Question

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

5p a) Arătați că det $A=4$.

$5 \mathbf{p}$ b) Arătați că $A \cdot A+3 A+4 I_{2}=O_{2}$.

$5 p$ c) Determinaţi numerele reale $x$ şi $y$ astfel incât $A \cdot A \cdot A=x A+y I_{2}$.

Answer 1

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

a)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=2-(1)×2=2+2=4

b)

$A\cdot A=\left(\begin{array}{cc}-2 & 2 \\ -1 & -1\end{array}\right)\cdot \left(\begin{array}{cc}-2 & 2 \\ -1 & -1\end{array}\right)=\left(\begin{array}{cc}2 & -6 \\ 3 & -1\end{array}\right)$

$A\cdot A+3A+4I_2=\left(\begin{array}{cc}2 & -6 \\ 3 & -1\end{array}\right)+\left(\begin{array}{cc}-6 & 6 \\ -3 & -3\end{array}\right)+\left(\begin{array}{cc}4 & 0 \\ 0 & 4\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)=O_2$

c)

$A\cdot A\cdot A=\left(\begin{array}{cc}2 & -6 \\ 3 & -1\end{array}\right)\cdot \left(\begin{array}{cc}-2 & 2 \\ -1 & -1\end{array}\right)=\left(\begin{array}{cc}2 & 10 \\ -5 & 7\end{array}\right)$

$\left(\begin{array}{cc}2 & 10 \\ -5 & 7\end{array}\right)=\left(\begin{array}{cc}-2x & 2x \\ -x& -x\end{array}\right)+\left(\begin{array}{cc}y & 0 \\ 0 & y\end{array}\right)$

2=-2x+y

10=2x

x=5

2=-10+y

y=12

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918986

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