## Linear Algebra and Its Applications (5th Edition)

By definition, $\vec{v}_{1}=(0,0,2),\ \vec{v}_{2}=(0,5,-8),\ \vec{v}_{3}$$=(-3,4,1)$ are linearly independent if and only if the vector equation $c_{1}\vec{v}_{1}+c_{2}\vec{v}_{2}+c_{3}\vec{v}_{3}=\vec{0}$ has only the trivial solution. In matrix form, this is equivalent to the condition that every column has a pivot, so that there are no free variables. Row reducing the coefficient matrix, we get: $\begin{bmatrix} 0&0&-3\\ 0&5&4\\ 2&-8&1 \end{bmatrix} \sim \begin{bmatrix} 2&-8&1\\ 0&5&4\\ 0&0&-3 \end{bmatrix}$. (Note that the only operations required are row-swaps.) Since every column has a pivot, it must be that the the vectors that form the matrix columns are linearly independent.