#### Answer

Yes, the columns of this matrix form a linearly independent set of vectors.

#### Work Step by Step

To see if the columns are linearly independent, we reduce the matrix to row echelon form:
$ \begin{bmatrix} -4&-3&0\\0&-1&4\\1&0&3\\5&4&6 \end{bmatrix}\sim \begin{bmatrix} 1&0&3\\0&-1&4\\-4&-3&0\\5&4&6 \end{bmatrix}\sim \begin{bmatrix} 1&0&3\\0&-1&4\\0&-3&12\\0&4&-9 \end{bmatrix} \sim \begin{bmatrix} 1&0&3\\0&-1&4\\0&0&7\\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.