Question

Find the inverse of the given group element.

(a) $\mathbf{\left(}\begin{array}{cc}\mathbf{2}& \mathbf{0}\\ \mathbf{2}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}in\hspace{0.17em}}}{\mathit{\mathbb{Z}}}_{\mathbf{3}}$

(b) $\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{2}\\ \mathbf{3}& \mathbf{4}\end{array}\mathbf{\right)}\mathbf{\text{\hspace{0.17em}in\hspace{0.17em}}}{\mathit{\mathbb{Z}}}_{\mathbf{5}}$

(c)$\left(\begin{array}{cc}\mathbf{3}& \mathbf{5}\\ \mathbf{4}& \mathbf{6}\end{array}\right)\mathbf{\text{\hspace{0.17em}in\hspace{0.17em}}}{\mathit{\mathbb{Z}}}_{\mathbf{7}}$

Short answer

(a) The required inverse is$\left(\begin{array}{cc}2& 0\\ 2& 1\end{array}\right)$.

(b) The required inverse is$\left(\begin{array}{cc}3& 1\\ 4& 2\end{array}\right)$.

(c) The required inverse is$\left(\begin{array}{cc}4& 6\\ 2& 2\end{array}\right)$

Step-by-step solution

Solution to Part (a) 

First find,$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

Thus,, $\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$which is in ${\mathbb{Z}}_3$.

Now, write the system of linear equation in${\mathbb{Z}}_3$.

$\{\begin{array}{l}2a=1\\ 2b=0\\ 2a+c=0\\ 2b+d=1\end{array}$

We get,.$a=2,b=0,c=2,d=1$

Therefore, the inverse of the given group element localid="1659416074770" $\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$in${\mathbb{Z}}_3$is as follows:

localid="1659416090335" $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 2& 1\end{array}\right)$

Hence, the required inverse is.localid="1659416100468" $\left(\begin{array}{cc}2& 0\\ 2& 1\end{array}\right)$

Solution to Part (b) 

First find,$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

Thus,$\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, which is in ${\mathbb{Z}}_5$.

Now, write the system of linear equation in ${\mathbb{Z}}_5$.

$\{\begin{array}{l}a+2c=1\\ b+2d=0\\ 3a+4c=0\\ 3b+4d=1\end{array}$

We get$a=3,b=1,c=4,d=2$,.

Therefore, the inverse of the given group elementlocalid="1659416304787" $\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$in${\mathbb{Z}}_5$is as follows:

localid="1659416295649" $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}3& 1\\ 4& 2\end{array}\right)$

Hence, the required inverse islocalid="1659416314711" $\left(\begin{array}{cc}3& 1\\ 4& 2\end{array}\right)$.

Solution to Part (c) 

First find,$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

Thus, $\left(\begin{array}{cc}3& 5\\ 4& 6\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$which is in${\mathbb{Z}}_7$.

Now, write the system of linear equation in${\mathbb{Z}}_7$.

$\{\begin{array}{l}3a+5c=1\\ 3b+5d=0\\ 4a+6c=0\\ 4b+6d=1\end{array}$

We get$a=3,b=1,c=4,d=2$,.

Therefore, the inverse of the given group elementlocalid="1659416364951" $\left(\begin{array}{cc}3& 5\\ 4& 6\end{array}\right)$in${\mathbb{Z}}_7$is as follows:

localid="1659416385206" $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}4& 6\\ 2& 2\end{array}\right)$

Hence, the required inverse is .localid="1659416395686" $\left(\begin{array}{cc}4& 6\\ 2& 2\end{array}\right)$