10. Prove that the set of matrices of the form $$ \left(\begin{array}{lll}1 & x & y \ 0 & 1 & z \ 0 & 0 & 1\end{array}\right) $$ is a group under matrix multiplication. This group, known as the Heisenberg group, is important in quantum physics. Matrix multiplication in the Heisenberg group is defined by $$ \left(\begin{array}{lll}1 & x & y \ 0 & 1 & z \ 0 & 0 & 1\end{array}\right)\left(\begin{array}{lll}1 & x^{\prime} & y^{\prime} \ 0 & 1 & z^{\prime} \ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & x+x^{\prime} & y+y^{\prime}+x z^{\prime} \ 0 & 1 & z+z^{\prime} \ 0 & 0 & 1\end{array}\right) $$

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To prove that the set of matrices of the form

$$
H = \left\{ \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right) : x, y, z \in \mathbb{R} \right\}
$$

is a group under matrix multiplication, we need to verify the following group properties:

1. **Closure**: For any two matrices in $$ H $$, their product must also be in $$ H $$.
2. **Associativity**: The operation of matrix multiplication must be associative.
3. **Identity Element**: There must be an identity element in $$ H $$.
4. **Inverses**: Every element in $$ H $$ must have an inverse in $$ H $$.

### Step 1: Closure

Let

$$
A = \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right) \quad \text{and} \quad B = \left(\begin{array}{ccc} 1 & x' & y' \ 0 & 1 & z' \ 0 & 0 & 1 \end{array}\right)
$$

be two matrices in $$ H $$. We compute their product $$ AB $$:

$$
AB = \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & x' & y' \ 0 & 1 & z' \ 0 & 0 & 1 \end{array}\right)
$$

Using the given multiplication rule, we find:

$$
AB = \left(\begin{array}{ccc} 1 & x + x' & y + y' + x z' \ 0 & 1 & z + z' \ 0 & 0 & 1 \end{array}\right)
$$

Since $$ x + x' $$, $$ y + y' + x z' $$, and $$ z + z' $$ are all real numbers, $$ AB $$ is also of the form in $$ H $$. Thus, $$ H $$ is closed under multiplication.

### Step 2: Associativity

Matrix multiplication is associative in general. Therefore, for any three matrices $$ A, B, C \in H $$:

$$
(A B) C = A (B C)
$$

This property holds for all matrices, so it holds for matrices in $$ H $$ as well.

### Step 3: Identity Element

The identity element in matrix multiplication is the identity matrix:

$$
I = \left(\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)
$$

This can be represented in the form of matrices in $$ H $$ as:

$$
E = \left(\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)
$$

For any matrix $$ A \in H $$:

$$
AE = A \quad \text{and} \quad EA = A
$$

Thus, $$ E $$ is the identity element in $$ H $$.

### Step 4: Inverses

To find the inverse of a matrix $$ A \in H $$:

$$
A = \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right)
$$

We need to find a matrix $$ B \in H $$ such that $$ AB = E $$. Let $$ B $$ be:

$$
B = \left(\begin{array}{ccc} 1 & -x & y' \ 0 & 1 & -z \ 0 & 0 & 1 \end{array}\right)
$$

We need to determine $$ y' $$ such that:

$$
AB = \left(\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)
$$

Calculating $$ AB $$:

$$
AB = \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & -x & y' \ 0 & 1 & -z \ 0 & 0 & 1 \end{array}\right)
$$

Using the multiplication rule:

$$
AB = \left(\begin{array}{ccc} 1 & 0 & y + y' - x z \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right)
$$

To satisfy $$ AB = E $$, we need:

$$
y + y' - x z = 0 \implies y' = x z - y
$$

Thus, the inverse of $$ A $$ is:

$$
A^{-1} = \left(\begin{array}{ccc} 1 & -x & x z - y \ 0 & 1 & -z \ 0 & 0 & 1 \end{array}\right)
$$

Since $$ -x $$, $$ -z $$, and $$ x z - y $$ are real numbers, $$ A^{-1} $$ is also in $$ H $$.

### Conclusion

Since all four group properties are satisfied, we conclude that the set of matrices of the form

$$
\left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right)
$$

is indeed a group under matrix multiplication, known as the Heisenberg group.
The set of matrices of the form $$ \left(\begin{array}{ccc} 1 & x & y \ 0 & 1 & z \ 0 & 0 & 1 \end{array}\right) $$ forms a group under matrix multiplication, known as the Heisenberg group. This group satisfies the closure, associativity, identity element, and inverse properties required for a group structure.

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