Hey I want to check my solutions for this problem.
Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the linear map with the following representation matrix in the canonical basis: $A=\begin{pmatrix} -1 &1 &2 \\ 3& 3 &4 \\ 2& 1 & 1 \end{pmatrix}$
Determine $M_B^B(T)$, where $B$ is the following base of $\mathbb{R}^3$:
$B:=(f_1=\begin{pmatrix}1\\1\\-1\end{pmatrix},f_2=\begin{pmatrix}1\\0\\1\end{pmatrix},f_3=\begin{pmatrix}1\\1\\0\end{pmatrix})$
First, I have calculated $T(f_1,f_2,f_3)$
The results are $T(f_1)=\left(\begin{matrix} -2 \\ 2 \\ 2 \end{matrix}\right) \\ T(f_2)=\left(\begin{matrix} 1 \\ 7 \\ 3 \end{matrix}\right) \\T(f_3)=\left(\begin{matrix} 0 \\ 6 \\ 3 \end{matrix}\right)$
So after that I solved the linear systems
$af_1+bf_2+cf_3=T(f_1),T(f_2),T(f_3)$ And I reached $-6f_1-4f_2+8f_3=T(f_1)$
$-9f_1-6f_2+16f_3=T(f_2)$
$-9f_1-6f_2+15f_3=T(f_3)$
Thus obtaining the matrix $M_B^B(T)=\begin{pmatrix} -6 &-9 &-9 \\ -4& -6 &-6 \\ 8& 16 &15 \end{pmatrix}$
Everything Right or have I done any mistake?
Yes, it is correct, directly of the definition $M_{B}^{B}(T):=\begin{bmatrix}\uparrow & \uparrow& \uparrow\\ [T(f_1)]_{B}&[T(f_2)]_{B}& [T(f_3)]_{B}\\\downarrow& \downarrow & \downarrow \end{bmatrix}$, with $[T(f_k)]_{B}$ the coordinates of the vector $T(f_k)$ respect to the bases $B$ and it is obtained as your solution.
Alternately, denote $\beta$ the canonical bases of $R^3$ and $\beta'$ other bases of $R^3$ (in this case $B$ as in the problem). Then we can write $[T]_{\beta'}=Q_{\beta'}^{-1}[T]_{\beta}Q_{\beta'}$, where $[T]_{\beta}=\begin{bmatrix} -1 &1 &2 \\ 3& 3 &4 \\ 2& 1 & 1 \end{bmatrix}$ and $Q_{\beta'}=\begin{bmatrix}1&1&1\\1&0&1\\-1&1&0 \end{bmatrix}$ (the change coordinates matrix of $\beta'$-coordinates into $\beta$-coordinates). Therefore (calculating $Q_{\beta'}^{-1}$ and then calculating $Q_{\beta'}^{-1}[T]_{\beta}Q_{\beta'}$) we have $[T]_{\beta'}=\begin{bmatrix}-6&-9&-9\\-4&-6&-6\\8&16&15\end{bmatrix}$ or $M_{B}^{B}(T)$ using your notation.