If `B_(0)=[(-4, -3, -3),(1,0,1),(4,4,3)], B_(n)=adj(B_(n-1), AA n in N` and I is an identity matrix of order 3, then `B_(1)+B_(3)+B_(5)+B_(7)+B_(9)` is equal to

Options

A. `B_(0)`
B. `5B_(0)`
C. `25B_(0)`
D. `5I`

Answer

To solve the problem, we need to calculate the sum \( B_1 + B_3 + B_5 + B_7 + B_9 \) where \( B_n \) is defined recursively as the adjoint of the previous matrix \( B_{n-1} \) starting from \( B_0 \).

### Step-by-Step Solution:

1. **Identify \( B_0 \)**:
\[ B_0 = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \]

2. **Calculate \( B_1 \)**:
Since \( B_1 = \text{adj}(B_0) \), we need to find the adjoint of \( B_0 \). The adjoint of a matrix is the transpose of its cofactor matrix.

- Calculate the cofactors:
- \( C_{11} = \begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = (0)(3) - (1)(4) = -4 \)
- \( C_{12} = -\begin{vmatrix} 1 & 1 \\ 4 & 3 \end{vmatrix} = -((1)(3) - (1)(4)) = 1 \)
- \( C_{13} = \begin{vmatrix} 1 & 0 \\ 4 & 4 \end{vmatrix} = (1)(4) - (0)(4) = 4 \)
- \( C_{21} = -\begin{vmatrix} -3 & -3 \\ 4 & 3 \end{vmatrix} = -((-3)(3) - (-3)(4)) = -(-9 + 12) = -3 \)
- \( C_{22} = \begin{vmatrix} -4 & -3 \\ 4 & 3 \end{vmatrix} = (-4)(3) - (-3)(4) = -12 + 12 = 0 \)
- \( C_{23} = -\begin{vmatrix} -4 & -3 \\ 1 & 1 \end{vmatrix} = -((-4)(1) - (-3)(1)) = -(-4 + 3) = 1 \)
- \( C_{31} = \begin{vmatrix} -3 & -3 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (-3)(0) = -3 \)
- \( C_{32} = -\begin{vmatrix} -4 & -3 \\ 1 & 1 \end{vmatrix} = -((-4)(1) - (-3)(1)) = -(-4 + 3) = 1 \)
- \( C_{33} = \begin{vmatrix} -4 & -3 \\ 1 & 0 \end{vmatrix} = (-4)(0) - (-3)(1) = 3 \)

- Form the cofactor matrix:
\[ C = \begin{pmatrix} -4 & 1 & 4 \\ -3 & 0 & 1 \\ -3 & 1 & 3 \end{pmatrix} \]

- Transpose the cofactor matrix to get the adjoint:
\[ B_1 = \text{adj}(B_0) = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 1 & 3 \end{pmatrix} \]

3. **Observe the pattern**:
From the properties of the adjoint, we find that:
\[ B_2 = \text{adj}(B_1) = B_0 \]
\[ B_3 = \text{adj}(B_2) = B_1 \]
\[ B_4 = \text{adj}(B_3) = B_0 \]
\[ B_5 = \text{adj}(B_4) = B_1 \]
\[ B_6 = \text{adj}(B_5) = B_0 \]
\[ B_7 = \text{adj}(B_6) = B_1 \]
\[ B_8 = \text{adj}(B_7) = B_0 \]
\[ B_9 = \text{adj}(B_8) = B_1 \]

Therefore, we can conclude:
- \( B_1 = B_1 \)
- \( B_3 = B_1 \)
- \( B_5 = B_1 \)
- \( B_7 = B_1 \)
- \( B_9 = B_1 \)

4. **Calculate the sum**:
\[ B_1 + B_3 + B_5 + B_7 + B_9 = 5B_1 \]

5. **Final Result**:
Since \( B_1 = B_0 \), we have:
\[ B_1 + B_3 + B_5 + B_7 + B_9 = 5B_0 \]