\begin{exercise}{V2}{Linear combinations}{0000}
\begin{exerciseStatement}
Consider the following statement.
\begin{itemize}
\item The vector \( \left[\begin{array}{c}
-2 \\
-6 \\
-6 \\
-4
\end{array}\right] \)is a linear combination of the vectors \( \left[\begin{array}{c}
3 \\
2 \\
-5 \\
-3
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
-3 \\
1
\end{array}\right] , \text{ and } \left[\begin{array}{c}
4 \\
-3 \\
-3 \\
0
\end{array}\right] \).
\end{itemize}
\begin{enumerate}[(a)]
\item Write an equivalent statement using a vector equation.
\item Explain why your statement is true or false.
\end{enumerate}
\end{exerciseStatement}
\begin{exerciseAnswer}\[\operatorname{RREF} \left[\begin{array}{ccc|c}
3 & 0 & 4 & -2 \\
2 & 1 & -3 & -6 \\
-5 & -3 & -3 & -6 \\
-3 & 1 & 0 & -4
\end{array}\right] = \left[\begin{array}{ccc|c}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right] \]
\begin{enumerate}[(a)]
\item The vector equation \( x_{1} \left[\begin{array}{c}
3 \\
2 \\
-5 \\
-3
\end{array}\right] + x_{2} \left[\begin{array}{c}
0 \\
1 \\
-3 \\
1
\end{array}\right] + x_{3} \left[\begin{array}{c}
4 \\
-3 \\
-3 \\
0
\end{array}\right] = \left[\begin{array}{c}
-2 \\
-6 \\
-6 \\
-4
\end{array}\right] \)has a solution.
\item
\( \left[\begin{array}{c}
-2 \\
-6 \\
-6 \\
-4
\end{array}\right] \) is not a linear combination of the vectors \( \left[\begin{array}{c}
3 \\
2 \\
-5 \\
-3
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
-3 \\
1
\end{array}\right] , \text{ and } \left[\begin{array}{c}
4 \\
-3 \\
-3 \\
0
\end{array}\right] \).
\end{enumerate}
\end{exerciseAnswer}
\end{exercise}
