Question

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

$5 p$ a) Arătați că det $A=0$.

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

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

Answer 1

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

a)

Facem diferenta dintre produsul diagonalelor

detA=2-2=0

b)

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

x=3

c)

det(A+I₂)+det(A-I₂)=det(aI₂)

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

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

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