so i have the the basises $$B=\{b_1,b_2,b_3\}$$ where: $$b_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix},b_2=\begin{bmatrix} 1\\0\\1 \end{bmatrix},b_3=\begin{bmatrix} 1\\1\\1\end{bmatrix}$$
and $$C=\{c_1,c_2,c_3\}$$ where: $$c_1=\begin{bmatrix} 1\\1\\1 \end{bmatrix},c_2=\begin{bmatrix} 1\\-1\\0 \end{bmatrix},c_3=\begin{bmatrix} 1\\1\\-2\end{bmatrix}$$
And i am trying to find the change of basis matrix to $B$ to $C$ So this is supposed to be :
$$A_{B\to C}=\begin{bmatrix}[b_1]_c &[b_2]_c & [b_3]_c\end{bmatrix}$$ which after calculating the columns is: $$A= \begin{bmatrix} 1/3 & 2/3 & 1 \\ 1/2 & 1/2 & 0 \\ 1/6 & -1/6 & 0 \end{bmatrix}$$ which but this is not correct. what am i doing wrog?
Are you sure that is not the right answer? It seems alright to me.
By definition, the "base change matrix" expresses a vector in the old basis in terms of the vectors in the new basis.
For example, with the $A$ you have computed
$$Ab_1 = \begin{bmatrix} 1/3 & 2/3 & 1 \\ 1/2 & 1/2 & 0 \\ 1/6 & -1/6 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1/3 \\ 1/2 \\ 1/6 \end{bmatrix}$$
What this means is that the "coordinates" of $b_1$ in the new basis is $(1/3,1/2,1/6)$, which should satisfy
$$b_1 = \frac 13 c_1 + \frac 12 c_2 + \frac 16 c_3$$
and it does.