TITLE: How is b not a linear combination of these vectors?
QUESTION [6 upvotes]: Determine if $\vec b$ is a linear combination of $\vec a_1,\vec a_2,\vec a_3$.
$\vec a_1 = \left[\begin{array}{c}
1     \\
-2   \\
0    \\
\end{array}\right], \vec a_2 = \left[\begin{array}{c}
0    \\
1   \\
2    \\
\end{array}\right], \vec a_3=\left[\begin{array}{c}
5     \\
-6   \\
8    \\
\end{array}\right], \vec{b} = \left[\begin{array}{c}
2    \\
-1   \\
6    \\
\end{array}\right]$
Okay, so I made my constants $x_{1}, x_{2}, x_{3}$ for $\vec a_1, \vec a_2,\vec a_3$, respectively. I end up getting the following consistent system:
$$\left[\begin{array}{cccc}
1 & 0 & 5 & 2    \\
0 & 1 & 4 & 3   \\
0 & 0 & 0 & 0    \\
\end{array}\right]$$
Which has the general solution: $$
\begin{cases}
x_{1} = 2 - 5x_{3} \\
x_{2} = 3 - 4x_{3} \\
x_{3} = \text{free}.
\end{cases}
$$
So $\vec b$ is equal to infinitely  many linear combinations of $\vec a_1,\vec a_2,\vec a_3$, right? Why does my book say that $\vec b$ is not a linear combination of these three vectors? Must the constants be a unique solution?

REPLY [6 votes]: The book is wrong.  $\vec{b}$ is a linear combination of those three vectors.  (e.g. $\vec{b} = 2\vec{a_1} + 3\vec{a_2}$, among other combinations.)