Question

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

5p a) Arătați că det $A=1$.

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

Answer 1

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

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

a)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=0-(-1)=1

b)

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

c)

$A\cdot B=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)\cdot \left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)$

$B\cdot A= \left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)$

AB-BA=O₂

$A+B=\left(\begin{array}{cc}1 & 1 \\ -1 & 0\end{array}\right)+\left(\begin{array}{cc}0 & -1 \\ 1 & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)$

$\left(\begin{array}{cc}1 & 0 \\ 0& 1\end{array}\right)=\left(\begin{array}{cc}log_2x& 0 \\ 0& log_3y\end{array}\right)$

$log_2x=1\\\\x=2\\\\log_3y=1\\\\y=3$

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

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