# Question # Let M_{2 times 2} (mathbb{Z}/mathbb{6Z}) be the set of 2 x 2 matrices with the entries in mathbb{Z}/mathbb{6Z} a) Can you find a matrix M_{2 times 2}

Matrices
ANSWERED Let $$M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})$$ be the set of 2 x 2 matrices with the entries in $$\mathbb{Z}/\mathbb{6Z}$$
a) Can you find a matrix $$M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})$$ whose determinant is non-zero and yet is not invertible?
b) Does the set of invertible matrices in $$M_{2 \times 2} (\mathbb{Z}/\mathbb{6Z})$$ form a group? 2021-02-09
Step 1
(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.