Question

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

5p 1. Calculaţi det $A$.

5 2. Arătaţi că $A^{2}-2 A+I_{2}=O_{2}$, unde $A^{2}=A \cdot A$.

$5 p$ 3. Determinaţi numerele reale $m$ pentru care $\operatorname{det}((m-1) A)=m+1$.

5p 4. Arătaţi că $A \cdot B=B \cdot A=I_{2}$.

5p 5. Demonstrați că, dacă $X \in \mathcal{M}_{2}(\mathbb{R})$ astfel încât $A \cdot X=X \cdot A$, atunci $X=\left(\begin{array}{ll}a & b \\ 0 & a\end{array}\right)$, unde $a$ și $b$ sunt numere reale

6. Determinaţi numerele reale $x$ şi $y$, ştiind că $x A+y B=\left(\begin{array}{cc}5 & -2 \\ 0 & 5\end{array}\right)$.

Answer 1

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

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

1)

Calculam detA, facem diferenta dintre produsul diagonalelor

detA=1×1-2×0=1-0=1

2)

$A^2-2A+I_2=O_2$

$A^2=\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)\cdot \left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right)\\\\A^2-2A+I_2=\left(\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right)-2\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)+\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)=O_2$

3)

det((m-1)A)=m+1

(m-1)²-0=m+1

m²-2m+1=m+1

m²-3m=0

m(m-3)=0

m=0 si m=3

4)

A·B=B·A=I₂

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

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

5)

$A\cdot X=\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)\cdot \left(\begin{array}{ll}a &b \\ 0 &a\end{array}\right)=\left(\begin{array}{ll}a & b+2a \\ 0 & a\end{array}\right)$

$X\cdot A=\left(\begin{array}{ll}a &b \\ 0 &a\end{array}\right)\cdot \left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}a & b+2a \\ 0 & a\end{array}\right)$

6)

$xA+yB=\left(\begin{array}{ll}x& 2x \\ 0 & x\end{array}\right)+ \left(\begin{array}{ll}y & -2y \\ 0 & y\end{array}\right)=\left(\begin{array}{ll}x+y & 2x-2y \\ 0 & x+y\end{array}\right)$

x+y=5

2x-2y=-2   |:2

x-y=-1

x+y=5

x-y=-1

Le adunam

2x=4

x=2 si y=3

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9919024

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