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

By definition, $\vec{v}_{1}=(5,0,0),\ \vec{v}_{2}=(7,2,-6),\ \vec{v}_{3}$$=(9,4,-8)$ 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: $\begin{bmatrix} 5&7&9\\ 0&2&4\\ 0&-6&-8 \end{bmatrix} \sim \begin{bmatrix} 5&7&9\\ 0&1&2\\ 0&3&4 \end{bmatrix} \sim \begin{bmatrix} 5&7&9\\ 0&1&2\\ 0&0&2 \end{bmatrix}$. Since every column has a pivot, it must be that the the vectors that form the matrix columns are linearly independent.