$ and $B = A^{-1}A'$, then $BB'$ equals:

Options:

(A) $B^{-1}$

(B) $(B^{-1})'$

(C) $I+B$

(D) $I$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a $3 	imes 3$ non-singular matrix $A$. The condition that $A$ is non-singular implies that its determinant is non-zero and its inverse, $A^{-1}$, exists.

We are also provided with the condition $AA' = A'A$. This identifies $A$ as a **Normal Matrix**, which serves as the foundational constraint for our derivation. Our objective is to determine the product $BB'$ where $B = A^{-1}A'$.

### Phase 1: The Transpose Reversal Law

To evaluate $BB'$, we first determine the expression for $B'$. Given $B = A^{-1}A'$, we take the transpose of both sides:
$$B' = (A^{-1}A')'$$

Applying the **Transpose Reversal Law**, which states that $(XY)' = Y'X'$, we obtain:
$$B' = (A')'(A^{-1})'$$

Since the transpose of a transpose returns the original matrix, $(A')' = A$. Furthermore, the transpose of an inverse is the inverse of the transpose, meaning $(A^{-1})' = (A')^{-1}$. Thus, our expression simplifies to:
$$B' = A(A')^{-1}$$

### Phase 2: The Associative Symphony

Now, we construct the product $BB'$ using our derived expressions:
$$BB' = (A^{-1}A')(A(A')^{-1})$$

Matrix multiplication is associative, allowing us to regroup the terms without changing their order. We focus on the central terms:
$$BB' = A^{-1}(A'A)(A')^{-1}$$

We now utilize the "golden ticket" provided in the problem statement: $A'A = AA'$. Substituting this into our equation allows us to rearrange the matrices to facilitate cancellation.

### Phase 3: The Grand Finale

Substituting the condition $A'A = AA'$ into the expression, we get:
$$BB' = A^{-1}(AA')(A')^{-1}$$

By applying the associative property once more, we group the terms as follows:
$$BB' = (A^{-1}A)(A'(A')^{-1})$$

Recognizing the definition of the Identity matrix $I$, where $A^{-1}A = I$ and $A'(A')^{-1} = I$, the expression simplifies to:
$$BB' = I \cdot I$$

The final result is:
**$BB' = I$**

This result demonstrates that $B$ is an **orthogonal matrix**. The logical progression confirms that the structural properties of $A$ dictate the behavior of $B$ throughout the transformation.

Animated Solution for Mathematics - Matrices and Determinants: If $A$ is a $3 \times 3$ non-singular matrix such that $A A^{'} = A^{'} A$ and $B = A^{- 1} A^{'}$ , then $B B^{'}$ equals:

Select Answer:

Visualized Solution

Given Conditions

Transpose Reversal Law

Calculating $B^{'}$

Setting up $B B^{'}$

Applying Matrix Associativity

Using the Given Condition

Simplifying to Identity

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A is a 3×3 non-singular matrix such that AA′=A′A and B=A−1A′, then BB′ equals:

Select Answer:

Visualized Solution

Given Conditions

Transpose Reversal Law

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B′

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BB′

Applying Matrix Associativity

Using the Given Condition

Simplifying to Identity

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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