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

Concept: When working with matrix conjugation identities of the type $ABA^{-1} = B^k$, computing subsequent matrix powers reveals an exponential pattern. This properties relies on the middle cancellation of adjacent inverse matrix units: $$(ABA^{-1})^2 = (ABA^{-1})(ABA^{-1}) = AB(A^{-1}A)BA^{-1} = AB^2A^{-1}$$ Step 1: Analyze the recursive power pattern.
We are given the base transformation rule: $$ABA^{-1} = B^2$$ Let us square both sides of this equation: $$\left(ABA^{-1}\right)^2 = (B^2)^2 \quad \Rightarrow \quad AB^2A^{-1} = B^4$$ Substitute our original definition for $B^2 = ABA^{-1}$ into the left side of this equation: $$A\left(ABA^{-1}\right)A^{-1} = B^4 \quad \Rightarrow \quad A^2BA^{-2} = B^4 = B^{2^2}$$

Step 2: Extend the relation using induction.
If we repeat this squaring sequence a third time: $$A^3BA^{-3} = B^8 = B^{2^3}$$ Extending this mathematical induction step to an arbitrary power exponent $k$: $$A^k B A^{-k} = B^{2^k} \quad \cdots (1)$$

Step 3: Apply the boundary condition $A^5 = I$.
The problem states that matrix $A$ has a power of 5, meaning $A^5 = I$ (which also implies $A^{-5} = I$). Let us substitute $k = 5$ into our general equation (1): $$A^5 B A^{-5} = B^{2^5}$$ Substitute the identity matrices into the equation: $$I \cdot B \cdot I = B^{32} \quad \Rightarrow \quad B = B^{32}$$

Step 4: Isolate the power of matrix $B$.
Multiply both sides of the equation by the matrix inverse $B^{-1}$: $$B \cdot B^{-1} = B^{32} \cdot B^{-1} \quad \Rightarrow \quad I = B^{31}$$ This confirms that the power exponent of matrix $B$ is exactly 31. Looking at the interval ranges given in our options, the number 31 lies strictly between 28 and 32, matching choice (B).