Question

Show that $\mathit{M}\mathbf{\left(}{\mathbf{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\right)}$(all 2 X 2 matrices with entries in ${\mathbf{\mathbb{Z}}}_{\mathbf{2}}$) is a 16-element noncommutative ring with identity.

Short answer

It is proved that$M\left({\mathrm{\mathbb{Z}}}_2\right)$ is a 16-element non-commutative ring with identity.

Step-by-step solution

Determine M(ℤ2) is a 16-element non-commutative ring with identity.

Consider the matrix,

$M\left({\mathrm{\mathbb{Z}}}_2\right)=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\overline{)a,b,c,d\in {\mathrm{\mathbb{Z}}}_2}\right\}$

Here, $\left|S\right|$= the number of elements in set. As,

$$\begin{gathered} \left|{\mathrm{\mathbb{Z}}}_2\right|=\left|\left\{0,1\right\}=2\right| \\ \left|M\left({\mathrm{\mathbb{Z}}}_2\right)\right|=\left|{\mathrm{\mathbb{Z}}}_2\right|·\left|{\mathrm{\mathbb{Z}}}_2\right|·\left|{\mathrm{\mathbb{Z}}}_2\right|·\left|{\mathrm{\mathbb{Z}}}_2\right|=2^4=16 \end{gathered}$$

The matrix $M\left({\mathrm{\mathbb{Z}}}_2\right)$ has elements and each element is or .

Obtain that ℤ2 is a ring

Let’s consider the matrix,

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$and

$\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)$is an arbitrary matrix of $M\left({\mathrm{\mathbb{Z}}}_2\right)$.

Using the conditions of a ring,

i. Matrix addition:

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)=\left(\begin{array}{cc}a+e& b+f\\ c+g& d+h\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

ii. Generally, matrix addition is associative.

iii. Now,

$\begin{array}{rcl}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)& =& \left(\begin{array}{cc}a+e& b+f\\ c+g& d+h\end{array}\right)\\ & =& \left(\begin{array}{cc}e+a& f+b\\ g+c& h+d\end{array}\right)\\ & =& \left(\begin{array}{cc}e& f\\ g& h\end{array}\right)+\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\end{array}$

iv. Consider that,

$0_{M\left({\mathrm{\mathbb{Z}}}_2\right)}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

Then,

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)+\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

v. Suppose,

$\left(\begin{array}{cc}-a& -b\\ -c& -d\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

Then,

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}-a& -b\\ -c& -d\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$

Matrix multiplication:

$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)=\left(\begin{array}{cc}ae+bg& af+bh\\ ce+dg& cf+dh\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

vi. Generally, matrix multiplication is associative.

vii. Generally, distribution law holds matrix.

From all the above conditions, it is proved that $M\left({\mathrm{\mathbb{Z}}}_2\right)$ is a ring.

Obtain that ℤ2 is identity element

Let’s consider, $1_{M\left({\mathrm{\mathbb{Z}}}_2\right)}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

Then,$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

Here,$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ is identity element in $M\left({\mathrm{\mathbb{Z}}}_2\right)$.

Now, consider the matrices,

role="math" localid="1648125457940" $\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)\in M\left({\mathrm{\mathbb{Z}}}_2\right)$

Then, notice that,

$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\ne \left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$

Form the above matrices,$M\left({\mathrm{\mathbb{Z}}}_2\right)$ is not commutative.

Therefore, it is shown that $M\left({\mathrm{\mathbb{Z}}}_2\right)$is a 16-element non-commutative ring with identity.