[tex]\displaystyle A(n)= \left(\begin{array}{ccc}1&0&0\\0&2^n&0\\0&2^n-1&1\end{array}\right) \\ \\ a).det(A(0))=1 \\ \\ det(A(0))= \left|\begin{array}{ccc}1&0&0\\0&2^0&0\\0&2^0-1&1\end{array}\right|= \left|\begin{array}{ccc}1&0&0\\0&1&0\\0&1-1&1\end{array}\right|= \left|\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right|= \\ \\ =1 \cdot 1 \cdot 1+0 \cdot 0 \cdot 0+0 \cdot 0 \cdot 0-0 \cdot 1 \cdot 0-0 \cdot 0 \cdot 1-1 \cdot 0 \cdot 0= \\ \\ =1+0+0-0-0-0=1 \Rightarrow det(A(0))=1[/tex]
[tex]\displaystyle b).A(n) \cdot A(1)=A(3) \\ \\ A(n) \cdot A(1)= \left(\begin{array}{ccc}1&0&0\\0&2^n&0\\0&2^n-1&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&0\\0&2^1&0\\0&2^1-1&1\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}1&0&0\\0&2^n&0\\0&2^n-1&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&0\\0&2&0\\0&2-1&1\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}1&0&0\\0&2^n&0\\0&2^n-1&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&0\\0&2&0\\0&1&1\end{array}\right)= \left(\begin{array}{ccc}1&0&0\\0&2^{n+1}&0\\0&2^{n+1}-1&1\end{array}\right) [/tex]
[tex]\left(\begin{array}{ccc}1&0&0\\0&2^{n+1}&0\\0&2^{n+1}-1&1\end{array}\right) =A(3) \\ \\ \left(\begin{array}{ccc}1&0&0\\0&2^{n+1}&0\\0&2^{n+1}-1&1\end{array}\right) = \left(\begin{array}{ccc}1&0&0\\0&2^3&0\\0&2^3-1&1\end{array}\right) \\ \\ \left(\begin{array}{ccc}1&0&0\\0&2^{n+1}&0\\0&2^{n+1}-1&1\end{array}\right) = \left(\begin{array}{ccc}1&0&0\\0&8&0\\0&8-1&1\end{array}\right)[/tex]
[tex]\displaystyle \left(\begin{array}{ccc}1&0&0\\0&2^{n+1}&0\\0&2^{n+1}-1&1\end{array}\right) = \left(\begin{array}{ccc}1&0&0\\0&8&0\\0&7&1\end{array}\right) \\ \\ 2^{n+1}=8 \Rightarrow 2^{n+1}=2^3 \Rightarrow n+1=3 \Rightarrow n=3-1 \Rightarrow n=2
[/tex]
[tex]\displaystyle c).A(p) \cdot A(q)=A(pq) \\ \\ A(p) \cdot A(q)= \left(\begin{array}{ccc}1&0&0\\0&2^p&0\\0&2^p-1&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&0\\0&2^q&0\\0&2^q-1&1\end{array}\right) = \\ \\ = \left(\begin{array}{ccc}1&0&0\\0&2^{p+q}&0\\0&2^{p+q}-1&1\end{array}\right) \\ \\ \left(\begin{array}{ccc}1&0&0\\0&2^{p+q}&0\\0&2^{p+q}-1&1\end{array}\right)=A(pq) [/tex]
[tex]\displaystyle \left(\begin{array}{ccc}1&0&0\\0&2^{p+q}&0\\0&2^{p+q}-1&1\end{array}\right)= \left(\begin{array}{ccc}1&0&0\\0&2^{pq}&0\\0&2^{pq}-1&1\end{array}\right) \\ \\ 2^{p+q}=2^{pq} \Rightarrow p+q=pq \Rightarrow p=0,~q=0 \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~p=2,~q=2[/tex]
