[tex]\displaystyle\bf\\A=\left(\begin{array}{cc}6&9\\-4&-6\end{array}\right)\\\\I_2=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)\\\\O_2=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)[/tex]
.
[tex]\displaystyle\bf\\Rezolvare:\\a)\\A^2=\left(\begin{array}{cc}6&9\\-4&-6\end{array}\right)^2=\left(\begin{array}{cc}6&9\\-4&-6\end{array}\right)\cdot\left(\begin{array}{cc}6&9\\-4&-6\end{array}\right)=\\\\=\left(\begin{array}{cc}6\cdot6-9\cdot4&9\cdot6-6\cdot9\\-4\cdot6-6\cdot(-4)&-4\cdot9-6\cdot(-6)\end{array}\right)=\left(\begin{array}{cc}0&0\\0&0\end{array}\right)=O_2[/tex]
.
[tex]\displaystyle\bf\\b)\\\textbf{Trebuie sa aratam ca inversa lui: }~I_2-3A~este~I_2+3A\\\\Daca~inversa~lui~B~este~C~atunci~~B\times C=I_n~~(Aici~n=2)\\\\\textbf{Putem calcula pe }I_{2}-3A~si~pe~I_{2}+3A~\textbf{si apoi sa le inmultim,}\\\textbf{dar obtinem un calcul complicat ca la punctul a).}\\\\\textbf{Propun sa profitam de calculul de la punctul a) unde }~A^2=O_2\\\\Folosim formula:~~(a-b)(a+b)=a^2-b^2\\\\(I_{2}-3A)(I_{2}+3A)=(I_{2})^2-(3A)^2=I_2-9A^2=\\\\=I_2-9\cdot O_2=I_2-O_2=I_2\\\\[/tex]
.
[tex]\displaystyle\bf\\c)\\\textbf{Rezolvam ecuatia:}\\\\det(2A-xI_2)=25\\\\2A=2\left(\begin{array}{cc}6&9\\-4&-6\end{array}\right)=\left(\begin{array}{cc}12&18\\-8&-12\end{array}\right)\\\\xI_2=\left(\begin{array}{cc}x&0\\0&x\end{array}\right)\\\\2A-xI_2=\left(\begin{array}{cc}12&18\\-8&-12\end{array}\right)-\left(\begin{array}{cc}x&0\\0&x\end{array}\right)=\left(\begin{array}{cc}12-x&18\\-8&-12-x\end{array}\right)[/tex]
.
[tex]\displaystyle\bf\\det(2A-xI_2)=det\left(\begin{array}{cc}12-x&18\\-8&-12-x\end{array}\right)=\\\\(-12-x)(12-x)-18\cdot(-8)=(-1)\cdot(12+x)(12-x)+18\cdot8=\\\\=(-1)\cdot(144-x^2)+144=-144+x^2+144=\boxed{\bf~x^2}\\\\det(2A-xI_2)=25\\\\x^2=25\\x_{12}=\pm\sqrt{25}=\pm5\\\\\boxed{\bf~x_1=5~~si~~x_2=-5}[/tex]
