# 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 ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ be the set of 2 x 2 matrices with the entries in $\mathbb{Z}/\mathbb{6}\mathbb{Z}$
a) Can you find a matrix ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ whose determinant is non-zero and yet is not invertible?
b) Does the set of invertible matrices in ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ 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 $A={M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ whose determinant is non-zero and yet it is not invertible. Then the matrix A is:
$A=\left[\begin{array}{cc}3& \beta 3\\ \beta 5& 5\end{array}\right]$
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:
$A=\left[\begin{array}{cc}3& \beta 3\\ \beta 5& 5\end{array}\right]$
$A=\left[\begin{array}{cc}3& \beta 3\\ \beta 5& 5\end{array}\right]$
$=\left[\begin{array}{cc}3& 3\\ 1& 5\end{array}\right]\begin{array}{c}\beta 3\beta  3mod\left(6\right)\\ \beta 5\beta  1mod\left(6\right)\end{array}$
Step 3
This shows that, the determinant of matrix A is 12 which is non- zero but itself matrix A is not invertible.
$|A|=|\begin{array}{cc}3& 3\\ 1& 5\end{array}|$
$=\left(15\beta 3\right)$
$=12$

Step 4
(b)
yes, the set of invertible matrices in ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ forms a group because it satisfies the following property of group:
1)For any two matrices $A,B\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then $A\beta  B\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then it satisfies closure property.
2) $\left(A\beta  B\right)\beta  C=A\beta  \left(B\beta  C\right)$, for any three matrices $A,B,C\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ then it satisfies the associative property.
Step 5
Identity: There exist an identity element κ² should belong to the set of matrices ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ for all matrix $A\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ so we have $A\beta  I=I\beta  A=I$ 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 ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ contains the invertible matrices. So that the inverse of every matrix exists. This implies that for every $A\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ there exist ${A}^{\beta 1}\beta {M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ such that $A\beta  {A}^{\beta 1}={A}^{\beta 1}\beta  A=I$ then,
Step 7
This shows that the set of invertible matrices in ${M}_{2\Gamma 2}\left(\mathbb{Z}/\mathbb{6}\mathbb{Z}\right)$ satisfies the all properties of a group. Hence, it forms a group.

Jeffrey Jordon