# 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}

Let ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ be the set of 2 x 2 matrices with the entries in $\mathbb{Z}/\mathbb{6}\mathbb{Z}$
a) Can you find a matrix ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ whose determinant is non-zero and yet is not invertible?
b) Does the set of invertible matrices in ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ form a group?
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Leonard Stokes

Step 1
(a)
According to given condition in the first part (a), Yes, we can construct a matrix $A={M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ whose determinant is non-zero and yet it is not invertible. Then the matrix A is:
$A=\left[\begin{array}{cc}3& -3\\ -5& 5\end{array}\right]$
Step 2
As the determinant of the matrix A is zero which is not invertible. But elements of matrix A from ($\mathbb{Z}/\mathbb{6}\mathbb{Z}$). So, after applying the modulo 6, the matrix A becomes:
$A=\left[\begin{array}{cc}3& -3\\ -5& 5\end{array}\right]$
$A=\left[\begin{array}{cc}3& -3\\ -5& 5\end{array}\right]$
$=\left[\begin{array}{cc}3& 3\\ 1& 5\end{array}\right]\begin{array}{c}-3\cong 3mod\left(6\right)\\ -5\cong 1mod\left(6\right)\end{array}$
Step 3
This shows that, the determinant of matrix A is 12 which is non- zero but itself matrix A is not invertible.
$|A|=|\begin{array}{cc}3& 3\\ 1& 5\end{array}|$
$=\left(15-3\right)$
$=12$
$\ne 0$
Step 4
(b)
yes, the set of invertible matrices in ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ forms a group because it satisfies the following property of group:
1)For any two matrices $A,B\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then $A\cdot B\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then it satisfies closure property.
2) $\left(A\cdot B\right)\cdot C=A\cdot \left(B\cdot C\right)$, for any three matrices $A,B,C\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then it satisfies the associative property.
Step 5
Identity: There exist an identity element ꓲ should belong to the set of matrices ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ for all matrix $A\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ so we have $A\cdot I=I\cdot A=I$ then the identity element of the set of matrices is:
$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Step 6
The existence of inverse: As the set of matrices ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ contains the invertible matrices. So that the inverse of every matrix exists. This implies that for every $A\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ there exist ${A}^{-1}\in {M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ such that $A\cdot {A}^{-1}={A}^{-1}\cdot A=I$ then,
Step 7
This shows that the set of invertible matrices in ${M}_{2×2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ satisfies the all properties of a group. Hence, it forms a group.

Jeffrey Jordon