Denote by S the global system with the origin $P$ and the normal $N$.
Denote by S' the local coordinate system of the tank with the origin $Q$ and z-axis parallel to $N$.
Then the transformation between the coordinates $x = (x, y, z)^T$ in the S system
and the coordinate $x' = (x', y', z')^T$ in the S' system is given by
$$x = M x' + Q$$
M is the given orthogonal matrix whose columns are the unit vectors of the S' system
expressed in the S system and $Q$ is the translation vector also expressed in the S system.
Now the tank rotates about its (local) z-axis by the angle $\theta = -45^\circ$ (ie. to the right) and moves k units ahead,
say in the direction of its positive (local) x-axis. This constitutes a new local
coordinate system S''. The relation between S' and S'' is given by the general formula:
$$x' = R(\theta) x'' + t$$
where
\begin{align}
R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}
\end{align}
is the rotation matrix and
$$t = t(\theta) = k (\cos(\theta), \sin(\theta), 0)^T$$
is the translation vector.
By substituting $x'$ into the first equation we get
$$x = M R(\theta) x'' + M t(\theta) + Q$$
It follows that the matrix
$M$ has to be multiplied by the matrix $R(\theta)$ and the vector $Q$ has to be shifted by by the the vector $M t(\theta)$, where $\theta = -45^\circ$.