Contradiction between RREF & Determinant in evaluating Linear Independence

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Working through Anton's Elementary Linear Algebra and there is an example that aims to determine the linear independence or dependence of a system, however I believe the RREF of the coefficients suggests it is linearly independent, whereas the determinant suggests it is linearly dependent.

Determine whether the vectors: $\vec{v}_1 = (1, 2, 2, -1), \vec{v}_2 = (4, 9, 9, -4), \vec{v}_3 = (5, 8, 9, -5)$ in $\mathbb{R}^4$ are linearly independent or linearly dependent.

For the system to be linearly independent the vector equation $k_1 \vec{v}_1 + k_2 \vec{v}_2 + k_3\vec{v}_3 = \vec{0}$ must have only the trivial solution, so:

$k_1 (1,2,2,-1) + k_2 (4,9,9,-4) + k_3 (5,8,9,-5) = (0,0,0,0)$. From this the coefficient matrix can be formed as such:

$\begin{bmatrix} 1 & 4 & 5 & 0 \\ 2 & 9 & 8 & 0 \\ 2 & 9 & 9 & 0 \\ -1 & -4 & -5 & 0 \end{bmatrix}$, which when converted into RREF can be expressed as $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.

From RREF it can be shown that $k_1, k_2, k_3 = 0$ and thus the system is linearly independent. However, because the final row are all $0$s, the determinant must be $0$, which implies linear dependence.

Does one evaluation take more weight than the other or am I looking at these concepts incorrectly? Any guidance would be greatly appreciated, cheers

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I think you are misunderstanding what it means to use the determinant to check whether a set of vectors is linearly independent. Your $4 \times 4$ matrix does indeed have a determinant of zero -- and what that means is that the four columns of the matrix are linearly dependent. That is,

$$\begin{bmatrix} 1 \\ 2 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} 4 \\ 9 \\ 9 \\ -4 \end{bmatrix}, \begin{bmatrix} 5 \\ 8 \\ 9 \\ -5 \end{bmatrix}, \textrm{ and } \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

are linearly dependent.

Another way to look at it: you are computing the determinant of an augmented matrix, in which the last column is all zeroes. Such a matrix will always have a zero determinant, regardless of what the other three columns are. So that can't possibly be a reliable way to find whether the other three columns are linearly dependent.