unit vector transformation in rotation of coordinate system The transformation of coordinate from $(x, y, z)$ to $(x', y', z')$ is given by
$$ x' = a_{11}x + a_{12}y + a_{13}z\\
y' = a_{21}x + a_{22}y + a_{23}z\\
z' = a_{31}x + a_{32}y + a_{33}z$$
If $(\hat i , \hat j, \hat k)$ and $(\hat i_1 , \hat j_1, \hat k_1)$  be unit vectors in two coordinates. Equating $ \vec r = \vec r'$ I got 
$$\begin{bmatrix}
\hat i\\ 
\hat j\\ 
\hat k
\end{bmatrix} =
 \begin{bmatrix}
a_{11} & a_{21} & a_{31}\\ 
a_{12} & a_{22} & a_{32}\\
a_{13} & a_{23} & a_{33}
\end{bmatrix} \times 
\begin{bmatrix}
\hat i_1\\ 
\hat j_1\\ 
\hat k_1
\end{bmatrix}$$
The help manual says 
$$\begin{bmatrix}
\hat i_1\\ 
\hat j_1\\ 
\hat k_1
\end{bmatrix} =
 \begin{bmatrix}
a_{11} & a_{21} & a_{31}\\ 
a_{12} & a_{22} & a_{32}\\
a_{13} & a_{23} & a_{33}
\end{bmatrix} \times 
\begin{bmatrix}
\hat i\\ 
\hat j\\ 
\hat k
\end{bmatrix}$$
I don't know if the manual is wrong or I am wrong. Please help.
The original problem is states as 

Show that the operator $\nabla$ remains invariant passing from one rectangular Cartesian system of axes to another.

 A: Which $(i,j,k)$ corresponds to which $(x,y,z)$? In general, if $A$ is a matrix whose row vectors are the basis vectors for coordinates $(i,j,k)$ and $B$ is a matrix whose row vectors are the basis vectors for coordinates $(i_1,j_1,k_1)$ then if they span the same space, there is an invertible 3x3 matrix $R$ so that $A=RB$ and for $\alpha$ in the basis given by $A$, its image $\beta$ in the basis given by $B$ is $β=R^Tα$.
This makes sense intuitively since $R$ expresses the vectors for $A$ as a linear combination of the vectors in $B$, so anything in terms of $A$ multiplied by $R$ will give you the vector as a linear combination in terms of $B$ i.e. under the basis given by $B$.  So if,
$$\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix}  $$
Then correspondingly by the explanation above:
$$ \begin{bmatrix} \hat{i} \\ \hat{j} \\ \hat{k} \end{bmatrix} = \begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}^T \begin{bmatrix} \hat{i}_1 \\ \hat{j}_1 \\ \hat{k}_1 \end{bmatrix} = \begin{bmatrix}a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33} \end{bmatrix} \begin{bmatrix} \hat{i}_1 \\ \hat{j}_1 \\ \hat{k}_1 \end{bmatrix}. $$
So unless I'm making the same mistake you are, I believe you're right.
