Question

cum calculez X^5=A, unde A este o matrice de forma 2 2
1 1
?

Answer 1

$X^5 = A \\ \\ A= \left(\begin{array}{cc}2&2\\1&1\end{array}\right) \\ A^2 = \left(\begin{array}{cc}2&2\\1&1\end{array}\right)\cdot \left(\begin{array}{cc}2&2\\1&1\end{array}\right) = \left(\begin{array}{cc}4+2&4+2\\2+1&2+1\end{array}\right) = \left(\begin{array}{cc}6&6\\3&3\end{array}\right) = \\ \\ ~~~~=3\cdot \left(\begin{array}{cc}2&2\\1&1\end{array}\right) = 3\cdot A$

$A^2 = 3\cdot A\Big|\cdot A^3 \\ A^2\cdot A^3 = 3\cdot A\cdot A^3 \\ A^5 = 3\cdot A^4 = 3\cdot A^2\cdot A^2 = 3\cdot 3\cdot A\cdot 3\cdot A = 27\cdot A^2 = 27\cdot 3\cdot A = 81\cdot A \\ \\ A^5 = 81\cdot A\Big|:81 \\ \dfrac{A^5}{81} = A \\ \\ \Big(\dfrac{A}{\sqrt[5]{81}}}\Big)^5 = A \\ \\ \Rightarrow X = \dfrac{A}{\sqrt[5]{81}}} = \dfrac{1}{\sqrt[5]{81}}}\cdot A \Rightarrow \boxed{X = \dfrac{1}{\sqrt[5]{81}}\cdot \left(\begin{array}{cc}2&2\\1&1\end{array}\right)}}$