Q:

Find the matrix of the linear transformation $f^*$ on the orthonormal basis on which all the vectors are given.

Linear transformation $f$ maps $a_1=(0,0,1), a_2=(0,1,1), a_3=(1,1,1)$ to $b_1=(1,2,1), b_2=(3,1,2), b_3=(7,-1,4)$. Find the matrix of $f^*$ on the orthonormal basis on which all the vectors are given. $f^*$ is conjugate transformation of $f$.

I got two matrices constructed by the coordinates of both $\langle a_1,a_2,a_3 \rangle$ and $\langle b_1,b_2,b_3 \rangle $
$$
A = \begin{bmatrix}
0 & 0 & 1 \\ 
0 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix} \text{ and } B  = \begin{bmatrix}
1 & 3 & 7 \\ 
2 & 1 & -1 \\
1 & 2 & 4 \\
\end{bmatrix} $$
To find the matrix of $f$ on this basis calculated $T_f=BA^{-1}$
$$ 
T_f = \begin{bmatrix}
-2 & -4 & 7 \\ 
1 & 2 & -1 \\
-1 & -2 & 4 \\
\end{bmatrix}$$
To find the matrix of $T_{f^*}$, I calculated $T_{f^*}=\big(\overline{T_f}\big)^T={T_f}^T$ (since it is real-valued).
But the answer $\textit{in the book}$ is quite different:
$$ 
T_{f^*} = \begin{bmatrix}
4 & -2 & 2 \\ 
2 & -1 & 1 \\
1 & 2 & 1 \\
\end{bmatrix}^T$$
Where is my mistake?

A:

Thanks to @amd, I found my mistake. It seems I missed an error in finding $A^{-1}$. I corrected that and now my answer is the same as in the book.
$$ 
T_{f^*} = \begin{bmatrix}
4 & 2 & 1 \\ 
-2 & -1 & 2 \\
2 & 1 & 1 \\
\end{bmatrix}$$
Thanks!

