Question

..........................​

Answer 1

Explicație pas cu pas:

$A^{2} = \left(\begin{array}{cc}1&0\\a-1&a\end{array}\right) \cdot \left(\begin{array}{cc}1&0\\a-1&a\end{array}\right) = \\ = \left(\begin{array}{cc}1+0&0+0\\a-1+a(a-1)&0+a \cdot a\end{array}\right) = \left(\begin{array}{cc}1&0\\a^{2}-1&a^{2}\end{array}\right)$

presupunem că este adevărat pentru k și demonstrăm pentru k + 1:

$A^{k} \cdot A = \left(\begin{array}{cc}1&0\\a^{k}-1&a^{k}\end{array}\right) \cdot \left(\begin{array}{cc}1&0\\a-1&a\end{array}\right) = \\ = \left(\begin{array}{cc}1+0&0+0\\a^{k}-1+a^{k}(a-1)&0+a^{k} \cdot k\end{array}\right) = \left(\begin{array}{cc}1&0\\a^{k+1}-1&a^{k}\end{array}\right)$

=>

$A^{n} = \left(\begin{array}{cc}1&0\\a^{n}-1&a^{n}\end{array}\right) \iff A^{n} = \left(\begin{array}{cc}1&0\\x_{n}-1&x_{n}\end{array}\right)$

$\implies (x_{n}) = a^{n}$

$B = A \cdot A^{2} \cdot A^{3} \cdot A^{4} \cdot ... \cdot A^{n} = A^{1 \cdot 2} \cdot A^{3} \cdot A^{4} \cdot ... \cdot A^{n} = A^{2!} \cdot A^{3} \cdot A^{4} \cdot ... \cdot A^{n} = A^{3!} \cdot A^{4} \cdot ... \cdot A^{n} = A^{4!} \cdot ... \cdot A^{n} = ... = A^{n!}$

$\implies B = \left(\begin{array}{cc}1&0\\a^{n!}-1&a^{n!}\end{array}\right)$