Finding a specific solution from a linearly dependant set of vectors

Question

So I have the matrix, $$\begin{pmatrix}1&4&-2\\0&0&0\end{pmatrix}$$ in the form $$\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}$$

If i want to find a specific solution showing dependance I need,

$$\lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3}=0$$ where we have, $$\begin{pmatrix}\lambda_{1}+4\lambda_{2}-2\lambda_{3}\\ 0\end{pmatrix}$$ so, $$ \lambda_{1}+4\lambda_{2}-2\lambda_{3} = 0$$ $$\lambda_{1} = -4\lambda_{2}+2\lambda_{3}$$ Where $\lambda_{2}$ and $\lambda_{3}$ are arbitrary

So i thought I'd just take $\lambda_{2} = 1$ and $\lambda_{3} = 1$ so $\lambda_{1}= -2$

But this doesnt work out. What is it im misunderstanding?

Answer 1

$$\lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3}=0$$

Careful with your notation: you have a linear combination of vectors on the left-hand side, so the right-hand side should be the zero vector:

$$\lambda_1 \begin{pmatrix}1 \\ 0\end{pmatrix}+\lambda_2 \begin{pmatrix}4 \\ 0\end{pmatrix}+\lambda_3 \begin{pmatrix}-2 \\ 0\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}$$ and from this you indeed have $\lambda_{1}+4\lambda_{2}-2\lambda_{3} = 0$.

So i thought I'd just take $\lambda_{2} = 1$ and $\lambda_{3} = 1$ so $\lambda_{1}= -2$

But this doesnt work out. What is it im misunderstanding?

What do you mean, "doesn't work out"? Notice that:

$$-2 \begin{pmatrix}1 \\ 0\end{pmatrix}+ 1\begin{pmatrix}4 \\ 0\end{pmatrix}+ 1\begin{pmatrix}-2 \\ 0\end{pmatrix}= \begin{pmatrix} -2+4-2 \\ 0\end{pmatrix}= \begin{pmatrix}0 \\ 0\end{pmatrix}$$