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

[tex]$5 p$[/tex] a) Arătați că det [tex]$A=0$[/tex].

[tex]$5 p$[/tex] b) Determinați numărul real [tex]$x$[/tex] pentru care [tex]$A \cdot A=x A$[/tex].

[tex]$5 p$[/tex] c) Determinați numerele reale a pentru care [tex]$\operatorname{det}\left(A+I_{2}\right)+\operatorname{det}\left(A-I_{2}\right)=\operatorname{det}\left(a I_{2}\right)$[/tex].

Răspuns :

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

a)

Facem diferenta dintre produsul diagonalelor

detA=2-2=0

b)

[tex]A\cdot A=\left(\begin{array}{ll}2 & 1 \\ 2 & 1\end{array}\right)\cdot \left(\begin{array}{ll}2 & 1 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}6 & 3 \\ 6 & 3\end{array}\right)\\\\\left(\begin{array}{ll}4 & 3 \\ 4 & 3\end{array}\right)=\left(\begin{array}{ll}2x & x \\ 2x & x\end{array}\right)[/tex]

x=3

c)

det(A+I₂)+det(A-I₂)=det(aI₂)

[tex]A+I_2=\left(\begin{array}{ll}2 & 1 \\ 2 & 1\end{array}\right)+\left(\begin{array}{ll}1 &0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}3& 1 \\ 2 & 2\end{array}\right)\\\\det(A+I_2)=6-2=4\\\\A-I_2=\left(\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right)\\\\det(A-I_2)=0-2=-2\\\\det(aI_2)=a^2-0=a^2\\\\4-2=a^2\\\\a^2=2\\\\a=\sqrt{2} \ sau\ a=-\sqrt{2}[/tex]

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928435

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