The problem described can be solved as follows. Let
M = m_11 m_12 m_13
m_21 m_22 m_23
m_31 m_32 m_33
denote the desired rotation matrix. We require
1 0 0 * M + t = x_x x_y x_z
0 1 0 y_x y_y y_z
0 0 1 z_x z_y z_y
where t denotes the translation; we see that this matrix equality can be solved by multiplying from the left with the identity matrix, which is the inverse of itself; hence we obtain the following equality.
M + t = x_x x_y x_z
y_x y_y y_z
z_x z_y z_y
This can be rearranged by subtracting t from both sides to obtain the desired matrix M as follows.
M = x_x x_y x_z - t = x_x-t_x x_y-t_y x_z-t_z
y_x y_y y_z y_x-t_x y_y-t_y y_z-t_z
z_x z_y z_y z_x-t_x z_y-t_y z_z-t_z
Note that this was relatively easy as the initial matrix consists out of the basic vectors of the standard base. In general it is more difficult and involves a basis transformation, which basically can be done by Gaussian elimination, but can be numerically difficult.