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

5p a) Arătați că [tex]$\operatorname{det}(A-B)=1$[/tex].

[tex]$5 p$[/tex] b) Demonstrați că matricea [tex]$C=A \cdot A+B \cdot B$[/tex] nu este inversabilă.

[tex]$5 p$[/tex] c) Determinaţi numerele reale [tex]$x$[/tex] şi [tex]$y$[/tex] pentru care [tex]$A \cdot X=X \cdot B$[/tex], unde [tex]$X=\left(\begin{array}{ll}1 & 2 \\ x & y\end{array}\right)$[/tex].

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

[tex]B=\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)[/tex]

a)

Calcula detA si detB, facand diferenta dintre produsul celor doua diagonale

[tex]detA=\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right|=1-0=1[/tex]

[tex]detB=\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|=1-0=1[/tex]

b)

Ca o matrice sa fie inversabila trebuie ca determinantul sau sa fie diferit de 0. Daca determinantul este egal cu 0, atunci matricea nu este inversabila

[tex]C=A\times A+B\times B[/tex]

Calculam mai intai A×A

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

Calculam B×B

[tex]B\times B=\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)\times \left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right)[/tex]

Calculam C

[tex]C=\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)+\left(\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right)[/tex]

Calculam detC

[tex]detC=\left|\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right|=4-4=0[/tex]

detC=0⇒ C nu este inversabila

c)

Calculam A×X

[tex]A\times X=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\times \left(\begin{array}{ll}1 & 2 \\ x & y\end{array}\right)=\left(\begin{array}{ll}1+x & 2+y \\ x & y\end{array}\right)[/tex]

Calculam X×B

[tex]X\times B=\left(\begin{array}{ll}1 & 2 \\ x &y\end{array}\right)\times \left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)=\left(\begin{array}{ll}3 & 2 \\ x+y & y\end{array}\right)[/tex]

Egalam egalitatile si obtinem:

[tex]\left(\begin{array}{ll}3 & 2 \\ x+y & y\end{array}\right)=\left(\begin{array}{ll}1+x & 2 +y\\ x& y\end{array}\right)[/tex]

Egalam termenii

2+y=2

y=0

1+x=3

x=2

Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/713615

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