Răspuns :
a) [tex]X(-1)=\left(\begin{array}{cc}1&-1-1\\0&-1\end{array}\right)=\left(\begin{array}{cc}1&-2\\0&-1\end{array}\right)\\\\X^2(-1)=x(-1)\cdot X(-1)=\left(\begin{array}{cc}1&-2\\0&-1\end{array}\right)\cdot\left(\begin{array}{cc}1&-2\\0&-1\end{array}\right)=\left(\begin{array}{cc}1&-4\\0&1\end{array}\right)\\\\det(X^2(-1))=\left|\begin{array}{cc}1&-4\\0&1\end{array}\right|=1[/tex]
b) [tex]X(a)\cdot X(b)=\left(\begin{array}{cc}1&a-1\\0&a\end{array}\right)\cdot\left(\begin{array}{cc}1&b-1\\0&b\end{array}\right)=\left(\begin{array}{cc}1&b-1+ab+b\\0&ab\end{array}\right)=\left(\begin{array}{cc}1&ab-1\\0&ab\end{array}\right)=X(ab), \forall a, b \in \mathbb{R}^*[/tex]
c) Folosim punctul b).
[tex]X(2^a)X(2^{a+1})\cdot...\cdot X(2^{a+9})=X(2)\\X(2^a\cdot2^{a+1}\cdot...\cdot2^{a+9})=X(2)\\\displaystyle 2^{a+a+1+...+a+9}=2^1\\2^{10a+1+...+9}=2^1\\\displaystyle 10a+\frac{9\cdot10}{2}=1\\10a+45=1\Rightarrow a=-\displaystyle\frac{22}{5}\in\mathbb{R}^*[/tex]