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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.

Show that G is a group with respect to matrix multiplication, given the set:
\[
G = \left\{ A = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix} \mid a, b \in \mathbb{R}, \det(A) \neq 0 \right\}.
\]


Explore the proof that the set G of 2x2 matrices of the form \( A = \begin{pmatrix} a & b i \\ b i & a \end{pmatrix} \), with real elements and non-zero determinant, forms a group under matrix multiplication. This includes checking closure, associativity, identity, and inverse elements, key concepts in group theory and linear algebra.

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