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
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