Răspuns :
[tex]A(a)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{a}\end{array}\right)[/tex]
1)
Aratati ca det(A(3))=125
Facem diferenta dintre produsul diagonalelor
det(A(3))=1×5³-0=125
2)
A(a)·A(b)=A(a+b)
[tex]A(a)\cdot A(b)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{a}\end{array}\right)\cdot \left(\begin{array}{cc}1 & 0 \\ 0 & 5^{b}\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{a}\cdot 5^b\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{a+b}\end{array}\right)=A(a+b)[/tex]
3)
A(1)·A(4)-A(2)·A(3)=O₂
[tex]A(1)\cdot A(4)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{1}\end{array}\right)\cdot \left(\begin{array}{cc}1 & 0 \\ 0 & 5^{4}\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{5}\end{array}\right)=A(5)[/tex]
[tex]A(2)\cdot A(3)=A(5)[/tex] (conform punctului 2)
A(5)-A(5)=O₂
4)
Matricea A(a) este inversabila daca det(A(a)) este diferit de zero
det(A(a))=5ᵃ-0=5ᵃ
5ᵃ≠0, a∈R⇒ matricea este inversabila
5)
Observam ca A(0)=I₂
Atunci X=A(2)⁻¹
[tex]A(2)=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{2}\end{array}\right)\\\\A(2)^t=\left(\begin{array}{cc}1 & 0 \\ 0 & 5^{2}\end{array}\right)\\\\A(2)^*=\left(\begin{array}{cc}5^2 & 0 \\ 0 & 1\end{array}\right)[/tex]
det(A(2))=25
[tex]X=A(2)^{-1}=\frac{1}{25}\cdot \left(\begin{array}{cc}25 & 0 \\ 0 & 1\end{array}\right) =\left(\begin{array}{cc}1& 0 \\ 0 & \frac{1}{25} \end{array}\right)[/tex]
6)
[tex]det(A(n))\leq \sqrt[3]{125}[/tex]
[tex]det(A(n))=5^n\\\\5^n\leq 5\\\\n\in N,\ n=0\ sau\ n=1[/tex]
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9835794
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