Va rog punctul b) (daca sunteti amabili si punctul c) )

[tex] \it A^2= A\cdot A = \begin{pmatrix} 3\ \ \ 3\ \ \ 3 \\
3\ \ \ 3\ \ \ 3
\\
3\ \ \ 3\ \ \ \ 3 \end {pmatrix} = 3A \Rightarrow A^2 = 3A \Rightarrow \dfrac{1}{3} A^2=A \Rightarrow
\\ \\ \\
\Rightarrow \left(\dfrac{1}{\sqrt3}\right)^2 A^2 = A \Rightarrow \left(\dfrac{\sqrt3}{ 3}\right)^2 A^2 = A \Rightarrow \left(\dfrac{\sqrt3}{3} A\right)^2 = A
\\ \\ \\
Matricea \ cerut\breve{a} \ este \ B = \left(\dfrac{\sqrt3}{3} A\right)^2 , \ pentru\ care\ B^2 = A [/tex]

Albastruverde12 Albastruverde12 · Answer 2 · 2018-08-05T13:36:39+03:00

[tex] \displaystyle La~b)~se~cere~intr-adevar~\bold{determinarea}~unei~matrice~B,~si~nu \\ \\ rezolvarea~ecuatiei,~asa~cum~am~crezut~initial.~Cel~mai~la~\\ \\ indemana~este~sa~cautam~matrici~de~forma \begin{pmatrix} a & a & a \\ a & a & a \\ a & a & a \end{pmatrix}. \\ \\ Gasim~usor~a= \pm \frac{1}{\sqrt{3}}. [/tex]

[tex] \displaystyle c)~Am~verificat,~iar~forma~corecta~este~urmatoarea: \\ \\ \det(A+xB)= (\det B)x^3+\text{tr}(AB^*)x^2+ \text{tr}(A^*B)x+\det A. \\ \\ (apropo,~sa~nu~uitam~ca~\text{tr}(UV)= \text{tr}(VU)~pentru~ca~UV~si~VU~au\\ \\ acelasi~polinom~caracteristic) \\ \\ Deci~\det(C+xA)=(\det A)x^3+ \text{tr}(CA^*)x^2+ \text{tr}(C^*A)x+ \det C \\ \\ \forall~x \in \mathbb{R}. \\ \\ Evident~\det A=0~si~A^*=O_3,~deci \\ \\ \det(C+xA)=qx+c~\forall~x \in \mathbb{C}.~Am~notat~q= \text{tr}(C^*A)~si~c= \det C. [/tex]

[tex] \displaystyle Relatia~de~demonstrat~devine~(qx+c)(-qx+c) \le c^2 \Leftrightarrow \\ \\ c^2-(qx)^2 \le c^2 \Leftrightarrow (qx)^2 \ge 0,~adevarat! [/tex]