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Let $\vec{v}_1 = \left<12,-4,-4\right>, \ \vec{v}_2 = \left<6,-2,-4\right>, \ \vec{v}_3 = \left<3,-1,-4\right>$. Are the vectors $\vec{v}_1, \vec{v}_2$ and $\vec{v}_3$ linearly independent?
v1 = (12,-4,-4); v2 = (6,-2,-4); v3= (3,-1,-4); A = column_matrix([v1,v2,v3]); A, A.rref()
Since the rref of $A$ has a free column (the 3rd column), its columns, i.e. $\vec{v}_1, \vec{v}_2, \vec{v}_3$ are linearly dependent. Moreover, $\bm{1/2 \\ -3/2 \\1}$ is a nontrivial solution of the system $A\vec{x} = \vec{0}$ (obtained by setting the free variable to be $1$), so $$ \frac{1}{2}\vec{v}_1 - \frac{3}{2}\vec{v}_2 + \vec{v}_3 = \vec{0}. $$
Note that, any nontrivial solution of the system give rise to a nontrivial linear relation of the $\vec{v}_i$'s
Let $\vec{v}_1 = \left<-3,-3\right>, \ \vec{v}_2 = \left<-4,-1\right>, \ \vec{v}_3 = \left<-3,6\right>$. Are the vectors $\vec{v}_1, \vec{v}_2$ and $\vec{v}_3$ linearly independent?
v1 = (-3,-3); v2 = (-4,-1); v3 = (-3,6); A = column_matrix([v1,v2,v3]); A, A.rref()
A linear independent set of $K^2$ cannot have more than 2 vectors, so we know that $\vec{v}_1, \vec{v}_2, \vec{v}_3$ must be linearly dependent. (This is reflected by the fact that a matrix with more columns than rows must have a free column in its rref (see above)).
Since $\bm{3\\-3\\1}$ is a nontrivial solution to $A\vec{x} = \vec{0}$. Hence $3\vec{v}_1-3\vec{v}_2+\vec{v}_3 = \vec{0}$.
Note that, any nontrivial solution of the system give rise to a nontrivial linear relation of the $\vec{v}_i$'s
Construct a $4 \times 3$ matrix $A = \lbrack \ \vec{a}_1 \ \ \vec{a}_2 \ \ \vec{a}_3 \ \rbrack$ such that $\lbrace \vec{a}_1, \vec{a}_2 \rbrace$ is a linearly independent set and $\vec{a}_3$ is not in the span of $\lbrace \vec{a}_1, \vec{a}_2 \rbrace$.
It is easy to produce two linearly independent vectors: take $\vec{a}_1 = \bm{1 \\ 0 \\ 0 \\ 0}$ and $\vec{v}_2 = \bm{0 \\1 \\ 0 \\ 0}$. If we take $\vec{v}_3 = \bm{0 \\ 0 \\ 1 \\0}$ then $\vec{v}_3$ is not in the span of $\vec{v}_1$ and $\vec{v}_2$ since any linear combination of them will have the 3rd entry 0 but the 3rd entry of $\vec{v}_3 =1$.