Question

Se consideră matricele $O_{2}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)$ şi $A(a)=\left(\begin{array}{cc}1 & a^{2} \\ a^{2} & 1\end{array}\right)$, unde $a$ este număr real pozitiv.

5p 1. Arătați că $\operatorname{det}(A(0))=1$.

5p 2. Determinaţi numărul real pozitiv a pentru care det $(A(a))=0$.

5p 3. Arătaţi că $A(1) \cdot A(1)-2 A(1)=O_{2}$.

5p 4. Determinaţi numărul real pozitiv a pentru care $A(\sqrt{2}) \cdot A(a)=3 A(1)$.

5p 5. Demonstrați că $\operatorname{det}(A(a)-A(0)) \leq 0$, pentru orice număr real pozitiv $a$.

5p 6. Determinaţi perechile $(a, b)$ de numere reale pozitive, ştiind că $A(\sqrt{a}) \cdot A(\sqrt{b})=A(2)+A\left(\frac{1}{2}\right)$.

Answer 1

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

1)

Calculam det(A(0)), inlocuim pe a cu 0 si facem diferenta dintre produsul diagonalelor

det(A(0))=1×1-(0×0)=1

2)

det(A(a))=0

1-a⁴=0

(1-a²)(1+a²)=0

1-a²=0

a²=1

a=1, a fiind numar real pozitiv

1+a²=0

a²=-1, nu se poate

3)

$A(1)\cdot A(1)=\left(\begin{array}{cc}1 & 1 \\1 & 1\end{array}\right)\cdot \left(\begin{array}{cc}1 & 1 \\1 & 1\end{array}\right)=\left(\begin{array}{cc}2& 2 \\2 & 2\end{array}\right)\\\\A(1)\cdot A(1)-2A(1)=\left(\begin{array}{cc}2& 2 \\2 & 2\end{array}\right)-\left(\begin{array}{cc}2& 2 \\2 & 2\end{array}\right)=\left(\begin{array}{cc}0& 0\\0 & 0\end{array}\right)=O_2$

4)

$A(\sqrt{2} )\cdot A(a)=\left(\begin{array}{cc}1& 2 \\2 & 1\end{array}\right)\cdot \left(\begin{array}{cc}1& a^2 \\a^2 & 1\end{array}\right)=\left(\begin{array}{cc}3& 3a^2 \\3a^2 & 3\end{array}\right)\\\\\left(\begin{array}{cc}1+2a^2& a^2+2 \\2+a^2 & 2a^2+1\end{array}\right)=\left(\begin{array}{cc}3& 3a^2 \\3a^2 & 3\end{array}\right)$

1+2a²=3

2a²=2

a²=1

a=1

5)

det(A(a)-A(0))≤0

$A(a)-A(0)=\left(\begin{array}{cc}1 & a^{2} \\ a^{2} & 1\end{array}\right)-\left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right)=\left(\begin{array}{cc}0 & a^{2} \\ a^{2} & 0\end{array}\right)\\\\\left|\begin{array}{cc}0 & a^{2} \\ a^{2} & 0\end{array}\right|=-a^4\leq 0$

6)

$A(\sqrt{a} )\cdot A(\sqrt{b}) =\left(\begin{array}{cc}1 & a \\ a & 1\end{array}\right)\cdot \left(\begin{array}{cc}1 & b\\ b& 1\end{array}\right)=\left(\begin{array}{cc}1+ab & b+a \\ a +b&ab+ 1\end{array}\right)$

$A(2)+A(\frac{1}{2} )=\left(\begin{array}{cc}1 & 4 \\ 4 & 1\end{array}\right)+\left(\begin{array}{cc}1 & \frac{1}{4} \\ \frac{1}{4} & 1\end{array}\right)=\left(\begin{array}{cc}2& \frac{1}{4}+4 \\\frac{1}{4}+4 & 2\end{array}\right)$

Egalam termenii si obtinem:

$1+ab=2\\\\ab=1\\\\a+b=\frac{1}{4}+4 =\frac{17}{4}$

$a=\frac{1}{b} \ inlocuim\ in\ a\ doua\ relatie$

$\frac{1}{b} +b=\frac{17}{4}\ aducem\ la\ acelasi\ numitor\ comun\\\\ 4+4b^2=17b\\\\4b^2-17b+4=0\\\\\Delta=289-64=225\\\\b_1=\frac{17+15}{8}=4\\\\ a=\frac{1}{4} \\\\b_2=\frac{17-15}{8} =\frac{1}{4}\\\\ a=4$

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928500

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