Solution
= Solution
The <Pauli Y gate> satisfies $Y^\dagger=Y$ and $Y^2=I$. With $c=\cos(\pi/6)=\sqrt3/2$ and $s=\sin(\pi/6)=1/2$, the given <matrix> is <Hermitian>, and
$$
M_2^\dagger M_2=(cI+sY)^2=(c^2+s^2)I+2csY=I+\frac{\sqrt3}{2}Y\ne I.
$$
Therefore \b[$M_2$ is not a <unitary matrix>]. Its <eigenvalues> are $c+s$ and $c-s$, neither of which has modulus one. An expression $\cos\theta\,I+i\sin\theta\,Y$ would behave differently, but the printed <matrix> has no such factor of $i$.