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a fost răspuns

Se consideră matricele [tex]$A=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right), B=\left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)$[/tex] şi [tex]$O_{2}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$[/tex].

5p a) Arătați că det [tex]$A=1$[/tex].

[tex]$5 p$[/tex] b) Arătați că [tex]$B \cdot B+A=O_{2}$[/tex].
[tex]$5 p$[/tex] c) Determinați [tex]$x, y \in(0,+\infty)$[/tex], pentru care [tex]$A \cdot B+B \cdot A-(A+B)=\left(\begin{array}{cc}\log _{2} x & 0 \\ 0 & \log _{3} y\end{array}\right)$[/tex].

Răspuns :

[tex]A=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)[/tex]

[tex]B=\left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)$[/tex]

a)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=0-(-1)=1

b)

[tex]B\cdot B+A=\left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)+\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)=\left(\begin{array}{cc}-1 & -1 \\ 1 & 0\end{array}\right)+\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)\\\\B\cdot B+A=\left(\begin{array}{cc}0 & 0 \\ 0& 0\end{array}\right)=O_2[/tex]

c)

[tex]A\cdot B=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)\cdot \left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)[/tex]

[tex]B\cdot A= \left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)[/tex]

AB-BA=O₂

[tex]A+B=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)+\left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)[/tex]

[tex]\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)=\left(\begin{array}{cc}log_2x& 0 \\ 0& log_3y\end{array}\right)[/tex]

[tex]log_2x=1\\\\x=2\\\\log_3y=1\\\\y=3[/tex]

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9853119

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