Completed math worksheet demonstrating the correct application of PEMDAS rules to solve ten different algebraic expressions.

It seems you've uploaded an image containing a mathematical problem related to the Center of Special Linear Group $ \text{SL}(n, \mathbb{Z}) $, specifically focusing on determining its center. The solution involves group theory and linear algebra concepts. Below, I will carefully analyze and solve the problem step by step.

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Problem Statement

The task is to determine the center of the group $ \text{SL}(n, \mathbb{Z}) $, which consists of all $ n \times n $ matrices with integer entries and determinant 1. The center of a group $ G $, denoted $ Z(G) $, is defined as:
\[
Z(G) = \{ g \in G \mid \forall h \in G, gh = hg \}.
\]
In this case, we are tasked with finding $ Z(\text{SL}(n, \mathbb{Z})) $.

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Solution

#### Step 1: Understanding the Group $ \text{SL}(n, \mathbb{Z}) $
- $ \text{SL}(n, \mathbb{Z}) $ is the group of $ n \times n $ matrices with integer entries and determinant 1.
- For a matrix $ A \in \text{SL}(n, \mathbb{Z}) $, we have $ \det(A) = 1 $.
- The center $ Z(\text{SL}(n, \mathbb{Z})) $ consists of all matrices $ A \in \text{SL}(n, \mathbb{Z}) $ that commute with every matrix in $ \text{SL}(n, \mathbb{Z}) $, i.e., $ AB = BA $ for all $ B \in \text{SL}(n, \mathbb{Z}) $.

#### Step 2: Key Properties of the Center
1. Commutativity: If $ A \in Z(\text{SL}(n, \mathbb{Z})) $, then $ A $ must commute with every matrix in $ \text{SL}(n, \mathbb{Z}) $.
2. Scalar Matrices: In general, the center of $ \text{SL}(n, k) $ (where $ k $ is a field) consists of scalar matrices of the form $ \lambda I_n $, where $ \lambda \in k $ and $ \det(\lambda I_n) = 1 $. However, since we are working over $ \mathbb{Z} $, the only possible scalars are integers.

#### Step 3: Analyzing Scalar Matrices in $ \text{SL}(n, \mathbb{Z}) $
- A scalar matrix in $ \text{SL}(n, \mathbb{Z}) $ has the form $ \lambda I_n $, where $ \lambda \in \mathbb{Z} $ and $ \det(\lambda I_n) = 1 $.
- The determinant of $ \lambda I_n $ is given by:
\[
\det(\lambda I_n) = \lambda^n.
\]
- For $ \lambda I_n \in \text{SL}(n, \mathbb{Z}) $, we require:
\[
\lambda^n = 1.
\]
- Since $ \lambda \in \mathbb{Z} $, the only integer solutions to $ \lambda^n = 1 $ are $ \lambda = 1 $ and $ \lambda = -1 $ (for even $ n $).

#### Step 4: Verifying Commutativity
- For $ \lambda = 1 $, the matrix $ I_n $ (the identity matrix) clearly commutes with every matrix in $ \text{SL}(n, \mathbb{Z}) $.
- For $ \lambda = -1 $, the matrix $ -I_n $ also commutes with every matrix in $ \text{SL}(n, \mathbb{Z}) $ because:
\[
(-I_n)B = -B = B(-I_n) \quad \text{for all } B \in \text{SL}(n, \mathbb{Z}).
\]

#### Step 5: Conclusion
- The center $ Z(\text{SL}(n, \mathbb{Z})) $ consists of the matrices $ I_n $ and $ -I_n $ if $ n $ is even. If $ n $ is odd, then $ -I_n \notin \text{SL}(n, \mathbb{Z}) $ because $ \det(-I_n) = (-1)^n = -1 $, so the center is just $ \{ I_n \} $.

Thus, the center of $ \text{SL}(n, \mathbb{Z}) $ is:
\[
Z(\text{SL}(n, \mathbb{Z})) =
\begin{cases}
\{ I_n \}, & \text{if } n \text{ is odd}, \\
\{ I_n, -I_n \}, & \text{if } n \text{ is even}.
\end{cases}
\]

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Final Answer

\[
\boxed{\{ I_n \} \text{ if } n \text{ is odd, and } \{ I_n, -I_n \} \text{ if } n \text{ is even}}
\]