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

To see if the columns are linearly independent, we reduce the matrix to row echelon form: $\begin{bmatrix} 0&-8&5\\ 3&-7&4\\ -1&5&-4\\ 1&-3&2 \end{bmatrix}\sim \begin{bmatrix} 1&-3&2\\ 3&-7&4\\ -1&5&-4\\ 0&-8&5 \end{bmatrix}\sim \begin{bmatrix} 1&-3&2\\ 0&2&-2\\ 0&2&-2\\ 0&-8&5 \end{bmatrix} \sim \begin{bmatrix} 1&-3&2\\ 0&2&-2\\ 0&0&-3\\ 0&0&0 \end{bmatrix}$ Since every column of the row-equivalent echelon matrix has a pivot, there are no free variables, so the columns are linearly independent.