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 $\theta = -45^\circ$.