Define a relation R over a class of n x n real matrices A and B as "ARB iff there exists a non-singu

Q: Wait, why do we insist that $P$ must be non-singular? What if $|P| = 0$?

That is a fantastic question! If $|P| = 0$, the matrix $P$ is singular, which means it does not have an inverse ($P^{-1}$ does not exist). Without the inverse, the definition $PAP^{-1}$ breaks down completely. We need $P^{-1}$ to 'undo' the transformation $P$ applies, effectively returning us to the original space. If $P$ were singular, we would lose information, and the relationship would not be reversible.

Q: Why can't we just say $A = B$ for symmetry? Why the $P^{-1}BP$ mess?

Because similarity is not equality! Two matrices can be similar without being identical. They share the same 'DNA' (eigenvalues, trace, determinant), but they look different. Symmetry asks if the relationship is reversible—if $A$ is similar to $B$, is $B$ similar to $A$? The algebra proves it is, provided we use the inverse transformation. We aren't proving $A=B$; we are proving that the 'similarity' link works in both directions.

Q: Is there a shortcut for transitivity? It looks like a lot of algebra.

The algebra is the proof! In JEE Advanced, you cannot skip steps. The beauty lies in the reversal law of inverses: $(QP)^{-1} = P^{-1}Q^{-1}$. Once you see that, the algebra collapses beautifully into $S = QP$. It's not just algebra; it's the structure of the group of invertible matrices. Embrace the derivation; it builds the intuition you need for harder problems.

Question

Define a relation R over a class of $n 	imes n$ real matrices A and B as "ARB iff there exists a non-singular matrix P such that $PAP^{-1} = B$". Then which of the following is true?

Options:

(A) R is symmetric, transitive but not reflexive,

(B) R is reflexive, symmetric but not transitive

(C) R is an equivalence relation

(D) R is reflexive, transitive but not symmetric

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The concept of **Matrix Similarity** is a fundamental pillar of Linear Algebra. It represents the idea that two matrices, while appearing distinct, represent the same linear transformation viewed from different coordinate systems.

To prove that similarity is an equivalence relation, we must demonstrate that it satisfies three specific properties: **Reflexivity**, **Symmetry**, and **Transitivity**.

### Phase 1: The Reflexive Mirror

A relation $R$ is reflexive if every element is related to itself. For a matrix $A$, we must determine if there exists a non-singular matrix $P$ such that $PAP^{-1} = A$.

Consider the Identity matrix $I$. Since $I$ is non-singular ($|I| = 1 
eq 0$) and $I^{-1} = I$, we can write:

$$IAI^{-1} = IAI = A$$

Because we have identified a valid $P$ (where $P=I$), the condition $A R A$ holds true. Thus, the relation is **reflexive**.

### Phase 2: The Symmetry of Transformation

Next, we examine symmetry. If $A$ is similar to $B$, we must prove that $B$ is similar to $A$. We assume $A R B$, which implies there exists a non-singular matrix $P$ such that:

$$PAP^{-1} = B$$

To isolate $A$, we must apply matrix inverses carefully, noting that matrix multiplication is non-commutative. We pre-multiply by $P^{-1}$ and post-multiply by $P$:

$$P^{-1}(PAP^{-1})P = P^{-1}BP$$

$$A = P^{-1}BP$$

Let $Q = P^{-1}$. Since $P$ is non-singular, $Q$ is also non-singular, and its inverse is $P$. Substituting this, we obtain $A = QBQ^{-1}$, which confirms that $B R A$. The relation is **symmetric**.

### Phase 3: The Transitive Chain

Finally, we address transitivity. If $A R B$ and $B R C$, we must prove $A R C$. We are given:

$$B = PAP^{-1} \quad 	ext{and} \quad C = QBQ^{-1}$$

Substituting the expression for $B$ into the equation for $C$, we get:

$$C = Q(PAP^{-1})Q^{-1}$$

Using the associative property of matrix multiplication, we group the terms:

$$C = (QP)A(P^{-1}Q^{-1})$$

Applying the reversal law of inverses, where $(QP)^{-1} = P^{-1}Q^{-1}$, we define $S = QP$:

$$C = SAS^{-1}$$

Since the product of two non-singular matrices is non-singular, $S$ is a valid transformation matrix. Thus, $A R C$, proving the relation is **transitive**.

### Conclusion: The Equivalence

We have rigorously demonstrated that the relation of matrix similarity is reflexive, symmetric, and transitive.

In the language of set theory, this confirms that similarity is an **equivalence relation**. This partitions the space of $n 	imes n$ matrices into equivalence classes, where all matrices within a class share identical fundamental properties, such as their **eigenvalues** and **trace**.

Animated Solution for Mathematics - Sets and Relations: Define a relation R over a class of $n \times n$ real matrices A and B as "ARB iff there exists a non-singular matrix P such that $P A P^{- 1} = B$ ". Then which of the following is true?

Select Answer:

Visualized Solution

Definition of Relation $R$

Reflexivity: $A R A$

Proving Reflexivity

Symmetry: If $A R B$ , then $B R A$ ?

Isolating Matrix $A$

Proving Symmetry

Transitivity: $A R B$ and $B R C$

Combining the Relations

Simplifying the Product

Proving Transitivity

Equivalence Relation

The Sigma Insight: Equivalence Relations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

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