Question

Va rog de aici sa ma ajutati cu 30 c) si cu 31 d), va rog mult chiar am nevoie acum urgent!!!​

Answer 1

Explicație pas cu pas:

$A^{2} = \left(\begin{array}{ccc}-1&0\\0&2\end{array}\right)\left(\begin{array}{ccc}-1&0\\0&2\end{array}\right) = \left(\begin{array}{ccc} {( - 1)}^{2} &0\\0&2^{2}\end{array}\right)$

$A^{3} = A \cdot A^{2} = \left(\begin{array}{ccc} - 1 &0\\0&2\end{array}\right)\left(\begin{array}{ccc} {( - 1)}^{2} &0\\0&2^{2}\end{array}\right)= \left(\begin{array}{ccc}(-1)^{3}&0\\0&2^{3}\end{array}\right)$

$A^{4} = A \cdot A^{3} = \left(\begin{array}{ccc} - 1 &0\\0&2\end{array}\right) \left(\begin{array}{ccc} {( - 1)}^{3} &0\\0&2^{3}\end{array}\right) = \left(\begin{array}{ccc}{( - 1)}^{4}&0\\0&2^{4}\end{array}\right)$

=> presupunem P(n) adevărată

$A^{n} = \left(\begin{array}{ccc}(-1)^{n}&0\\0&2^{n}\end{array}\right)$

și demonstrăm că P(n+1) este adevărată:

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

q.e.d.