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

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

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

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


Răspuns :

[tex]A(a)=\left(\begin{array}{ccc}1 & 0 & \ln a \\ 0 & a & 0 \\ 0 & 0 & 1\end{array}\right)[/tex]

a)

Aratati ca det(A(e))=e

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

[tex]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|[/tex]

                                                 1    0   1

                                                 0   e   0

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

b)

[tex]det(A(a^2))=det(A(a)A(a))[/tex]

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

[tex]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)[/tex]

[tex]\left|\begin{array}{ccc}1 & 0 & 2\ln a \\ 0 & a^2 & 0 \\ 0 & 0 & 1\end{array}\right|[/tex]

  1     0    2lna

  0    a²    0

=(a²+0+0)-(0+0+0)=a²

[tex]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|[/tex]

                                                      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)

[tex]A(ab)=\left(\begin{array}{ccc}1 & 0 & \ln ab \\ 0 & ab & 0 \\ 0 & 0 & 1\end{array}\right)[/tex]

[tex]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)[/tex]

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

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

[tex]\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)[/tex]

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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