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ubuntu2004
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\begin{exercise}{V2}{Linear combinations}{0001}

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\begin{exerciseStatement}

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Consider the following statement.

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\begin{itemize}

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\item  The vector \( \left[\begin{array}{c}

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-6 \\

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1 \\

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7 \\

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-2

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\end{array}\right] \)is not a linear combination of the vectors \( \left[\begin{array}{c}

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4 \\

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-2 \\

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1 \\

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0

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\end{array}\right] , \left[\begin{array}{c}

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-5 \\

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0 \\

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4 \\

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-3

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\end{array}\right] , \left[\begin{array}{c}

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-3 \\

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1 \\

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-4 \\

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-1

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\end{array}\right] , \text{ and } \left[\begin{array}{c}

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-2 \\

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-2 \\

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2 \\

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-2

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\end{array}\right] \). 

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\end{itemize}

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\begin{enumerate}[(a)]

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\item  Write an equivalent statement using a vector equation.

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\item  Explain why your statement is true or false.

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\end{enumerate}

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    \end{exerciseStatement}

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\begin{exerciseAnswer}\[\operatorname{RREF}  \left[\begin{array}{cccc|c}

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4 & -5 & -3 & -2 & -6 \\

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-2 & 0 & 1 & -2 & 1 \\

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1 & 4 & -4 & 2 & 7 \\

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0 & -3 & -1 & -2 & -2

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\end{array}\right] = \left[\begin{array}{cccc|c}

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1 & 0 & 0 & 0 & -1 \\

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0 & 1 & 0 & 0 & 1 \\

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0 & 0 & 1 & 0 & -1 \\

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0 & 0 & 0 & 1 & 0

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\end{array}\right] \]

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\begin{enumerate}[(a)]

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\item  The vector equation \( x_{1} \left[\begin{array}{c}

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4 \\

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-2 \\

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1 \\

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0

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\end{array}\right] + x_{2} \left[\begin{array}{c}

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-5 \\

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0 \\

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4 \\

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-3

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\end{array}\right] + x_{3} \left[\begin{array}{c}

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-3 \\

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1 \\

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-4 \\

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-1

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\end{array}\right] + x_{4} \left[\begin{array}{c}

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-2 \\

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-2 \\

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2 \\

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-2

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\end{array}\right] = \left[\begin{array}{c}

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-6 \\

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1 \\

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7 \\

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-2

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\end{array}\right] \)has no solutions.

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\item 

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\( \left[\begin{array}{c}

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-6 \\

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1 \\

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7 \\

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-2

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\end{array}\right] \) is a linear combination of the vectors \( \left[\begin{array}{c}

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4 \\

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-2 \\

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1 \\

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0

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\end{array}\right] , \left[\begin{array}{c}

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-5 \\

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0 \\

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4 \\

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-3

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\end{array}\right] , \left[\begin{array}{c}

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-3 \\

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1 \\

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-4 \\

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-1

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\end{array}\right] , \text{ and } \left[\begin{array}{c}

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-2 \\

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-2 \\

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2 \\

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-2

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\end{array}\right] \). 

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\end{enumerate}

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    \end{exerciseAnswer}

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\end{exercise}

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