[tex]\frac{\alpha(1)}{1}=\frac{\alpha(2)}{2}=...=\frac{\alpha(n)}{n}=\frac{\alpha(1)+\alpha(2)+...+\alpha(n)}{1+2+...+n}=1,
\\ \text{Deoarece }\alpha(1)+\alpha(2)+...+\alpha(n)=1+2+...+n\\
\text{Rezulta asadar, }\\
\alpha(1)=1\cdot1=1,\\
\alpha(2)=2\cdot1=2,\\
...\\
\alpha(n)=n\cdot1=n\\
\text{Deci }\\
\alpha=\left(\begin{array}{cccc}1&2&...&n\\1&2&...&n\end{array}\right)\\
\text{Pentru $\beta$ se procedeaza analog si se gaseste: }\\
[/tex]
[tex]\beta=\left(\begin{array}{cccc}1&2&...&n\\n&n-1&...&1\end{array}\right)\\[/tex]
