Question

Se consideră matricea $A(a)=\left(\begin{array}{cc}1 & \ln a \\ 0 & 1\end{array}\right)$, unde $a \in(0,+\infty)$.

$5 \mathbf{a}$ a) Arătați că $\operatorname{det}(A(a))=1$, pentru orice $a \in(0,+\infty)$.

$5 \mathbf{p}$ b) Demonstrați că $A(a) \cdot A(b)=A(a b)$, pentru orice $a, b \in(0,+\infty)$.

$5 p$ c) Determinați $a \in(0,+\infty)$, astfel incât $A(a) \cdot A(a) \cdot A(a)=\left(\begin{array}{cc}1 & 2020 \\ 0 & 1\end{array}\right)$.

Answer 1

$A(a)=\left(\begin{array}{cc}1 & \ln a \\ 0 & 1\end{array}\right)$

a)

Aratati ca det(A(a))=1

Facem diferenta dintre produsul diagonalelor

det(A(a))=1-0=1

b)

$A(a)\cdot A(b)=\left(\begin{array}{cc}1 & \ln a \\ 0 & 1\end{array}\right)\cdot \left(\begin{array}{cc}1 & \ln b \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}1 & \ln a+lnb \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}1 & \ln ab \\ 0 & 1\end{array}\right)=A(ab)$

c)

Ne folosim de punctul b

A(a)·A(a)=A(a²)

A(a²)·A(a)=A(a³)

$\left(\begin{array}{cc}1 & \ln a^3 \\ 0 & 1\end{array}\right)=A(a)=\left(\begin{array}{cc}1 & 2020 \\ 0 & 1\end{array}\right)$

$lna^3=2020\\\\3lna=2020\\lna=\frac{2020}{3}\\\\ a=e^{\frac{2020}{3}}$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918945

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