Question

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

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

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

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

Answer 1

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

$B=\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)$

a)

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

$detA=\left|\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right|=1-0=1$

$detB=\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|=1-0=1$

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

$C=A\times A+B\times B$

Calculam mai intai A×A

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

Calculam B×B

$B\times B=\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)\times \left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right)$

Calculam C

$C=\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)+\left(\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right)=\left(\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right)$

Calculam detC

$detC=\left|\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right|=4-4=0$

detC=0⇒ C nu este inversabila

c)

Calculam A×X

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

Calculam X×B

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

Egalam egalitatile si obtinem:

$\left(\begin{array}{ll}3 & 2 \\ x+y & y\end{array}\right)=\left(\begin{array}{ll}1+x & 2 +y\\ x& y\end{array}\right)$

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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