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The image contains a few mathematical questions related to group theory and matrices. Let's break them down:

### Question 6:
The problem asks whether the set of matrices

\[
G = \left\{ \begin{bmatrix} a & a & a \\ a & a & a \\ a & a & a \end{bmatrix} \mid a \in \mathbb{R} \setminus \{0\} \right\}
\]

forms a group under matrix multiplication, which is associative.

#### Requirements for a group:
1. **Closure**: For any two elements \( A \) and \( B \) in \( G \), the product \( AB \) should also be in \( G \).
2. **Associativity**: Matrix multiplication is already associative, so this property is naturally satisfied.
3. **Identity element**: There should exist an identity matrix \( I \) such that for any matrix \( A \in G \), \( A \cdot I = I \cdot A = A \).
4. **Inverse**: For each matrix \( A \in G \), there should exist a matrix \( B \in G \) such that \( A \cdot B = B \cdot A = I \).

#### Analysis:
- **Associativity**: As the answer says, associativity is guaranteed because matrix multiplication is associative.
- **Identity element**: The identity matrix in standard matrix multiplication is the identity matrix:

  \[
  I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
  \]

  But no matrix of the form \( \begin{bmatrix} a & a & a \\ a & a & a \\ a & a & a \end{bmatrix} \) can act as an identity matrix unless \( a = 0 \), which is not allowed because \( a \in \mathbb{R} \setminus \{0\} \). Therefore, the set \( G \) lacks an identity element.
  
- **Inverse element**: Since there is no identity element, it's also impossible to find an inverse matrix for each element in \( G \).

Thus, the set \( G \) **does not form a group** because it fails to satisfy the identity and inverse conditions.

---

### Question 7:
This seems to be related to group elements and the property of orders of elements in a group. It says:

- Let \( a \) and \( b \) be elements of a group \( G \).
- Prove that \( |a| = |bab^{-1}| \), where \( |a| \) denotes the order of the element \( a \).

#### Proof Sketch:
Let the order of \( a \) be \( n \), meaning \( a^n = e \) (where \( e \) is the identity element in the group).

1. Consider the element \( bab^{-1} \).
2. Raise \( bab^{-1} \) to the power \( n \):

   \[
   (bab^{-1})^n = ba^n b^{-1}
   \]

3. Since \( a^n = e \), we have:

   \[
   ba^n b^{-1} = beb^{-1} = bb^{-1} = e
   \]

Thus, the element \( bab^{-1} \) also has order \( n \), and therefore \( |a| = |bab^{-1}| \).

---

Do you want further details or have any specific questions on these explanations?

Here are 5 follow-up questions you might explore:
1. How can we generalize the condition for identity elements in different matrix sets?
2. What happens if the matrices in \( G \) are allowed to contain zeroes?
3. Can we consider subgroups within the set \( G \)? What are their properties?
4. How does the property \( |a| = |bab^{-1}| \) relate to conjugation in group theory?
5. What are some examples of sets that form groups under matrix multiplication?

**Tip**: When working with group theory, always check for the identity element first, as it simplifies verifying the other group properties!

Apakah himpunan G = {[a a a] [a a a] [a a a] | a ∈ R - {0}} membentuk suatu grup terhadap operasi kali matriks yang asosiatif? Jelaskan syarat-syarat yang dipenuhi dan juga syarat-syarat yang tidak dipenuhi.

Explore whether the set of matrices G = {[a a a] [a a a] [a a a] | a ∈ R - {0}} forms a group under matrix multiplication, analyzing the fulfillment of associativity, identity, and inverse element conditions. This problem is suitable for university students studying group theory and abstract algebra.

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