= Solution
Set $c=\cos(2\alpha_1)$ and $s=\sin(2\alpha_1)$. Multiplying the <matrices>, or using the <Pauli matrix multiplication law>, gives
$$
A_1^\dagger\tau_1A_1=\tau_1,\qquad
A_1^\dagger\tau_2A_1=c\tau_2+s\tau_3,\qquad
A_1^\dagger\tau_3A_1=c\tau_3-s\tau_2.
$$
Collecting the coefficients of $i\tau_j$ in $M'$ yields
$$
\boxed{\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}
=\begin{pmatrix}1&0&0\\0&c&-s\\0&s&c\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}.}
$$
This is the right-handed active <rotation> by $2\alpha_1$ about the positive $x_1$ axis. Likewise, with $c_j=\cos(2\alpha_j)$ and $s_j=\sin(2\alpha_j)$,
$$
\boxed{A_2:\ (x_1,x_2,x_3)\mapsto(c_2x_1+s_2x_3,\ x_2,\ -s_2x_1+c_2x_3),}
$$
$$
\boxed{A_3:\ (x_1,x_2,x_3)\mapsto(c_3x_1-s_3x_2,\ s_3x_1+c_3x_2,\ x_3).}
$$
Thus the other axes also have <rotation> angle $2\alpha_j$. The factor of two is essential: replacing $A_j$ by $-A_j$ leaves the <conjugation> unchanged, illustrating the two-to-one relation between <SU(2)> and three-dimensional <rotations>.
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