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
asked 2021-02-08
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?

Answers (1)

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.
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