Solution
= Solution
\b[Yes.] Suppose $\alpha_1a_1+\alpha_2a_2+\alpha_3a_3=0$. Taking its <dot product> with $a_2\times a_3$ eliminates the last two terms and gives
$$
\alpha_1\,a_1\cdot(a_2\times a_3)=0.
$$
The nonzero <scalar triple product> forces $\alpha_1=0$. Cyclically using $a_3\times a_1$ and $a_1\times a_2$ forces $\alpha_2=\alpha_3=0$ as well. This is the definition of <linear independence>. Equivalently the three-column <matrix> has nonzero <determinant>.