We need to determine whether the given set of $3 \times 3$ matrices forms an abelian group under matrix multiplication. The set $G$ is given as:

$G = \left\{ \begin{bmatrix} 1 & x & x \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix} \mid x, y \in \mathbb{R} \right\}$

Step 1: Check Group Properties

A set forms a group under matrix multiplication if it satisfies the following properties:

1. Closure: If $A, B \in G$, then their product $AB$ must also be in $G$.
2. Identity: There must exist an identity element $I \in G$.
3. Inverse: Each element must have an inverse in $G$.
4. Associativity: Matrix multiplication is always associative.

Closure Check

Take two matrices from $G$:

$A = \begin{bmatrix} 1 & x_1 & x_1 \\ 0 & 1 & y_1 \\ 0 & 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & x_2 & x_2 \\ 0 & 1 & y_2 \\ 0 & 0 & 1 \end{bmatrix}$

Compute their product:

$AB = \begin{bmatrix} 1 & x_1 & x_1 \\ 0 & 1 & y_1 \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & x_2 & x_2 \\ 0 & 1 & y_2 \\ 0 & 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 1 & x_1 + x_2 & x_1 + x_2 \\ 0 & 1 & y_1 + y_2 \\ 0 & 0 & 1 \end{bmatrix}$

Since this matrix is still in $G$ (the form remains the same), closure holds.

Identity Element

The identity matrix:

$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

is clearly in $G$ when $x = 0, y = 0$, so the identity exists.

Inverse Element

For an element:

$A = \begin{bmatrix} 1 & x & x \\ 0 & 1 & y \\ 0 & 0 & 1 \end{bmatrix}$

we need to find $A^{-1}$. Using matrix inversion formulas, the inverse is:

$A^{-1} = \begin{bmatrix} 1 & -x & -x + xy \\ 0 & 1 & -y \\ 0 & 0 & 1 \end{bmatrix}$

which is still in $G$. So inverses exist.

Associativity

Matrix multiplication is associative, so this holds.

Step 2: Check for Commutativity

For $G$ to be abelian, matrix multiplication must be commutative, meaning $AB = BA$ for all $A, B \in G$.

Using the earlier matrices $A$ and $B$, we compute:

$BA = \begin{bmatrix} 1 & x_2 + x_1 & x_2 + x_1 \\ 0 & 1 & y_2 + y_1 \\ 0 & 0 & 1 \end{bmatrix}$

Since matrix multiplication is not necessarily commutative, we check whether $AB = BA$:

$\begin{bmatrix} 1 & x_1 + x_2 & x_1 + x_2 \\ 0 & 1 & y_1 + y_2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & x_2 + x_1 & x_2 + x_1 \\ 0 & 1 & y_2 + y_1 \\ 0 & 0 & 1 \end{bmatrix}$

Since the sum of real numbers is commutative ($x_1 + x_2 = x_2 + x_1$ and $y_1 + y_2 = y_2 + y_1$), we see that $AB = BA$ always holds.

Conclusion

Since $G$ satisfies all group properties and matrix multiplication within $G$ is commutative, $G$ is an abelian group under matrix multiplication.

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Do you want further details or have any questions?

Here are five related questions to explore:

1. How do you compute the inverse of a general $3 \times 3$ matrix?
2. What happens if we modify the definition of $G$ to allow nonzero values in other positions?
3. Can we generalize this concept to $n \times n$ matrices?
4. How does this group relate to upper triangular or lower triangular matrices?
5. What are some real-world applications of groups of matrices?

Tip: Always check if matrix multiplication is commutative before assuming a group is abelian!