Problemas Resueltos - Razonamiento matemático
Problema 43
Si \( x = (-2, -2) \), \( y = (-1, \, 3) \) y se sabe que \( -\vec{x} + 2 \vec{y} + 3 \vec{z} = (6, \, 5) \), entonces necesariamente \( \vec{z} \) es igual a:
-
\( (-6, \, 3) \)
-
\( (2, -1) \)
-
\( \left( \frac{2}{3}, \, 1 \right) \)
-
\( (-2, \, 1) \)
-
\( (6, \, 1) \)
Intenta resolverlo antes de ver la respuesta...
-
\( (2, -1) \)
En efecto:
\[
\begin{aligned}
&-\vec{x} + 2\vec{y} + 3\vec{z} = (6, \, 5)
\\[.5em]
&\hspace{.2em}\Rightarrow
– (-2, -2) + 2 (-1, \, 3) + 3( z_1, \, z_2 )
\\
&\hspace{5em} = (6, \, 5)
\\[.5em]
&\hspace{.2em}\Rightarrow
3 (z_1, \, z_2)
\\
&\hspace{2em} = (6, \, 5) + (-2, -2) -2(-1, \, 3)
\\[.5em]
&\hspace{.2em}\Rightarrow
(3 z_1, \, 3z_2)
\\
&\hspace{2em} = (6, \, 5) + (-2, -2) + (2, -6)
\\[.5em]
&\hspace{.2em}\Rightarrow
(3z_1, \, 3z_2) = (6-2 + 2, \, 5 – 2 – 6)
\\[.5em]
&\hspace{.2em}\Rightarrow
(3z_1 , \, 3z_2) = (6, -3)
\\[.5em]
&\hspace{.2em}\Rightarrow
3z_1 = 6 \quad \text{ y } \quad 3z_2 = -3
\\[.5em]
&\hspace{.2em}\Rightarrow
z_1 = 2 \quad \text{ y } \quad z_2 = -1
\\[.5em]
&\hspace{.2em}\Rightarrow
\vec{z} = (z_1, \, z_2) = \boldsymbol{ (2, -1) }
\end{aligned}
\]
\[
\begin{aligned}
-\vec{x} + 2\vec{y} + 3\vec{z} = (6, \, 5)
&\Rightarrow
– (-2, -2) + 2 (-1, \, 3) + 3( z_1, \, z_2 ) = (6, \, 5)
\\[.5em]
&\Rightarrow
3 (z_1, \, z_2) = (6, \, 5) + (-2, -2) -2(-1, \, 3)
\\[.5em]
&\Rightarrow
(3 z_1, \, 3z_2) = (6, \, 5) + (-2, -2) + (2, -6)
\\[.5em]
&\Rightarrow
(3z_1, \, 3z_2) = (6-2 + 2, \, 5 – 2 – 6)
\\[.5em]
&\Rightarrow
(3z_1 , \, 3z_2) = (6, -3)
\\[.5em]
&\Rightarrow
3z_1 = 6 \quad \text{ y } \quad 3z_2 = -3
\\[.5em]
&\Rightarrow
z_1 = 2 \quad \text{ y } \quad z_2 = -1
\\[.5em]
&\Rightarrow
\vec{z} = (z_1, \, z_2) = \boldsymbol{ (2, -1) }
\end{aligned}
\]