What value of n makes the span of vectors equal to R3

Question

Im asked the question what values of $n$ makes

$$ span\begin{pmatrix} {\begin{bmatrix} 1\\0\\1 \end{bmatrix},\begin{bmatrix} 1\\1\\0 \end{bmatrix},\begin{bmatrix} 1\\1\\n \end{bmatrix}} \end{pmatrix} = \mathbb{R}^3 $$

What I did is:

$$ \begin{bmatrix} 1&1&1&|&b_{1} \\ 0&1&1&|&b_{2} \\ 1&0&n&|&b_{3} \end{bmatrix} \\ $$ So when I solve I get: $$ \begin{bmatrix} 1&0&0&|&b_{1}-b_{2} \\ 0&1&1&|&b_{2} \\ 0&0&n&|&b_{3}+b_{2}-b_{1} \end{bmatrix} $$ So that $n=b_{3}+b_{2}-b_{1}$ Is this correct or am I completely wrong?

Answer 1

You're not completely wrong, but you're making it unnecessary complicated.

You have $3$ vectors $v_1,v_2,v_3\in\mathbb R^3$. As $\dim(\mathbb R^3)=3$ we have: $$\operatorname{span}(v_1,v_2,v_3)=\mathbb R^3 \iff v_1,v_2,v_3~\text{are linearly independent}.$$

To check for which $n\in\mathbb N$ these vectors are linearly independent the easiest way would involve calculating the determinant of the matrix $\begin{pmatrix} v_1 & v_2 & v_3\end{pmatrix}$...can you take it from here?

Answer 2

Solving system of linear equations \begin{align} a_1 +a_2+a_3&=0\\ a_2+a_3&=0\\ a_1+a_3n&=0 \end{align} indicates that $a_1=0, a_2=-a_3$ and $a_3n=0$. Third equality imply that all $n \neq 0$ will make the given vectors L.I.