A quick counter-example to the question as stated is $a_0=0$, $a_1=1$ $a_2=-1$. Since $2arccos(0)-arccos(b)-arccos(-b)=0$ for all $b$, we have $2M_1-M_2-M_3=0$ where $M_1, M_2, M_3$ are the rows of the matrix.

So it is better to assume that the $a_i$ are nonnegative. In this case, the answer is yes. More generally, consider this problem for a function $f$. You want to choose $n$ linearly independent vectors from the set
$$
X := \{ (f(a_1x), f(a_2x),\dots, f(a_nx)) \mid x\in\mathbb R \} \subset \mathbb R^n
$$
This is not possible if and only if $X$ lies in an $(n-1)$-dimensional subspace. This means that there are constants $c_1,\dots,c_n$ (not all zero) such that $c_1v_1+\dots+c_nv_n=0$ for all $v\in X$. Or, equivalently,
$$
c_1 f(a_1x) + \dots c_n f(a_nx) = 0
$$
for all $x\in\mathbb R$ (such that all $a_ix$ belong to the domain of $f$). In other words, the functions $x\mapsto f(a_ix)$ are linearly dependent over $\mathbb R$.

This cannot happen if $f=arccos$. Indeed, assuming that $a_n$ is the maximum of $a_i$ such that $c_i\ne 0$, the above sum is well-defined and analytic on $[0,1/a_n)$ but its derivative goes to infinity as $x$ approaches $1/a_n$. Hence it is not constant on $[0,1/a_n)$, and hence non-constant in any neighborhood of 0.

UPDATE. A similar argument shows that the answer is the same for any non-polynomial function analytic near the origin (assuming $a_i>0$). Indeed, if $f(x)=\sum_{j=1}^\infty q_j x^{k_j}$ where $q_j\ne 0$, then Taylor expansion of the identity
$$
c_1 f(a_1x) + \dots c_n f(a_nx) = 0
$$
implies that $\sum_i c_i a_i^{k_j}=0$ for all $j$. This cannot happen because, if $a_n$ is the maximum of $a_i$, the term $a_i^{k_j}$ grows faster (or decays slower) than all other terms as $k_j\to\infty$.