If `A_(0) = [[2 ,-2,-4],[-1,3,4],[1,-2,-3]] and B_(0)= [[-4,-4,-4],[1,0,1],[4,4,3]]` and `B_(n) = adj (B_(n-1)),n inN` and `I` is an dientity matrix of order 3. `B_(2) + B_(3) + B_(4) +...+ B_(50) ` is equal to

Options

A. `B_(0)`
B. `7 B_(0)`
C. `49 B_(0)`
D. 49I

Explanation

To solve the problem, we need to analyze the matrices \( B_n \) defined by the recurrence relation \( B_n = \text{adj}(B_{n-1}) \) for \( n \in \mathbb{N} \).

### Step 1: Define the matrices \( A_0 \) and \( B_0 \)

Given:
\[ A_0 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}, \quad B_0 = \begin{bmatrix} -4 & -4 & -4 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} \]

### Step 2: Calculate the adjoint of \( B_0 \)

The adjoint of a matrix \( B \) is given by the transpose of its cofactor matrix. We need to calculate the cofactor matrix of \( B_0 \).

The cofactor matrix \( C \) of \( B_0 \) is calculated as follows:

1. For each element \( b_{ij} \) of \( B_0 \), calculate the determinant of the \( 2 \times 2 \) matrix that remains after removing the \( i \)-th row and \( j \)-th column from \( B_0 \).
2. Apply the sign factor \( (-1)^{i+j} \) to each determinant to get the cofactor.

Calculating the cofactors for \( B_0 \):

- \( C_{11} = \text{det} \begin{bmatrix} 0 & 1 \\ 4 & 3 \end{bmatrix} = (0)(3) - (1)(4) = -4 \)
- \( C_{12} = -\text{det} \begin{bmatrix} 1 & 1 \\ 4 & 3 \end{bmatrix} = -((1)(3) - (1)(4)) = 1 \)
- \( C_{13} = \text{det} \begin{bmatrix} 1 & 0 \\ 4 & 4 \end{bmatrix} = (1)(4) - (0)(4) = 4 \)
- Continuing this for all elements, we find the cofactor matrix \( C \).

After calculating all cofactors, we find:
\[ C = \begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & -3 \end{bmatrix} \]

Now, the adjoint \( B_1 = \text{adj}(B_0) = C^T \):
\[ B_1 = \begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & -3 \end{bmatrix} \]

### Step 3: Check if \( B_1 = B_0 \)

We need to check if \( B_1 \) is equal to \( B_0 \). If \( B_1 = B_0 \), then:
\[ B_1 = B_0 \implies B_2 = \text{adj}(B_1) = B_0 \]
Continuing this process, we find:
\[ B_2 = B_0, B_3 = B_0, \ldots, B_n = B_0 \text{ for all } n \geq 1 \]

### Step 4: Calculate \( B_2 + B_3 + B_4 + \ldots + B_{50} \)

Since \( B_n = B_0 \) for all \( n \geq 1 \), we can express the sum as:
\[ B_2 + B_3 + B_4 + \ldots + B_{50} = 49 B_0 \]

### Step 5: Final Result

Now, substituting \( B_0 \):
\[ 49 B_0 = 49 \begin{bmatrix} -4 & -4 & -4 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix} = \begin{bmatrix} -196 & -196 & -196 \\ 49 & 0 & 49 \\ 196 & 196 & 147 \end{bmatrix} \]

Thus, the final answer is:
\[ \boxed{\begin{bmatrix} -196 & -196 & -196 \\ 49 & 0 & 49 \\ 196 & 196 & 147 \end{bmatrix}} \]