(a)

According to given condition in the first part (a), Yes, we can construct a matrix \(A=M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) whose determinant is non-zero and yet it is not invertible. Then the matrix A is:

\(A=\begin{bmatrix} 3& -3 \\ -5 & 5 \end{bmatrix}\)

Step 2

As the determinant of the matrix A is zero which is not invertible. But elements of matrix A from (\(\mathbb{Z}/\mathbb{6Z}\)). So, after applying the modulo 6, the matrix A becomes:

\(A=\begin{bmatrix} 3& -3 \\ -5 & 5 \end{bmatrix}\)

\(A=\begin{bmatrix} 3& -3 \\ -5 & 5 \end{bmatrix}\)

\(=\begin{bmatrix} 3& 3 \\ 1 & 5 \end{bmatrix} \begin{matrix} -3 \cong 3 \bmod(6) \\ -5\cong1 \bmod(6) \end{matrix}\)

Step 3

This shows that, the determinant of matrix A is 12 which is non- zero but itself matrix A is not invertible.

\(|A|=\begin{vmatrix} 3& 3 \\ 1 & 5 \end{vmatrix}\)

\(=(15-3)\)

\(=12\)

\(\neq 0\)

Step 4

(b)

yes, the set of invertible matrices in \(M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) forms a group because it satisfies the following property of group:

1)For any two matrices \(A, B \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) then \(A \cdot B \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) then it satisfies closure property.

2) \((A \cdot B) \cdot C=A \cdot (B \cdot C), for any three matrices \(A, B ,C \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) then it satisfies the associative property.

Step 5

Identity: There exist an identity element ꓲ should belong to the set of matrices \(M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) for all matrix \(A \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) so we have \(A \cdot I=I \cdot A=I\) then the identity element of the set of matrices is:

\(I=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}\)

Step 6

The existence of inverse: As the set of matrices \(M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) contains the invertible matrices. So that the inverse of every matrix exists. This implies that for every \(A \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) there exist \(A^{-1} \in M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})\) such that \(A \cdot A^{-1}=A^{-1} \cdot A=I\) then, \(A^{-1} \in M_{2 \times 2}(\mathbb{Z}/\mathbb{6Z}) \text{ for all } A \in M_{2 \times 2}(\mathbb{Z}/\mathbb{6Z})\)

Step 7

This shows that the set of invertible matrices in \(M_{2 \times 2}(\mathbb{Z}/\mathbb{6Z})\) satisfies the all properties of a group. Hence, it forms a group.