Solution
= Solution
Writing the <matrix> explicitly gives
$$
M=\begin{pmatrix}ix_3&ix_1+x_2\\ix_1-x_2&-ix_3\end{pmatrix}.
$$
Its <determinant> is
$$
\det M=x_3^2-(ix_1+x_2)(ix_1-x_2)
=\boxed{x_1^2+x_2^2+x_3^2.}
$$
Since $\det A_1^\dagger=(\det A_1)^*=1$,
$$
\boxed{\det(A_1^\dagger MA_1)=\det M=|\mathbf x|^2.}
$$
The <conjugation> therefore preserves the squared Euclidean length encoded in $M$.