To determine the nature of the relation \( R \) defined as \( R = \{(A, B) : A = P^{-1}BP \text{ for some invertible matrix } P\} \), we need to check if it is reflexive, symmetric, and transitive.
### Step 1: Check Reflexivity
A relation is reflexive if every element is related to itself. In this case, we need to show that for any matrix \( A \), the pair \( (A, A) \) belongs to \( R \).
- Let \( P \) be the identity matrix \( I \). Then:
\[
A = I^{-1} A I = A
\]
- Since this holds for any matrix \( A \), we conclude that \( (A, A) \in R \).
**Conclusion**: The relation \( R \) is reflexive.
### Step 2: Check Symmetry
A relation is symmetric if whenever \( (A, B) \in R \), then \( (B, A) \in R \) as well.
- Assume \( (A, B) \in R \). This means:
\[
A = P^{-1}BP \text{ for some invertible matrix } P
\]
- To show symmetry, we need to express \( B \) in terms of \( A \):
\[
B = P A P^{-1}
\]
- Since \( P \) is invertible, we can take \( P^{-1} \) as the new matrix \( Q \):
\[
B = Q^{-1} A Q \text{ where } Q = P
\]
- Thus, \( (B, A) \in R \).
**Conclusion**: The relation \( R \) is symmetric.
### Step 3: Check Transitivity
A relation is transitive if whenever \( (A, B) \in R \) and \( (B, C) \in R \), then \( (A, C) \in R \).
- Assume \( (A, B) \in R \) and \( (B, C) \in R \):
\[
A = P^{-1}BP \text{ for some invertible matrix } P
\]
\[
B = Q^{-1}CQ \text{ for some invertible matrix } Q
\]
- Substitute \( B \) in the first equation:
\[
A = P^{-1}(Q^{-1}CQ)P
\]
- Rearranging gives:
\[
A = (PQ)^{-1} C (PQ)
\]
- Let \( R = PQ \) (which is also invertible since the product of invertible matrices is invertible):
\[
A = R^{-1} C R
\]
- Hence, \( (A, C) \in R \).
**Conclusion**: The relation \( R \) is transitive.
### Final Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation.
### Answer
The relation \( R \) is an equivalence relation.
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