<s>[INST] Are the vectors linearly dependent or independent? Problem: Let $\vec{v}=(5,9), \ \vec{u}=(3,-2)$ and $\vec{w}=(2,1)$. Determine the nature of their linear dependency.
Attempt: So, we are looking for constants $k_1,k_2,k_3$ such that $k_1\vec{v}+k_2\vec{u}+k_3\vec{w}=0.$ We can write this as 
$$k_1\left[\begin{matrix} 5 \\ 9 \end{matrix}\right]+k_2\left[\begin{matrix} 3 \\ -2 \end{matrix}\right]+k_3\left[\begin{matrix} 2 \\ 1 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
which in canonical form is a system of linear equations in terms of $k_1,k_2$ and $k_3$.
$$
M\vec{k}=\left[ {\begin{array}{cc}
   5 & 3 & 1 \\
   9 & -2 & 1 \\
\end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right].$$
Row echelon form on $M$ gives 
$$
\left[ {\begin{array}{cc}
   5 & 3 & 1 \\
   0 & -\frac{37}{5} & -\frac{4}{5} \\
\end{array} } \right],$$
This means that there are infinite solutions, but in order for them to be linearly independant, there should only exist one unique solution. Thus we have shown that the vectors $\vec{v},\vec{u},\vec{w}$ are linearly dependant. This means that $\text{span}\{\vec{v},\vec{u},\vec{w}\}=\mathbb{R}^2,$ because the addition of one of the vectors does not add another dimension to the span of the other two. The third vector that is linearly dependant of the other two, lies in their span.
Have I understood the concept of linear dependancy and span correctly? Any constructive input is very welcome!
 [/INST] You have made a mistake, it should be:
$$M\vec{k}=\left[ {\begin{array}{cc}
   5 & 3 & 2 \\
   9 & -2 & 1 \\
\end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
The solutions are of the form $\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right]$ for $\alpha \in \mathbb{R}$.
Thus, $$-7\vec{v}-13\vec{u}+37\vec{w} = 0$$
so $\{\vec{v}, \vec{u}, \vec{w}\}$ is linearly dependent in $\mathbb{R}^2$.
In fact, any three distinct vectors in a two-dimensional space must be linearly dependent.
Edit:
I found the solutions by reducing the matrix to row echelon form:
$$\left[ {\begin{array}{cc}
   5 & 3 & 2 \\
   9 & -2 & 1 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
   5 & 3 & 2 \\
   -1 & -8 & -3 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
   1 & 8 & 3 \\
   5 & 3 & 2 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
   1 & 8 & 3 \\
   0 & -37 & -13 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
   1 & 0 & \frac{7}{37} \\
   0 & 1 & \frac{13}{37} \\
\end{array} } \right]$$
The first row gives $k_1 = -\frac{7}{37}k_3$, and the second row gives $k_2 = -\frac{13}{37}k_3$.
Thus, the solutions are given by $k_3\cdot \left[\begin{matrix} -\frac{7}{37} \\ -\frac{13}{37} \\ 1 \end{matrix}\right]$ for $k_3 \in \mathbb{R}$, which we can also write as $$\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right] \text{    for } \alpha\in\mathbb{R}$$ by setting $k_3 = 37\alpha$.
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