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Pe: 2021-01-08T17:22:14+02:00 In: In:

Va rog frumos, ma poate ajuta cineva …

Va rog frumos, ma poate ajuta cineva la aceasta tema?

Multumesc!

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  1. Suport

    1.a)V este definit peste numerele complexe. Luand scalarul i si vectorul \bigl(\begin{smallmatrix} 1\\i\\i \end{smallmatrix}\bigr) care apartine lui V, obtinem ca i\bigl(\begin{smallmatrix} 1\\i\\ i \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix} i\\-1\\-1 \end{smallmatrix}\bigr), care nu apartine lui V deoarece i nu este numar real.
    b)V=\left \{ \begin{pmatrix} x_1\\x_2 \\x_3 \end{pmatrix}|x_1=0\right\}=\left \{ \begin{pmatrix} 0\\x_2 \\x_3 \end{pmatrix}\right\}=\left \{ x_2\begin{pmatrix} 0\\ 1\\0 \end{pmatrix}+x_3\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right\}=span\left \{ \begin{pmatrix} 0\\ 1\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right\}
    Dimensiunea acestui spatiu este 2, o baza fiind cea data de cei 2 vectori din span.
    c)V=\left \{ x=\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} | x_1+x_2=0 \right \}=\left \{ x=\begin{pmatrix} x_1\\-x_1\\x_3 \end{pmatrix}\right \}=\left \{ x=x_1\begin{pmatrix} 1\\-1\\0 \end{pmatrix}+x_3\begin{pmatrix} 0\\0\\1 \end{pmatrix}\right \}=span\left \{ \begin{pmatrix} 1\\-1\\0 \end{pmatrix}, \begin{pmatrix} 0\\0\\1 \end{pmatrix}\right \}
    Dimensiunea lui V este 2, o baza este formata din cei 2 vectori din span.
    d)V nu este spatiu vectorial deoarece vectorul nul nu se afla in V, pentru ca nu respecta conditia x_1+x_2=1.

    2.v_3=bv_1+cv_2+dv_4
    \begin{pmatrix}1\\-1\\2\\0\end{pmatrix}=b\begin{pmatrix}2\\a\\1\\1\end{pmatrix}+c\begin{pmatrix}-1\\1\\0\\1\end{pmatrix}+d\begin{pmatrix}0\\1\\-1\\1\end{pmatrix}
    \begin{pmatrix}1\\-1\\2\\0\end{pmatrix}=\begin{pmatrix}2b\\ab\\b\\b\end{pmatrix}+\begin{pmatrix}-c\\c\\0\\c\end{pmatrix}+\begin{pmatrix}0\\d\\-d\\d\end{pmatrix}
    \begin{pmatrix}1\\-1\\2\\0\end{pmatrix}=\begin{pmatrix}2b-c\\ab+c+d\\b-d\\b+c+d\end{pmatrix}
    \left\{\begin{matrix} 2b-c=1\\ab+c+d=-1\\b-d=2\\b+c+d=0 \end{matrix}\right.
    Din a 3-a ecuatie, avem ca b=d+2. Inlocuind in ultima, d+2+c+d=0, deci c=-2d-2. Inlocuind in prima, 2d+4-c=1, deci c=2d+3. Atunci -2d-2=2d+3, adica -5=4d, d=-\frac{5}4. Rezulta ca b=-\frac54+2=-\frac54+\frac84=\frac34  si c=-2d-2=\frac52-2=\frac12.In final, din prima ecuatie obtinem ca \frac34a+\frac12-\frac54=-1, adica 3a+2-5=-4, 3a=-1 si deci a=-\frac13.

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