# Chapter 4 - Vector Spaces - 4.4 Spanning Sets and Linear Independence - 4.4 Exercises - Page 178: 48

$S$ is not linearly independent .

#### Work Step by Step

Consider the combination $$a \left[\begin{array}{ll}{2} & {0} \\ {-3} & {1}\end{array}\right]+b\left[\begin{array}{cc}{-4} & {-1} \\ {0} & {5}\end{array}\right]+c\left[\begin{array}{ll}{-8} & {-3} \\ {-6} & {17}\end{array}\right]=\left[\begin{array}{ll}{0} & {0} \\ {0} & {0}\end{array}\right], \quad a,b,c\in R.$$ Which yields the following system of equations \begin{align*} 2a-4b-8c&=0\\ - b-3c&=0\\ -3a-6c&=0\\ a+5b+17c&=0. \end{align*} The coefficient matrix of the above system is given by $$\left[ \begin {array}{cccc} 2&-4&-8\\ 0&-1&-3 \\ -3&0&-6\\ 1&5&17 \end {array} \right].$$ Using Gauss elimination, the coefficient matrix has the row-echelon form $$\left[ \begin {array}{ccc} 2&-4&-8\\ 0&-1&-3 \\ 0&0&0\\ 0&0&0 \end {array} \right] .$$ Hence the system has non trivial solutions. Consequently, the set of vectors is not linearly independent

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