Determining whether a vector is an element of the row space. I've been having trouble with determining whether a vector is an element of the row space. Specifically, for the matrix:
$$\begin{bmatrix}-2 & 6 & 7 & -1 \\ 7 & -3 & 0 & -3 \\ 8 & 0 & 6 & 7\end{bmatrix}$$
Whether the vector:
$$\begin{bmatrix} 2 \\ 1 \\ 3 \\ -2 \end{bmatrix}$$ is an element.
My first attempt was to row reduce the original matrix to:
$$\begin{bmatrix}1 & 0 & 0 & \frac{-13}{4} \\ 0 & 1 & 0 & -\frac{79}{12} \\ 0 & 0 & 0 & \frac{32}{7}\end{bmatrix}$$.
Then, using the fact that the row-spaces of row-equivalent matrices are equivalent, I equated:
$$\begin{bmatrix} 2 \\ 1 \\ 3 \\ -2 \end{bmatrix} = \alpha _1\begin{bmatrix} 1 \\ 0 \\ 0 \\ -\frac{13}{4} \end{bmatrix} + \alpha _2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ -\frac{79}{12}\end{bmatrix} + \alpha _3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ \frac{32}{7}\end{bmatrix}$$
You thus get $$ \alpha _1 = 2, \alpha _2 = 1, \alpha _3 = 3$$ which gives $$2\frac{-13}{4} + \frac{-79}{12} + 3\frac{32}{7} = -2$$ by equating the final rows of the vector.
But in fact, $$2\frac{-13}{4} + \frac{-79}{12} + 3\frac{32}{7} = \frac{53}{84}$$, and so the vector is not an element of the row space.
However, by using a different method, consider the matrix $$\begin{bmatrix}8 & 7 & -2 & 2 \\ 0 & -3 & 6 & 1 \\ 6 & 0 & 7 & 3 \\ 7 & -3 & -1 & -2\end{bmatrix}$$, which is the augmented matrix of the transpose of the original and the vector, and noting that it row-reduces to the identity matrix, you can saw that the system of equations is consistent and hence the vector is an element of the row space.
But clearly these cannot both be true. Where is the error in my thinking?

If your reduced row echelon form (including the augmented part) reduces to the identity matrix, it is not consistent.
For example $$\left[\begin{array}{c|c} 1 & 0 \\ 0 & 1\end{array}\right]$$
implies that $x=0$ and $\color{blue}{0=1}$ which is not consistent.
