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?
