Question

(a) Show that$\mathit{a}=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right)$ has order 3 in$\mathit{G}\mathit{L}\left(2,R\right)$ and$\mathit{b}=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ has order 4.

Short answer

It is shown that$a=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right)$ has order 3 and$b=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ has order 4.

Step-by-step solution

Determine given matrices

Consider the given matrices,

$a=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right)$has order 3 and

$b=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$has order 4.

Determine a=(01-1-1)  has order 3 and b=(0-110) has order 4

To verify this, Use computation method, thus,

$$\begin{gathered} {\left[\begin{array}{cc}0& 1\\ -1& -1\end{array}\right]}^2=\left[\begin{array}{cc}-1& -1\\ 1& 0\end{array}\right] \\ {\left[\begin{array}{cc}0& 1\\ -1& -1\end{array}\right]}^3=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Also,

$$\begin{gathered} {\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}^2=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right] \\ {\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}^3=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right] \\ {\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]}^4=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Hence,$a=\left(\begin{array}{cc}0& 1\\ -1& -1\end{array}\right)$ has order 3 and$b=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ has order 4.