Question

Se consideră matricele $I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ şi $A(a)=\left(\begin{array}{cc}1 & a \\ a^{2} & a^{3}\end{array}\right)$, unde $a$ este număr real.

5p a) Arătați că $\operatorname{det}(A(a))=0$, pentru orice număr real $a$.

$5 p$ b) Determinați numerele reale $x$ pentru care $\operatorname{det}\left(A(2)+x I_{2}\right)=0$.

5 p c) Arătați că, dacă $A(a) \cdot A(b)=A(b) \cdot A(a)$, atunci $a=b$.

Answer 1

$A(a)=\left(\begin{array}{cc}1 & a \\ a^{2} & a^{3}\end{array}\right)$

a)

Calculam det(A(a)), facem diferenta dintre produsul diagonalelor

det(A(a))=a³-a³=0

b)

$det(A(2)+xI_2)=0$

$det(A(2)+xI_2)=\left|\begin{array}{cc}1+x & 2 \\ 4 & 8+x\end{array}\right|=(1+x)(8+x)-8\\\\(1+x)(8+x)-8=0\\\\8+x+8x+x^2-8=0\\\\x^2+9x=0\\\\x(x+9)=0\\\\x=0\\\\x=-9$

c)

$A(a)\cdot A(b)=\left(\begin{array}{cc}1 & a \\ a^{2} & a^{3}\end{array}\right)\cdot \left(\begin{array}{cc}1 & b \\ b^{2} & b^{3}\end{array}\right)=\left(\begin{array}{cc}1+ab^2 & b+ab^3 \\ a^{2}+a^3b^2 & a^2b+a^3b^3\end{array}\right)$

$A(b)\cdot A(a)=\left(\begin{array}{cc}1 & b \\ b^{2} &b^{3}\end{array}\right)\cdot \left(\begin{array}{cc}1 & a \\ a^{2} & a^{3}\end{array}\right)=\left(\begin{array}{cc}1+ba^2 & a+ba^3 \\ b^{2}+b^3a^2 & b^2a+b^3a^3\end{array}\right)$

Le egalam si obtinem:

1+ab²=1+ba²

ab²=ba²   |:ab

b=a

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928369

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