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 find the sum \( B_1 + B_3 + B_5 + B_7 + B_9 \) where \( B_n \) is defined as the adjoint of \( B_{n-1} \) and \( B_0 \) is given as:

\[ B_0 = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \]

### Step 1: Calculate \( B_1 \)

We start by calculating \( B_1 \), which is defined as \( B_1 = \text{adj}(B_0) \).

To find the adjoint of a matrix, we need to calculate the cofactor matrix and then take its transpose.

#### Cofactor Calculation

1. **Cofactor of \( -4 \)** (Element at (1,1)):
- Minor matrix after removing the first row and first column:
\[ \begin{pmatrix} 0 & 1 \\ 4 & 3 \end{pmatrix} \]
- Determinant: \( (0)(3) - (1)(4) = -4 \)
- Cofactor: \( (-1)^{1+1} \cdot (-4) = -4 \)

2. **Cofactor of \( -3 \)** (Element at (1,2)):
- Minor matrix:
\[ \begin{pmatrix} 1 & 1 \\ 4 & 3 \end{pmatrix} \]
- Determinant: \( (1)(3) - (1)(4) = -1 \)
- Cofactor: \( (-1)^{1+2} \cdot (-1) = 1 \)

3. **Cofactor of \( -3 \)** (Element at (1,3)):
- Minor matrix:
\[ \begin{pmatrix} 1 & 0 \\ 4 & 4 \end{pmatrix} \]
- Determinant: \( (1)(4) - (0)(4) = 4 \)
- Cofactor: \( (-1)^{1+3} \cdot 4 = 4 \)

Continuing this process for all elements, we find the cofactor matrix:

\[ \text{Cofactor}(B_0) = \begin{pmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{pmatrix} \]

#### Transpose of Cofactor Matrix

Now we take the transpose of the cofactor matrix to get \( B_1 \):

\[ B_1 = \text{adj}(B_0) = \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} \]

### Step 2: Calculate \( B_2 \)

Next, we calculate \( B_2 \):

\[ B_2 = \text{adj}(B_1) = B_0 \]

### Step 3: Generalize for \( B_n \)

Continuing this process, we find that:

- \( B_3 = \text{adj}(B_2) = B_1 = B_0 \)
- \( B_4 = \text{adj}(B_3) = B_2 = B_0 \)
- \( B_5 = \text{adj}(B_4) = B_3 = B_0 \)
- \( B_6 = \text{adj}(B_5) = B_4 = B_0 \)
- \( B_7 = \text{adj}(B_6) = B_5 = B_0 \)
- \( B_8 = \text{adj}(B_7) = B_6 = B_0 \)
- \( B_9 = \text{adj}(B_8) = B_7 = B_0 \)

### Step 4: Calculate the Final Sum

Now we can calculate the sum:

\[ B_1 + B_3 + B_5 + B_7 + B_9 = B_0 + B_0 + B_0 + B_0 + B_0 = 5B_0 \]

### Step 5: Final Result

Calculating \( 5B_0 \):

\[ 5B_0 = 5 \begin{pmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{pmatrix} = \begin{pmatrix} -20 & -15 & -15 \\ 5 & 0 & 5 \\ 20 & 20 & 15 \end{pmatrix} \]

Thus, the final answer is:

\[ \begin{pmatrix} -20 & -15 & -15 \\ 5 & 0 & 5 \\ 20 & 20 & 15 \end{pmatrix} \]