Find \(\vec{z} = {2} \vec{u} - {2} \vec{v} + {2} \vec{w} \text{,}\) where
\begin{equation*}
\vec{u} = \left[ \begin{array}{r} {-4} \\ {3} \\ {-4} \end{array}\right], \hspace{0.5cm}
\vec{v} = \left[ \begin{array}{r} {4}\\ {4} \\ {-1} \end{array}\right], \hspace{0.5cm} \text{and} \hspace{0.5cm}
\vec{w} = \left[ \begin{array}{r} {4}\\ {-3} \\ {2} \end{array}\right].
\end{equation*}
\(\displaystyle{}\)
\(\vec{z} = \left[ \right.\) , , \(\left. \right]^{\T}\)
Answer 1.
Answer 2.
Answer 3.
Solution.
\begin{equation*}
\vec{z} = \left[ \begin{array}{r}
({2} \cdot {-4}) - ({2} \cdot {4}) + ({2} \cdot {4})\\
({2} \cdot {3}) - ({2} \cdot {4}) + ({2} \cdot {-3})\\
({2} \cdot {-4}) - ({2} \cdot {-1}) + ({2} \cdot {2})\\
\end{array} \right] =
\left[ \begin{array}{r}
({-8}) - ({8}) + ({8})\\
({6}) - ({8}) + ({-6})\\
({-8}) - ({-2}) + ({4})\\
\end{array} \right] =
\left[ \begin{array}{r}
{-8}\\ {-8} \\ {-2}
\end{array} \right]
\end{equation*}