Hw7 Problem 1

### Problem 1

Let $\alpha =\begin{pmatrix} 1&2&3&4&5&6\\ 2&1&3&5&4&6\end{pmatrix}\ \ \ \ \beta=\begin{pmatrix} 1&2&3&4&5&6\\6&1&2&4&3&5\end{pmatrix}$ Compute the following
(a) $\alpha^{-1}$
(b) $\alpha\beta$
(c) $\beta\alpha$
(d) $\alpha^2$

Solution
(a) The inverse of a 1-1 onto function defines $\alpha^{-1}(y)=x \iff \alpha(x)=y$. Thus, $\alpha^{-1}=\begin{pmatrix} 1&2&3&4&5&6\\ 2&1&3&5&4&6\end{pmatrix}$

(b) $\alpha\beta=\begin{pmatrix}1&2&3&4&5&6\\ 6&2&1&5&3&4\end{pmatrix}$

(c) $\beta\alpha=\begin{pmatrix}1&2&3&4&5&6\\1&6&2&3&4&5\end{pmatrix}$

(d) $\alpha^2=\begin{pmatrix} 1&2&3&4&5&6\\ 1&2&3&4&5&6\end{pmatrix}$