How do I find all values $x$ such that a vector is a linear combination of a nonempty set of vectors in vector space $\mathbb{R^3}$? In vector space $\mathbb{R^3}$.
Find all values $x$ such that $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$.
My solution:
I used the equation:
$r_1\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}$ + $r_2\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$ + $r_3\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$ = $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$
which can be represented with the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
  1&0&1&2x^2\\
  1&1&2&-3x\\
  3&1&4&1\\
\end{array}
\right] $$
which has an RREF of:
$$ \left[
\begin{array}{ccc|c}
  1&0&1&2x^2\\
  0&1&1&-3x-2x^2\\
  0&0&0&1-4x^2+3x\\
\end{array}
\right] $$
I understood this as the system could only be consistent iff:
$1-4x^2+3x = 0$
So then, I solved for $x$ which yields $x = 1,\frac {-1}{4} $.
Is it correct to say, then, that $1$ and $\frac {-1}{4}$ are all the values of $x$ that will make $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}$ $\in$ span $\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \}$?
 A: Yes. You approach is correct. Also (if you are allowed) you can remove the vector $\begin{bmatrix}1\\2\\4\end{bmatrix}$ by writing
$$
\text{span}\left\{\begin{bmatrix}1\\1\\3\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\4\end{bmatrix}\right\}
=
\text{span}\left\{\begin{bmatrix}1\\1\\3\end{bmatrix},\begin{bmatrix}0\\1\\1\end{bmatrix}\right\}
$$
since
$$
\begin{bmatrix}1\\2\\4\end{bmatrix}=
\begin{bmatrix}1\\1\\3\end{bmatrix}
+
\begin{bmatrix}0\\1\\1\end{bmatrix}
.
$$
A: Here's another approach. Just for your own entertainment.
Using techniques from multivariable calculus you can write $$\text{span}\Bigg\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\Bigg\}=\big\{(x,y,z)\in \mathbb{R}^3:2x+y-z=0\big\}$$ Therefore $\begin{bmatrix} 2x^2 \\ -3x \\ 1 \end{bmatrix}\in \text{span}\Bigg\{\begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}\Bigg\}$ if and only if $$2(2x^2)-3x-1=0 \iff (4x+1)(x-1)=0$$ So $x=-1/4,1$.
