[tex]A^2=A\cdot A=\left(\begin{array}{cc}4 &-2\\3 &-1\end{array}\right)\cdot\left(\begin{array}{cc}4 &-2\\3 &-1\end{array}\right)=\left(\begin{array}{cc}10 &-6\\9 &-5\end{array}\right)[/tex]
Vrem să demonstrăm : [tex]\displaystyle P(n):A^n=(2^n-1)A-(2^n-2)I_2, \forall n\in\mathbb{N}, n\geq2.[/tex]
Etapa de verificare:
[tex]P(2):A^2=(2^2-1)A-(2^2-2)I_2=3A-2I_2=3\left(\begin{array}{cc}4 &-2\\3 &-1\end{array}\right)-2\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=\left(\begin{array}{cc}10 &-6\\9 &-5\end{array}\right)(A)[/tex]
Etapa de demonstrație:
Presupunem [tex]P(k):A^k=(2^k-1)A-(2^k-1)I_2,\: k>2[/tex] adevărată.
Demonstrăm [tex]P(k+1):A^{k+1}=(2^{k=1}-1)A-(2^{k+1}-2)I_2[/tex]
[tex]A^{k+1}=A^k\cdot A=\\=\left[(2^k-1)A-(2^k-2)I_2\right]\cdot A=(2^k-1)A^2-(2^k-2)I_2\cdot A=\\=(2^k-1)(3A-2I_2)-(2^k-2)A=\\=3(2^k-1)A-2(2^k-1)I_2-(2^k-2)A=\\=(3\cdot2^k-3-2^k+2)A-2(2^k-1)I_2=\\=(2\cdot2^k-1)A-(2\cdot2^k-2)I_2=(2^{k+1}-1)A-(2^{k+1}-2)I_2[/tex]
Avem că [tex]P(k)\rightarrow P(k+1)[/tex]. Deci [tex]P(n)\text{ adev\u arat\u a},\:\forall n\geq2[/tex]
