Question

(Mackenzie 1997) (x - y)/z = (x + z)/y = (z - y)/x = k Supondo k real não nulo, então o sistema anterior tem solução única:? alguém sabe? por favor ;)

Answer 1

$\mathrm{\dfrac{x-y}{z}=k\ \to\ x-y=kz\ \to\ x-y-kz=0}\\\\ \mathrm{\dfrac{x+z}{y}=k\ \to\ x+z=ky\ \to\ x-ky+z=0}\\\\ \mathrm{\dfrac{z-y}{x}=k\ \to\ z-y=kx\ \to\ kx+y-z=0}\\\\\\ \begin{cases}\ \mathrm{x-y-kz=0}\\ \ \mathrm{x-ky+z=0}\\ \ \mathrm{kx+y-z=0}\\ \end{cases}$

$\textbf{SPD}\ \to\ \mathbf{D\neq0}\\\\ \mathrm{D=\left[\begin{array}{ccc}1&-1&\mathrm{-k}\\1&\mathrm{-k}&1\\\mathrm{k}&1&-1\end{array}\right]\neq0}\\\\\\ \mathrm{1.(-k).(-1)+(-1).1.k+(-k).1.1-}\\ \mathrm{-(-k).(-k).k-1.1.1-(-1).1.(-1)\neq0}\\\\ \mathrm{k-k-k-k^3-1-1\neq0}\\\\ \mathrm{-k^3-k-2\neq0\ \to\ k^3+k+2\neq0}$

$\mathrm{*\ Poss\'iveis\ ra\'izes\ \to\ \pm\dfrac{divisores\ de\ a_0}{divisores\ de\ a_n}}}}\\\\ \mathrm{*\ Div.\ de\ 2\ \to\ 1,2\ \ \|\ \ *\ Div.\ de\ 1=1}\\\\ \mathrm{*\ Poss\'iveis\ ra\'izes\ \to\ \pm\dfrac{1,2}{1}=\pm1,\pm2}}}\\\\ \mathrm{*\ Para\ k=-1\ \to\ (-1)^3+(-1)+2=}\\ \mathrm{=-1-1+2=0\ \to\ -1\ \'e\ raiz}$

$\mathrm{*\ Fatorando\ a\ express\~ao:}\\\\ \mathrm{k^3+k+2=(k+1)\bigg(\dfrac{k^3+k+2}{k+1}\bigg)=}\\\\ \mathrm{=(k+1)(k^2-k+2)\ \to\ k^2-k+2\neq0}\\\\ \mathrm{*\ Aplicando\ a\ f\'ormula\ quadr\'atica:}\\\\ \mathrm{k\neq\bigg[\dfrac{-(-1)\pm\sqrt{(-1)^2-4.1.2}}{2.1}=}\\\\ \mathrm{=\dfrac{1\pm\sqrt{1-8}}{2}=\dfrac{1\pm\sqrt{-7}}{2}=\dfrac{1\pm i\sqrt{7}}{2}\bigg]}\\\\ \boxed{\mathrm{k\neq\bigg\{-1;\dfrac{1}{2}+\dfrac{\sqrt{7}}{2}i;\dfrac{1}{2}-\dfrac{\sqrt{7}}{2}i\bigg\}}}$

$\mathrm{*\ Como\ k\in\mathbb{R}\ \to\ \boxed{\mathbf{k\neq-1}}}\\\\ \textbf{Resposta:}\ \textrm{o sistema admite solu\c{c}\~ao \'unica}\\ \textrm{se (e somente se) k for diferente de -1. Ou}\\ \textrm{seja, teremos um SPD}\ \mathrm{\forall\ k\neq-1.}$