Răspuns :
[tex]A(a)=\left(\begin{array}{cc}1 & a \\ a & 0\end{array}\right)[/tex]
1)
Calculam det(A(0)), inlocuim pe a cu 0 si facem diferenta dintre produsul diagonalelor
det(A(0))=0-0=0
2)
[tex]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)[/tex]
Egalam termenii
2a+1=-2
2a=-3
[tex]a=-\frac{3}{2}[/tex]
3)
[tex]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)[/tex]
Dam factor comun pe 2020
[tex]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})[/tex]
4)
det(A(a)A(b))-det(A(a)+A(b))≥0
[tex]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\\\\[/tex]
[tex]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[/tex]
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)
[tex]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)[/tex]
[tex]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)[/tex]
[tex]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[/tex]
6)
det(A(a))=-a²
[tex]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[/tex]
-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
#BAC2022
#SPJ4
Vă mulțumim că ați ales să vizitați site-ul nostru dedicat Matematică. Sperăm că informațiile disponibile v-au fost utile. Dacă aveți întrebări sau aveți nevoie de asistență suplimentară, vă rugăm să ne contactați. Revenirea dumneavoastră ne va bucura, iar pentru acces rapid, nu uitați să ne salvați la favorite!