Let the matrix $$A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ and the matrix $$B_0 = A^{49} + 2A^{98}$$. If $$B_n = \text{Adj}(B_{n-1})$$ for all $$n \geq 1$$, then $$\det(B_4)$$ is equal to

We are given $$A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ and $$B_0 = A^{49} + 2A^{98}$$, with $$B_n = \text{Adj}(B_{n-1})$$ for $$n \geq 1$$.

Step 1: Find powers of A.

Matrix A swaps the first two rows (it is a permutation matrix). Therefore:

$$A^2 = I$$ (the identity matrix)

For any integer $$n$$: $$A^{\text{odd}} = A$$ and $$A^{\text{even}} = I$$.

Since 49 is odd: $$A^{49} = A$$

Since 98 is even: $$A^{98} = I$$

Step 2: Compute $$B_0$$.

$$B_0 = A + 2I = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$

Step 3: Find $$\det(B_0)$$.

Expanding along the third row:

$$\det(B_0) = 3 \cdot (4 - 1) = 3 \times 3 = 9 = 3^2$$

Step 4: Use the adjugate determinant formula.

For an $$n \times n$$ matrix M: $$\det(\text{Adj}(M)) = [\det(M)]^{n-1}$$

For our $$3 \times 3$$ matrices ($$n = 3$$): $$\det(B_n) = [\det(B_{n-1})]^2$$

Step 5: Compute $$\det(B_4)$$ iteratively.

$$\det(B_0) = 3^2$$

$$\det(B_1) = (3^2)^2 = 3^4$$

$$\det(B_2) = (3^4)^2 = 3^8$$

$$\det(B_3) = (3^8)^2 = 3^{16}$$

$$\det(B_4) = (3^{16})^2 = 3^{32}$$

The correct answer is Option C: $$3^{32}$$