 # Let \alpha = (1 7 3)(5 4 2 9)\ and\ Joseph Krupa 2021-12-10 Answered
Let . Then, ${\alpha }^{2}{\beta }^{-1}$ is
$\left(12794\right)\left(385\right)$
$\left(12749\right)\left(358\right)$
$\left(12794\right)\left(358\right)$
$\left(12974\right)\left(358\right)$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it scoollato7o
Step 1
elfa $=\left(173\right)\left(5429\right)$
${\left(elfa\right)}^{2}=\left(173\right)\left(5429\right).\left(173\right)\left(5429\right)$
that is (137)(52)(49) and beeta inverse will be (32)(47)(815)
that is ${\left(elfa\right)}^{2}.{\left(\beta \right)}^{-1}=\left(137\right)\left(52\right)\left(49\right).\left(32\right)\left(47\right)\left(815\right)$
(12794)(358)
Step 2
so here option (b) is true
###### Not exactly what you’re looking for? Melinda McCombs
$\alpha =\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 7& 9& 7& 2& 4& 7& 8& 9& 10& 1\end{array}\right]$
$\beta =\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 8& 3& 2& 7& 1& 7& 4& 5& 10& 1\end{array}\right]$
${\alpha }^{2}=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 7& 9& 1& 2& 4& 7& 3& 9& 5& 1\end{array}\right]$
$\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 7& 9& 1& 2& 4& 7& 3& 9& 5& 1\end{array}\right]$
${\alpha }^{2}=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 3& 5& 7& 9& 2& 3& 1& 5& 4& 7\end{array}\right]$
${\beta }^{1}=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 8& 3& 2& 7& 1& 7& 4& 5& 10& 1\end{array}\right]$
${\beta }^{1}=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 5& 2& 3& 4& 8& 4& 7& 1& 1& 8\end{array}\right]$
${\alpha }^{2}{\beta }^{1}=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 3& 5& 7& 9& 2& 3& 1& 5& 4& 7\end{array}\right]$
$x=\left[\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 5& 2& 3& 4& 5& 4& 7& 1& 1& 8\end{array}\right]$
$⇒\left[\begin{array}{cccccccc}1& 2& 3& 4& 5& 7& 9& 8\\ 2& 7& 8& 1& 3& 9& 4& 5\end{array}\right]$