To prove that $G$ is a group with respect to matrix multiplication, we need to show that the set $G$ satisfies the four group axioms: closure, associativity, existence of an identity element, and existence of inverses. The set $G$ is defined as:

$G = \left\{ A = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix} \mid a, b \in \mathbb{R}, \det(A) \neq 0 \right\}.$

Let's analyze each property:

1. Closure

To show closure under multiplication, take two arbitrary elements $A, B \in G$ where:

$A = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} c & d i \\ d i & c \end{pmatrix}.$

The product $A \cdot B$ is:

$A \cdot B = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix} \begin{pmatrix} c & d i \\ d i & c \end{pmatrix}.$

Calculating the elements of $A \cdot B$:

• The element at position (1,1): $a \cdot c + (b i) \cdot (d i) = a c - b d$.
• The element at position (1,2): $a \cdot (d i) + (b i) \cdot c = (a d + b c) i$.
• The element at position (2,1): $(b i) \cdot c + a \cdot (d i) = (b c + a d) i$.
• The element at position (2,2): $(b i) \cdot (d i) + a \cdot c = a c - b d$.

Thus,

$A \cdot B = \begin{pmatrix} a c - b d & (a d + b c) i \\ (a d + b c) i & a c - b d \end{pmatrix}.$

Since $a c - b d$ and $a d + b c$ are real numbers, the resulting matrix is of the same form as the elements in $G$. Therefore, $A \cdot B \in G$, proving closure.

2. Associativity

Matrix multiplication is associative, so $(A \cdot B) \cdot C = A \cdot (B \cdot C)$ for all $A, B, C \in G$. This satisfies the associativity requirement.

3. Identity Element

To find an identity element, we need a matrix $I \in G$ such that $A \cdot I = I \cdot A = A$ for all $A \in G$. The identity matrix for multiplication is:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$

This matrix has the required form with $a = 1$ and $b = 0$, and it has a non-zero determinant. Therefore, $I \in G$, and it acts as the identity element.

4. Inverses

For each $A \in G$, we need to find an inverse $A^{-1} \in G$ such that $A \cdot A^{-1} = A^{-1} \cdot A = I$. The inverse of a matrix

$A = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix}$

exists if $\det(A) \neq 0$. The determinant is calculated as:

$\det(A) = a^2 - (b i)^2 = a^2 + b^2.$

Since $\det(A) \neq 0$ (by definition of $G$), the matrix $A$ is invertible. The inverse matrix $A^{-1}$ will also be of the same form and thus belong to $G$.

Conclusion

Since $G$ satisfies closure, associativity, the existence of an identity element, and the existence of inverses, $G$ is a group under matrix multiplication.

Would you like more details on any step, or do you have specific questions about this proof?

Further Questions:

1. How do we find the explicit form of the inverse of matrices in $G$?
2. What happens to the structure of $G$ if we include matrices with $\det(A) = 0$?
3. Can you show that $G$ is also closed under addition, making it a ring structure?
4. How does the determinant condition $\det(A) \neq 0$ affect the group properties?
5. Is there a geometric interpretation of the matrices in $G$ regarding transformations?

Tip

When proving that a set forms a group, verifying the identity and inverse elements is crucial, as these are often where unique properties of the structure emerge.