Question

Let us define the power of a matrix \(A\) as the maximum \(m\in\mathbb{Z}^{+}\) such that \(A^{m}=I\). For two matrices \(A\) and \(B\) if \(A^{5}=I\) and \(ABA^{-1}=B^{2}\), then the power of the matrix \(B\) is between:

• A. 20 and 24
• B. 28 and 32
• C. 36 and 40
• D. 4 and 8

Hint:

The general rule for this type of matrix conjugation is $B^{(k^n - 1)} = I$, where $k$ is the power of the right-hand term ($B^2 \implies k=2$) and $n$ is the power index of the conjugating matrix ($A^5 \implies n=5$). Substituting these values gives $2^5 - 1 = 32 - 1 = 31$ instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves matrix similarity and exponentiation. The relation \( ABA^{-1} = B^2 \) implies that \( B^2 \) is similar to \( B \). When we perform this transformation multiple times, the exponent of \( B \) will grow exponentially. We use the property that \( A(B^k)A^{-1} = (ABA^{-1})^k = (B^2)^k = B^{2k} \).
Step 2: Key Formula or Approach:
1. Iterate the relation \( k \) times: \( A^k B A^{-k} = B^{2^k} \).
2. Use the given \( A^5 = I \).
3. Substitute \( k=5 \) into the iterated relation to find the power of \( B \).
Step 3: Detailed Explanation:
Given \( ABA^{-1} = B^2 \).
Perform similarity transform again:
\( A(ABA^{-1})A^{-1} = A B^2 A^{-1} \).
\( A^2 B A^{-2} = (ABA^{-1})(ABA^{-1}) = B^2 B^2 = B^4 = B^{2^2} \).
Generalizing this pattern for any integer \( k \):
\[ A^k B A^{-k} = B^{2^k} \]
We are given \( A^5 = I \). Substitute \( k = 5 \) into the formula:
\[ A^5 B A^{-5} = B^{2^5} \]
\[ (I) B (I^{-1}) = B^{32} \]
\[ B = B^{32} \]
Multiply both sides by \( B^{-1} \) (assuming \( B \) is non-singular, or by observing the group structure):
\[ I = B^{31} \]
This means that \( B^{31} = I \). The "power" \( m \) of matrix B is therefore 31.
Comparing 31 with the given ranges:
(A) 20 and 24: No.
(B) 28 and 32: Yes (31 is between 28 and 32).
(C) 36 and 40: No.
(D) 4 and 8: No.
Step 4: Final Answer:
The power of matrix B is 31, which lies between 28 and 32.