Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$. I am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 7, I was puzzled at some of it. Here is the problem description:

Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$.

I don't understand its description well.
I think it wants to get a span set like {$(0,1,1)$, $(1,0,0)$, $(0,0,1)$} which is a basis of $R^3$.
And I check the reference answer, which is as followings:

$(0,1,1)$, $(1,0,0)$, $(0,1,0)$ is one choice among many.

I think what I have done is what question wants.
So can anyone tell me am I right or wrong? 
Thanks sincerely.
 A: Note that 
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$ has independent rows. Hence you have found $3$ independent vectors in $\mathbb{R}^3$, that is it spans $\mathbb{R}^3$ and it forms a basis.
A: There is a kind of 'procedure' for dealing with questions of this kind, namely to consider the spanning set $\left\{ w_1, e_1, e_2, e_3\right\}$. Consider each vector from left to right. If one of these vectors is in the span of the previous one/s, then throw it out. If not, keep it. So in this case, we start by keeping $w_1$. Moving to the next vector, $e_1$ is not in the span of $w_1$, so we keep it as well. Moving to the next, $e_2$ is not in the span of the previous two vectors so we keep it as well. Now, considering the vector $e_3$ we see that it is in fact in the span of the previous three vectors, since $e_3 = w_1 - e_2.$ So we throw out the vector $e_3$ and end up with the basis $\left\{ w_1, e_1, e_2\right\}$. This explains the solution in the reference answer. Your solution is also correct, however. 
A: You are correct.  {$(0,1,1),(1,0,0),(0,0,1)$} is a basis of $\mathbb R^3$.  
Any element $(a,b,c)$ in $\mathbb R^3$ can be expressed as $a(1,0,0)+b(0,1,1)+(c-b)(0,0,1).$ 
