Let `A=[[1, 2],[0, 1]]` and `B = l + adj(A) + (adj A)^2 + …+ (adj A)^(10)`.
Then, the sum of all the elements of the matrix B is:

Options

A. 22
B. `-88`
C. `-124`
D. `-110`

Answer

To solve the problem, we need to find the sum of all elements of the matrix \( B \), which is defined as:

\[ B = I + \text{adj}(A) + \text{adj}(A)^2 + \ldots + \text{adj}(A)^{10} \]

where \( A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \).

### Step 1: Calculate the adjoint of \( A \)

For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the adjoint is given by:

\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

Applying this to our matrix \( A \):

\[ \text{adj}(A) = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Calculate powers of the adjoint

Next, we need to compute \( \text{adj}(A)^n \) for \( n = 1, 2, \ldots, 10 \).

- **For \( n = 1 \)**:

\[ \text{adj}(A)^1 = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \]

- **For \( n = 2 \)**:

\[ \text{adj}(A)^2 = \text{adj}(A) \cdot \text{adj}(A) = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} \]

- **For \( n = 3 \)**:

\[ \text{adj}(A)^3 = \text{adj}(A)^2 \cdot \text{adj}(A) = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 0 & 1 \end{pmatrix} \]

Continuing this pattern, we can observe that for \( n \geq 1 \):

\[ \text{adj}(A)^n = \begin{pmatrix} 1 & -2n \\ 0 & 1 \end{pmatrix} \]

### Step 3: Calculate \( B \)

Now we can express \( B \):

\[ B = I + \text{adj}(A) + \text{adj}(A)^2 + \ldots + \text{adj}(A)^{10} \]

Where \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

Thus,

\[ B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sum_{n=1}^{10} \begin{pmatrix} 1 & -2n \\ 0 & 1 \end{pmatrix} \]

Calculating the sum:

1. The sum of the first components (the diagonal elements) will be:

\[ 1 + 10 = 11 \]

2. The sum of the second components (the off-diagonal elements):

\[ 0 + (-2 \cdot 1) + (-2 \cdot 2) + \ldots + (-2 \cdot 10) = -2(1 + 2 + \ldots + 10) = -2 \cdot \frac{10 \cdot 11}{2} = -110 \]

Thus, we have:

\[ B = \begin{pmatrix} 11 & -110 \\ 0 & 11 \end{pmatrix} \]

### Step 4: Find the sum of all elements of \( B \)

The sum of all elements of \( B \):

\[ 11 + (-110) + 0 + 11 = 11 - 110 + 11 = -88 \]

### Final Answer

The sum of all the elements of the matrix \( B \) is:

\[ \boxed{-88} \]