# A trivial solution vs. a non-trivial solution - involving vectors

I'm not entirely sure I understood this question in my text book, but it said the following:

The zero vector $0 = \left(0,0,0\right)$ can be written as a linear combination of the vectors $v_1$, $v_2$, and $v_3$ because $0 = 0v_1+0v_2+0v_3$. This is called the trivial solution. Can you find a nontrivial way of writing $0$ as a linear combination of the three vectors?

$v_1=\left(1,0,1\right)$,

$v_2=\left(-1,1,2\right)$

$v_3=\left(0,1,4\right)$

Having no idea how to solve it I did it this way:

$a\begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix}$ + $b\begin{bmatrix} -1 \\ 1 \\ 2 \\ \end{bmatrix}$ + $c\begin{bmatrix} 0 \\ 1 \\ 4 \\ \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 2 & 4 & 0 \\ \end{bmatrix}$

Reducing this to $rref$ I got: $\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 2 & 5 & 0 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}$

Does this satisfy the question?

Assuming your row reduction is correct, what you have is the system of equations $a-c=0$, $2b+5c=0$ and $c=0$. This tells you $a=b=c=0$.