Explicație pas cu pas:
[tex]A=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right)[/tex]
Calculam A².
[tex]A^2=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right)[/tex]
Calculam A³.
[tex]A^3=A^2*A=\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3^3&0\\0&1\end{array}\right)[/tex]
Observam ca Aⁿ are forma:
[tex]A^n=\left(\begin{array}{ccc}3^n&0\\0&1\end{array}\right)[/tex]
Demonstram prin inductie ca:
[tex]P(n): A^n=\left(\begin{array}{ccc}3^n&0\\0&1\end{array}\right) ~este~adevarata.[/tex]
1) Etapa de verificare:
[tex]P(1): A=\left(\begin{array}{ccc}3^1&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3&0\\0&1\end{array}\right)[/tex]
[tex]P(2): A^2=\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) ~verificata~anterior[/tex]
2) Etapa inductiva:
[tex]Presupunem~P(k): A^k=\left(\begin{array}{ccc}3^k&0\\0&1\end{array}\right) ~adevarata[/tex]
[tex]Demonstram~P(k+1): A^{k+1}=\left(\begin{array}{ccc}3^{k+1}&0\\0&1\end{array}\right) ~adevarata.[/tex]
[tex]A^{k+1}=A^k*A=\left(\begin{array}{ccc}3^k&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^k*3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^{k+1}&0\\0&1\end{array}\right)[/tex]
Deci, avem ca P(k+1) este adevarata.
Din 1) si 2), conform principiului I al inductiei matematice, P(n) este adevarata.
Determinam B.
[tex]B=A+A^2+....+A^{2016}=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) +\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) +...+\left(\begin{array}{ccc}3^{2016}&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3+3^2+...+3^{2016}&0\\0&1+1+...+1\end{array}\right) =\left(\begin{array}{ccc}3*\frac{3^{2016}-1}{3-1} &0\\0&2016\end{array}\right) =\left(\begin{array}{ccc}\frac{3^{2017}-3}{2} &0\\0&1\end{array}\right)[/tex]
Determinam transpusa matricii B.
[tex]B^t=\left(\begin{array}{ccc}\frac{3^{2017}-3}{2} &0\\0&1\end{array}\right)[/tex]
