If `A=[{:(2,3),(-1,-2):}] and B=sum_(r=1)^(10)A^r` , then the value of det (B)is equal to

Answer

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 \( A^2 \)
We start by calculating \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \]
Calculating the elements:
- 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:
\[ A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

**Hint for Step 1:** Use matrix multiplication rules carefully to compute each element of the resulting matrix.

### Step 2: Identify the pattern for higher powers of \( A \)
Since \( A^2 = I \), we can deduce:
- \( A^3 = A^2 \cdot A = I \cdot A = A \)
- \( A^4 = A^2 \cdot A^2 = I \cdot I = I \)
- \( A^5 = A^4 \cdot A = I \cdot A = A \)
- Continuing this pattern, we find:
- \( A^{2n} = I \) for even \( n \)
- \( A^{2n+1} = A \) for odd \( n \)

### Step 3: Calculate \( B \)
Now we can calculate \( B \):
\[ B = A + A^2 + A^3 + A^4 + A^5 + A^6 + A^7 + A^8 + A^9 + A^{10} \]
Substituting the powers:
- Odd powers (1, 3, 5, 7, 9): \( 5A \)
- Even powers (2, 4, 6, 8, 10): \( 5I \)

Thus,
\[ B = 5A + 5I = 5(A + I) \]

### 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 the determinant of \( B \)
Now we can find the determinant of \( B \):
\[ B = 5 \begin{pmatrix} 3 & 3 \\ -1 & -1 \end{pmatrix} \]
The determinant of a scalar multiple of a matrix is the scalar raised to the power of the dimension of the matrix times the determinant of the matrix:
\[ \det(B) = 5^2 \cdot \det\begin{pmatrix} 3 & 3 \\ -1 & -1 \end{pmatrix} \]
Calculating the determinant:
\[ \det\begin{pmatrix} 3 & 3 \\ -1 & -1 \end{pmatrix} = (3 \cdot -1) - (3 \cdot -1) = -3 + 3 = 0 \]
Thus,
\[ \det(B) = 25 \cdot 0 = 0 \]

### Final Answer
The value of \( \det(B) \) is \( 0 \).

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