Question

Sabendo que o sistema abaixo é possível e não admite solução trivial, determine k.

Por favor me ajudem!! urgente!!!

Answer 1

$\begin{cases} 2x+y+4z=k \\ 3x+2y+5z=k \\ x+\dfrac{y}{2}+2z=k^2 \end{cases}\Longrightarrow\mathrm{O\ sistema\ \acute{e}\ poss\acute{\i}vel.}$

$D=\left|\begin{array}{ccc}2&1&4\\3&2&5\\1&\frac{1}{2}&2\end{array}\right|\Longrightarrow D=0\Longrightarrow\mathrm{O\ sistema\ \acute{e}\ poss\acute{\i}vel\ e\ indeterminado.}$

$\Longrightarrow D_x=0,\ D_y=0\ \text{e}\ D_z=0.$

$D_x=0\Longrightarrow\left|\begin{array}{ccc}k&1&4\\k&2&5\\k^2&\frac{1}{2}&2\end{array}\right|=0\Longrightarrow\dfrac{-6k^2+3k}{2}=0$

$\Longrightarrow 3k(-2k+1)=0\Longrightarrow\boxed{k=0}\ \text{ou}\ \boxed{k=\dfrac{1}{2}}$

$D_y=0\Longrightarrow\left|\begin{array}{ccc}2&k&4\\3&k&5\\1&k^2&2\end{array}\right|=0\Longrightarrow 2k^2-k=0$

$\Longrightarrow k(2k-1)=0\Longrightarrow\boxed{k=0}\ \text{ou}\ \boxed{k=\dfrac{1}{2}}$

$D_z=0\Longrightarrow\left|\begin{array}{ccc}2&1&k\\3&2&k\\1&\frac{1}{2}&k^2\end{array}\right|=0\Longrightarrow\dfrac{2k^2-k}{2}=0$

$\Longrightarrow k(2k-1)=0\Longrightarrow\boxed{k=0}\ \text{ou}\ \boxed{k=\dfrac{1}{2}}$

$\boxed{k\neq0}\ \because\ \mathrm{A\ solu\c{c}\tilde{a}o\ n\tilde{a}o\ \acute{e}\ trivial.}\Longrightarrow\boxed{k=\dfrac{1}{2}}$