Question

Se consideră matricea $A(a)=\left(\begin{array}{ll}2 & a \\ 0 & 2\end{array}\right)$, unde $a$ este număr real.

5p 1. Arătați că $\operatorname{det}(A(a))=4$, pentru orice număr real $a$.

5p 2. Arătați că [tex]$A(0) \cdot A(2020)=2 A(2020)$/[tex].

5p 3. Demonstrați că $A(-a) \cdot A(a)=4 I_{2}$, pentru orice număr real $a$, unde $I_{2}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$.

5p 4. Determinaţi numerele naturale nenule $m$ şi $n$ pentru care $A(m) \cdot A(n)=2 A(2)$.

$5 \mathbf{p}$ 5. Determinați numerele reale $a$ pentru care $A\left(a^{2}\right)-2 A(a)+A(-3)=O_{2}$, unde $O_{2}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$.

5p 6. Demonstrați că există o infinitate de perechi de numere reale $(x, y)$ pentru care $A(-3) \cdot\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}-2 y \\ 2 x+y\end{array}\right)$

Answer 1

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

1)

Aratati ca det(A(1))=4

Inlocuim pe a cu 1 si apoi facem diferenta dintre produsul diagonalelor

$det(A(1))=\left|\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right|=4-0=4$

2)

Calculam A(0)·A(2020)

$A(0)\cdot A(2020)=\left(\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right)\cdot \left(\begin{array}{ll}2 & 2020 \\ 0 & 2\end{array}\right)=\left(\begin{array}{ll}4& 4040\\ 0 & 4\end{array}\right)=2A(2020)$

3)

Calculam A(-a)·A(a)

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

4)

Calculam A(m)·A(n) si egalam cu 2A(2)

$A(m)\cdot A(n)=\left(\begin{array}{ll}2 & m \\ 0 & 2\end{array}\right)\cdot \left(\begin{array}{ll}2 &n \\ 0 & 2\end{array}\right)=\left(\begin{array}{ll}4 & 2(m+n) \\ 0 & 4\end{array}\right)$

$\left(\begin{array}{ll}4 & 2(m+n) \\ 0 & 4\end{array}\right)=\left(\begin{array}{ll}4 & 4 \\ 0 & 4\end{array}\right)$

2(m+n)=4

m+n=2

m=1 si n=1

5)

Calculam A(a²)-2A(a)+A(-3)=O₂

$\left(\begin{array}{ll}2 & a^2 \\ 0 & 2\end{array}\right)-\left(\begin{array}{ll}4& 2a \\ 0 & 4\end{array}\right)+\left(\begin{array}{ll}2 & -3 \\ 0 & 2\end{array}\right)=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)\\\\\left(\begin{array}{ll}0 & a^2-2a-3 \\ 0 & 0\end{array}\right)=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$

a²-2a-3=0

Δ=b²-4ac

Δ=4+12=16

$a_1=\frac{-b+\sqrt{\Delta} }{2a} =\frac{2+4}{2} =3\\\\a_2=\frac{-b-\sqrt{\Delta} }{2a} =\frac{2-4}{2}=-1$

6)

$\left(\begin{array}{ll}2 & -3 \\ 0 & 2\end{array}\right)\cdot \left(\begin{array}{ccc}x\\y\\\end{array}\right)= \left(\begin{array}{ccc}-2y\\2x+y\\\end{array}\right)\\\\\left(\begin{array}{ll}2x-3y \\ 2y\end{array}\right)= \left(\begin{array}{ccc}-2y\\2x+y\\\end{array}\right)$

2x-3y=-2y

2y=2x+y

2x=y⇒exista o infinitate de numere x si y care verifica relatia

Un exercitiu cu calculul determinantului gasesti aici: https://brainly.ro/tema/1009507

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