I have a set of vectors in $\mathbb{R}^{3}$: $\Biggl{\{}$$\begin{bmatrix}1\\2\\3\end{bmatrix}$,$\begin{bmatrix}-2\\2\\-3\end{bmatrix}$,$\mathbf{\vec{v}}$$\Biggl{\}}$
I want to be able to find the values for $\mathbf{\vec{v}}$ so that the set is linearly independent, but I am not sure what I need to do to find that. I know that the first two vectors are independent so I just need to find a single vector that is not in any relation to the first two so that it could make the set dependent.
What I have tried is turning the set into an augmented matrix to get:
$$ \left[\begin{array}{rrr|r} 1 & -2 & a & 0 \\ 2& 2& b& 0\\ 3& 3& c& 0 \end{array}\right] $$ Which becomes: $$ \left[\begin{array}{rrr|r} 1 & 0& \frac{(b+a)}{3} & 0 \\ 0& 1& \frac{3(b-2a)}{18}& 0\\ 0& 0& 2(c-3a)-3(B-2A)& 0 \end{array}\right] $$
But I do not know where to go after this, is there an easier way to solve this sort of question that I am missing?
How could I find $\mathbf{\vec{v}}$ so that this set is independent?
The idea is very good; the last column can be omitted, as it stays zero. I use a different elimination method, top down. \begin{align} \begin{bmatrix} 1 & 2 & a \\ 2 & 2 & b \\ 3 & -3 & c \end{bmatrix} &\to \begin{bmatrix} 1 & 2 & a \\ 0 & -2 & b-2a \\ 0 & -9 & c-3a \end{bmatrix}\\ &\to \begin{bmatrix} 1 & 2 & a \\ 0 & 1 & (2a-b)/2 \\ 0 & -9 & c-3a \end{bmatrix}\\ &\to \begin{bmatrix} 1 & 2 & a \\ 0 & 1 & (2a-b)/2 \\ 0 & 0 & c-3a+\frac{9}{2}(2a-b) \end{bmatrix} \end{align} The condition that the vectors are linearly independent is that $$ c-3a+\frac{9}{2}(2a-b)\ne0 $$ that can be rewritten $$ 12a-9b+2c\ne0 $$ Just find three numbers with this property and you're done.