1. Key Concepts and Formulas

• Powers of a Matrix: For a matrix $A$, its powers $A^n$ can often exhibit cyclic behavior, especially for permutation matrices. Identifying $A^k = I$ (the identity matrix) simplifies calculations for high powers.
• Determinant of an Adjoint Matrix: For any square matrix $M$ of order $k \times k$, the determinant of its adjoint, $\text{Adj}(M)$, is given by the formula: $\det(\text{Adj}(M)) = (\det(M))^{k-1}$ In this problem, all matrices are of order $3 \times 3$, so $k=3$. Thus, $\det(\text{Adj}(M)) = (\det(M))^{3-1} = (\det(M))^2$.
• Determinant of a Scalar Multiple: For a scalar $c$ and a $k \times k$ matrix $M$, $\det(cM) = c^k \det(M)$.

2. Step-by-Step Solution

Step 1: Analyze Matrix $A$ and its Powers We are given the matrix $A$ and need to simplify $B_0 = A^{49} + 2A^{98}$. To do this, we first investigate the powers of $A$. $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ Let's compute $A^2$: $A^2 = A \times A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ Now, let's compute $A^3$: $A^3 = A^2 \times A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ We found that $A^3 = I$, the $3 \times 3$ identity matrix. This is a cyclic property that greatly simplifies higher powers of $A$.

Step 2: Simplify the Expression for $B_0$ Using $A^3 = I$, we can simplify $A^{49}$ and $A^{98}$: For $A^{49}$: Divide $49$ by $3$. $49 = 3 \times 16 + 1$. $A^{49} = A^{3 \times 16 + 1} = (A^3)^{16} \cdot A^1 = I^{16} \cdot A = I \cdot A = A$ For $A^{98}$: Divide $98$ by $3$. $98 = 3 \times 32 + 2$. $A^{98} = A^{3 \times 32 + 2} = (A^3)^{32} \cdot A^2 = I^{32} \cdot A^2 = I \cdot A^2 = A^2$ Now substitute these back into the expression for $B_0$: $B_0 = A^{49} + 2A^{98} = A + 2A^2$ Let's compute the matrix $B_0$: $B_0 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + 2 \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$

Step 3: Calculate $\det(B_0)$ We compute the determinant of $B_0 = \begin{bmatrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$ using cofactor expansion along the first row: $\det(B_0) = 0 \cdot \det\begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} + 2 \cdot \det\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$ $\det(B_0) = 0 \cdot (0 \cdot 0 - 1 \cdot 2) - 1 \cdot (2 \cdot 0 - 1 \cdot 1) + 2 \cdot (2 \cdot 2 - 0 \cdot 1)$ $\det(B_0) = 0 - 1 \cdot (-1) + 2 \cdot (4)$ $\det(B_0) = 0 + 1 + 8 = 9$

Step 4: Establish the Recursive Relation for Determinants We are given $B_n = \text{Adj}(B_{n-1})$ for all $n \geq 1$. Using the property $\det(\text{Adj}(M)) = (\det(M))^2$ for a $3 \times 3$ matrix $M$, we can write: $\det(B_n) = \det(\text{Adj}(B_{n-1})) = (\det(B_{n-1}))^2$ This means the determinant of each subsequent matrix is the square of the determinant of the previous matrix.

Step 5: Calculate $\det(B_4)$ Now we apply this recursive relation starting from $\det(B_0) = 9$:

• For $n=1$: $\det(B_1) = (\det(B_0))^2 = 9^2$
• For $n=2$: $\det(B_2) = (\det(B_1))^2 = (9^2)^2 = 9^{2 \times 2} = 9^4$
• For $n=3$: $\det(B_3) = (\det(B_2))^2 = (9^4)^2 = 9^{4 \times 2} = 9^8$
• For $n=4$: $\det(B_4) = (\det(B_3))^2 = (9^8)^2 = 9^{8 \times 2} = 9^{16}$

Finally, we express $9^{16}$ in terms of powers of $3$, since $9 = 3^2$: $\det(B_4) = (3^2)^{16} = 3^{2 \times 16} = 3^{32}$

3. Common Mistakes & Tips

• Exponent Rules: A common error is to incorrectly apply exponent rules, for example, writing $(x^a)^b$ as $x^{a+b}$ instead of $x^{ab}$. Be careful when dealing with nested exponents.
• Matrix Powers: Always check for cyclic patterns for powers of special matrices (like permutation matrices or nilpotent matrices). This can significantly simplify calculations.
• Determinant Calculation: Double-check the signs and terms during cofactor expansion for determinants, especially for $3 \times 3$ matrices.
• Adjoint Formula: Remember the correct formula for the determinant of an adjoint matrix, $\det(\text{Adj}(M)) = (\det(M))^{k-1}$, where $k$ is the order of the matrix.

4. Summary

This problem required a systematic application of matrix properties. First, we identified the cyclic nature of matrix $A$ ($A^3=I$) to simplify the expression for $B_0$. Then, we explicitly calculated $B_0$ and its determinant, $\det(B_0)=9$. The crucial step was applying the property of the adjoint determinant, $\det(\text{Adj}(M)) = (\det(M))^2$ for $3 \times 3$ matrices, to establish a recursive relation for the determinants: $\det(B_n) = (\det(B_{n-1}))^2$. Iterating this relation four times, we found $\det(B_4) = (\det(B_0))^{16} = 9^{16}$. Finally, converting this to a power of $3$ yielded $3^{32}$.

5. Final Answer

The final answer is $\boxed{3^{32}}$, which corresponds to option (C).