Question

b si c !Mulțumesc anticipat ​

Answer 1

b)

$A^2=\left[\begin{array}{ccc}6&-10\\3&-5\end{array}\right] *\left[\begin{array}{ccc}6&-10\\3&-5\end{array}\right] =\left[\begin{array}{ccc}6&-10\\3&-5\end{array}\right]=A$

$M(a)*M(b)=(I_2+aA)(I_2+bA)=I_2+bA+aA+abA^2\\M(a)*M(b)=I_2+aA+bA+abA=I_2+(a+b+ab)A=M(a+b+ab)$

c)

$M(1)+M(2)=2I_2+A+2A=I_2+I_2+(1+2)A=I_2+M(1+2)\\\\M(1)+M(2)+M(3)=3I_2+A+2A+3A=2I_2+I_2+(1+2+3)A=I_2+M(1+2+3)\\\\....\\M(1)+M(2)+...+M(n)=(n-1)I_2+M(1+2+...+n)$

se demonstreaza prin inductie corectitudinea in poza

$M(1)+M(2)+...+M(2019)=2018I_2+M(1+2+...+2019)=2018I_2+M(N)\\\\2019M(a)=2019I_2+2019aA=2018I_2+I_2+2019aA=2018I_2+M(2019a)$

se egaleaza relatiile:

$2018I_2+M(2019a)=2018I_2+M(N)\\M(2019a)=M(N)\\2019a=N\\a=\frac{N}{2019}\\\\ N=\frac{2019*2020}{2}\\a=\frac{2019*2020}{2*2019}=2010$