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

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

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

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


Răspuns :

[tex]A=\left(\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right)[/tex]

a)

Facem diferenta dintre produsul diagonalelor

detA=1-0=1

b)

[tex]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[/tex]

c)

[tex]\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)\\\\[/tex]

[tex]\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)[/tex]

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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