Question

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

5p a) Arătaţi că det $A=1$.

$5 p$ b) Arătați că $2 A-A \cdot A=I_{2}$.

$5 p$ c) Determinaţi numerele reale $x, y$ şi $z$, pentru care $A \cdot\left(\begin{array}{cc}x-2 & y \\ z+1 & 1\end{array}\right)-I_{2}=O_{2}$.

Answer 1

$A=\left(\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right)$

a)

Facem diferenta dintre produsul diagonalelor

detA=1-0=1

b)

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

c)

$\left(\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right)\cdot \left(\begin{array}{cc}x-2& y\\ z+1 & 1\end{array}\right)-\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)\\\\\left(\begin{array}{cc}x-2& y\\ -2x+4+z+1 &-2y+ 1\end{array}\right)-\left(\begin{array}{cc}1& 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)$

$\left(\begin{array}{cc}x-3& y\\ -2x+5-z&-2y\end{array}\right)=\left(\begin{array}{cc}0& 0 \\ 0 & 0\end{array}\right)$

x-3=0

x=3

y=0

-2x+5-z=0

-6+5-z=0

z=1

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9835794

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