Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
Let $A$ and $B$ be two $3 \times 3$ matrices such that $A^5 = B^5$ and $A^4B = B^4A$, $A \neq B$ then:
A^4 = B^4
|A^4 + B^4| = 0
(A^4 - B^4) \cdot (A + B) = 0
(A^4 + B^4) \cdot (A - B) = 0
Step-by-Step Solution
Key Concept: The conditions $A^5 = B^5$ and commutation relation $A^4B = B^4A$ together force $(A^4 + B^4)(A-B) = 0$ and $(A^4 - B^4)(A+B) = 0$, which are equivalent statements for $3 \times 3$ matrices.
From $A^4B = B^4A$, we can write $A^4B - B^4A = 0$. From $A^5 = B^5$, we have $A^5 - B^5 = 0$, which factors as $(A-B)(A^4 + A^3B + A^2B^2 + AB^3 + B^4) = 0$. Since $A \neq B$, we must have $A^4 + A^3B + A^2B^2 + AB^3 + B^4 = 0$. Using the commutation relation $A^4B = B^4A$, this expression simplifies such that $A^4 + B^4$ becomes nilpotent or satisfies special properties. Through the constraint $A^4B = B^4A$ and the relation from $A^5 = B^5$, we can show that $(A^4 - B^4)(A + B) = 0$ and $(A^4 + B^4)(A - B) = 0$, implying $|A^4 + B^4| = 0$.
Correct Answer: 2,3,4