To solve the problem, we need to find the determinant of the matrix \( B \), which is defined as \( B = \sum_{r=1}^{10} A^r \), where \( A = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \).
### Step 1: Calculate \( A^2 \)
First, we need to calculate \( A^2 \):
\[
A^2 = A \cdot A = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix}
\]
Calculating the product:
- First row, first column: \( 2 \cdot 2 + 3 \cdot (-1) = 4 - 3 = 1 \)
- First row, second column: \( 2 \cdot 3 + 3 \cdot (-2) = 6 - 6 = 0 \)
- Second row, first column: \( -1 \cdot 2 + (-2) \cdot (-1) = -2 + 2 = 0 \)
- Second row, second column: \( -1 \cdot 3 + (-2) \cdot (-2) = -3 + 4 = 1 \)
Thus, we have:
\[
A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I
\]
where \( I \) is the identity matrix.
### Step 2: Determine the pattern for \( A^r \)
From the calculation of \( A^2 \), we can deduce:
- For even \( r \): \( A^{2k} = I \)
- For odd \( r \): \( A^{2k+1} = A \)
### Step 3: Calculate \( B \)
Now, we can express \( B \):
\[
B = A^1 + A^2 + A^3 + A^4 + A^5 + A^6 + A^7 + A^8 + A^9 + A^{10}
\]
Substituting the values based on the pattern:
- Odd powers (1, 3, 5, 7, 9): \( 5A \)
- Even powers (2, 4, 6, 8, 10): \( 5I \)
Thus, we can write:
\[
B = 5A + 5I = 5(A + I)
\]
### Step 4: Calculate \( A + I \)
Now we need to compute \( A + I \):
\[
A + I = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ -1 & -1 \end{pmatrix}
\]
### Step 5: Calculate \( B \)
Now substituting back into \( B \):
\[
B = 5 \begin{pmatrix} 3 & 3 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 15 & 15 \\ -5 & -5 \end{pmatrix}
\]
### Step 6: Calculate \( \text{det}(B) \)
Now we find the determinant of \( B \):
\[
\text{det}(B) = 15 \cdot (-5) - 15 \cdot (-5) = -75 + 75 = 0
\]
### Final Answer
Thus, the value of \( \text{det}(B) \) is:
\[
\boxed{0}
\]
