Question

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

5p a) Arătați că $\operatorname{det}(A(e))=e$.

5p b) Demonstrați că $\operatorname{det}\left(A\left(a^{2}\right)\right)=\operatorname{det}(A(a) \cdot A(a))$, pentru orice $a \in(0,+\infty)$.

$5 p$ c) Determinați numerele $a, b \in(0,+\infty)$ pentru care $A(a)+A(b)=2 A(a) \cdot A(b)$.

Answer 1

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

a)

Aratati ca det(A(e))=e

Inlocuim pe a cu e si adaugam primele doua linii ale determinantului

$det(A(e))=\left|\begin{array}{ccc}1 & 0 & \ln e \\ 0 & e & 0 \\ 0 & 0 & 1\end{array}\right|=\left|\begin{array}{ccc}1 & 0 &1 \1 \\ 0 & e & 0 \\ 0 & 0 & 1\end{array}\right|$

                                                 1    0   1

                                                 0   e   0

det(A(e))=(e+0+0)-(0+0+0)=e

b)

$det(A(a^2))=det(A(a)A(a))$

Calculam A(a)A(a), iar apoi ii calculam determinantul

$A(a)\cdot A(a)=\left(\begin{array}{ccc}1 & 0 & \ln a \\ 0 & a & 0 \\ 0 & 0 & 1\end{array}\right)\cdot \left(\begin{array}{ccc}1 & 0 & \ln a \\ 0 & a & 0 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & 0 & 2\ln a \\ 0 & a^2 & 0 \\ 0 & 0 & 1\end{array}\right)$

$\left|\begin{array}{ccc}1 & 0 & 2\ln a \\ 0 & a^2 & 0 \\ 0 & 0 & 1\end{array}\right|$

  1     0    2lna

  0    a²    0

=(a²+0+0)-(0+0+0)=a²

$det(A(a^2))=\left|\begin{array}{ccc}1 & 0 & \ln a^2 \\ 0 & a^2 & 0 \\ 0 & 0 & 1\end{array}\right|=\left|\begin{array}{ccc}1 & 0 & 2\ln a \\ 0 & a^2 & 0 \\ 0 & 0 & 1\end{array}\right|$

                                                      1    0     2lna

                                                      0    a²    0

=(a²+0+0)-(0+0+0)=a²

Observam ca sunt egali cei doi determinanti calculati mai sus

c)

A(a)+A(b)=2A(a)A(b)

Ne folosim de punctul b

A(a)A(b)=A(ab)

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

$A(a)+A(b)=\left(\begin{array}{ccc}1 & 0 & \ln a \\ 0 & a & 0 \\ 0 & 0 & 1\end{array}\right)+\left(\begin{array}{ccc}1 & 0 & \ln b \\ 0 & b & 0 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}2 & 0 & \ln ab \\ 0 & a+b & 0 \\ 0 & 0 & 2\end{array}\right)$

2A(a)A(b)=2A(ab)

A(a)+A(b)=2A(ab)

$\left(\begin{array}{ccc}2 & 0 & \ln ab \\ 0 & a+b & 0 \\ 0 & 0 & 2\end{array}\right)=\left(\begin{array}{ccc}2 & 0 & 2\ln ab \\ 0 & 2ab & 0 \\ 0 & 0 & 2\end{array}\right)$

Egalam termenii si obtinem:

a+b=2ab

ln(ab)=2ln(ab)

ln(ab)=0

ab=1

a+b=2ab

ab=1

a+b=2

ab=1

S=2 si P=1

t²-St+P=0

t²-2t+1=0

(t-1)²=0

t=1⇒ a=1 si b=1

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9685683

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