Question

Se consideră matricea $A(a)=\left(\begin{array}{cc}1 & a \\ a & 0\end{array}\right)$, unde $a$ este număr real.

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

5p 2. Determinați numărul real $a$, ştiind că $A(a)+A(a+1)=2 A(-1)$.

$5 p$ 3. Arătați că $A(1)+A(2)+A(3)+\ldots+A(2020)=2020 \cdot A\left(\frac{2021}{2}\right)$.

5p 4. Arătaţi că $\operatorname{det}(A(a) \cdot A(b))-\operatorname{det}(A(a)+A(b)) \geq 0$, pentru orice numere reale $a$ şi $b$.

5p 5. Demonstrați că $\operatorname{det}(A(x) \cdot A(y)-A(y) \cdot A(x)) \geq 0$, pentru orice numere reale $x$ şi $y$.

5p $\mid$ 6. Determinaţi numerele reale $a$ pentru care $\operatorname{det}(A(a))+\operatorname{det}(A(a) \cdot A(a))=0$.

Answer 1

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

1)

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

det(A(0))=0-0=0

2)

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

Egalam termenii

2a+1=-2

2a=-3

$a=-\frac{3}{2}$

3)

$A(1)+A(2)+...+A(2020)=\left(\begin{array}{cc}2020 & 1+2+...+2020 \\ 1+2+...+2020 & 0\end{array}\right)=\left(\begin{array}{cc}2020& \frac{2020\cdot 2021}{2} \\ \frac{2020\cdot 2021}{2} & 0\end{array}\right)$

Dam factor comun pe 2020

$A(1)+A(2)+...+A(2020)=2020\left(\begin{array}{cc}1& \frac{2021}{2} \\ \frac{2021}{2} & 0\end{array}\right)=2020A( \frac{2021}{2})$

4)

det(A(a)A(b))-det(A(a)+A(b))≥0

$A(a)\cdot A(b)=\left(\begin{array}{cc}1 & a \\ a & 0\end{array}\right)\cdot \left(\begin{array}{cc}1 &b \\ b& 0\end{array}\right)=\left(\begin{array}{cc}1+ab & b \\ a & ab\end{array}\right)\\\\det(A(a)\cdot A(b))=(1+ab)ab-ab=ab+a^2b^2-ab=a^2b^2$

$A(a)+ A(b)=\left(\begin{array}{cc}1 & a \\ a & 0\end{array}\right)+ \left(\begin{array}{cc}1 &b \\ b& 0\end{array}\right)=\left(\begin{array}{cc}2 &a+ b \\ a+b & 0\end{array}\right)\\\\det(A(a)+A(b))=-(a+b)^2$

det(A(a)A(b))-det(A(a)+A(b))=a²b²-[-(a+b)²]=a²b²+(a+b)² ≥0 pentru ca avem suma de numere cu putere para

5)

$A(x)\cdot A(y)=\left(\begin{array}{cc}1 & x \\ x & 0\end{array}\right)\cdot \left(\begin{array}{cc}1 &y \\ y& 0\end{array}\right)=\left(\begin{array}{cc}1+xy &y \\ x & xy\end{array}\right)$

$A(y)\cdot A(x)=\left(\begin{array}{cc}1 & y \\ y & 0\end{array}\right)\cdot \left(\begin{array}{cc}1 &x \\ x& 0\end{array}\right)=\left(\begin{array}{cc}1+xy &x \\ y & xy\end{array}\right)$

$A(x)A(y)-A(y)A(x)=\left(\begin{array}{cc}1+xy &y \\ x & xy\end{array}\right)-\left(\begin{array}{cc}1+xy &x \\ y & xy\end{array}\right)=\left(\begin{array}{cc}0 &y-x \\ x-y &0\end{array}\right)\\\\det(A(x)A(y)-A(y)A(x))=-(x-y)(y-x)=(x-y)^2\geq 0$

6)

det(A(a))=-a²

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

-a²+a⁴=0

a²(-1+a²)=0

a=0

-1+a²=0

a²=1

a=1 si a=-1

Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9928531

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