Question

Prove que os vetores $V_{1}=(1, \ 2, \ 1), \ \ V_{2}=(2, \ 9, \ 0), \ \ V_{3}=(3, \ 3, 4)$ formam uma base para R³.

Answer 1

$\boxed{\begin{array}{l}\sf(x,y,z)=a(1,2,3)+b(2,9,0)+c(3,3,4)\\\sf(0,0,0)=(a+2b+3c,2a+9b+3c,3a+4c)\\\begin{cases}\sf a+2b+3c=0\\\sf2a+9b+3c=0\\\sf3a+4c=0\end{cases}\\\begin{cases}\sf a+2b+3c=0\\\sf2a+9b+3c=0\\\sf c=-\dfrac{3}{4}a\end{cases}\\\sf 2a+9b+3\bigg(-\dfrac{3}{4}a\bigg)=0\\\sf 2a+9b-\dfrac{9}{4}a=0\cdot4\\\sf8a+36b-9a=0\\\sf 36b=9a-8a\\\sf 36b=a\\\sf b=\dfrac{a}{36}\\\sf a+2b+3c=0\\\sf a+2\cdot\bigg(\dfrac{a}{36}\bigg)+3\cdot\bigg(-\dfrac{3}{4}a\bigg)=0\cdot36\\\sf36a+2a-81a=0\\\sf-43a=0\implies a=0\\\sf b=\dfrac{0}{36}=0\\\sf c=-\dfrac{3}{4}\cdot0=0\\\sf o~conjunto~de~vetores~\acute e~LI.\end{array}}$

$\boxed{\begin{array}{l}\sf (a+2b+3c,2a+9b+3c,3a+4c)=(x,y,z)\\\begin{cases}\sf a+2b+3c=x\\\sf 2a+9b+3c=y\\\sf 3a+4c=z\end{cases}\\\\\begin{cases}\sf a+2b+3c=x\\\sf5b-3c=-2x+y\\\sf-6b-5c=-3x+z\end{cases}\\\\\begin{cases}\sf a+2b+3c=x\\\sf30b-18c=-12x+6y\\\sf-30b-25c=-15x+5z\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\begin{cases}\sf a+2b+3c=x\\\sf30b-18c=-12x+6y\\\sf-43c=-27x+6y+5z\end{cases}\\\sf -43c=-27x+6y+5z\cdot(-1)\\\sf 43c=27x-6y-5z\\\sf c=\dfrac{27x-6y-5z}{43}\\\sf 5b-3c=-2x+y\\\sf 5b-3\cdot\bigg(\dfrac{27x-6y-5z}{43}\bigg)=-2x+y\\\cdot(43)\\\sf215b-81x+18y+15z=-86x+43y\\\sf 215b=-86x+81x+43y-18y-15z\\\sf 215b=-5x+25y-15z\\\sf b=\dfrac{-5x+25y-15z}{215}\\\\\sf b=\dfrac{-x+5y-3z}{43}\end{array}}$

$\boxed{\begin{array}{l}\sf a+2b+3c=x\\\sf a+2\bigg(\dfrac{-x+5y-3z}{43}\bigg)+3\bigg(\dfrac{27x-6y-5z}{43}\bigg)=x\cdot(43)\\\\\sf43a-2x+10y-6z+81x-18y-15z=43x\\\sf 43a=43x+2x-81x-10y+18y+6z+15z\\\sf 43a=-36x+8y+21z\\\sf a=\dfrac{-36x+8y+21z}{43}\\\\\sf o~conjunto~gera~o~espac_{\!\!,}o\\\sf portanto~\acute e~base~de~\mathbb{R}^3\end{array}}$