Let $V$ be a vector space over the field $R$. Let the following bases of $V$ be given: $$\begin{align} \mathcal{B}&=(v_1,v_2,v_3,v_4)\\ \mathcal{C}&=(v_1,2v_1+v_2-v_4,3v_1+v_3+2v_4,-2v_1+v_2+2v_4) \\ \mathcal{D}&=(2v_2+v_3,2v_1-v_3+v_4,v_1+v_2,3v_1-2v_4)\end{align}$$ Specify the matrices for the following coordinate changes: a) from $\mathcal{B}$ to $\mathcal{C}$
I'm not really sure if it's right to just take every vector of $\mathcal{B}$ and represent $\mathcal{B}$ by using $\mathcal{C}$:
$\begin{align} v_1&=1v_1+0v_2+0v_3+0v_4\\ v_2&=2v_1+v_2+0v_3-v_4\\ v_3&=\dots\\ \vdots& \end{align}$
Therefore, $M^B_C=\begin{pmatrix}1 & 2 & 3 & -2\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 2 & 2\end{pmatrix}$
Is this correct?
Just take the vectors in $\mathcal B$ and put them as a linear combination of the elements of $\mathcal C$. For example, for $v_1$ it's easy to see that $$v_1=1v_1+0(2v_1+v_2-v_4)+0(3v_1+v_3+2v_4)+0(-2v_1+v_2+2v_4)$$ So $[v_1]_\mathcal{C}=(1,0,0,0)^t$. For $v_2$ we need to find some scalars $a_1,a_2,a_3,a_4$ such that $$v_2=a_1v_1+a_2(2v_1+v_2-v_4)+a_3(3v_1+v_3+2v_4)+a_4(-2v_1+v_2+2v_4)$$ i.e. $$\;v_2=(a_1+2a_2+3a_3-2a_4)v_1+(a_2+a_4)v_2+a_3v_3+(-a_2+2a_3+2a_4)v_4$$ From here, we can conclude that $$ \left\{ \begin{align} a_1+2a_2+3a_3-2a_4=0 \\ a_2+a_4=1 \\ a_3=0 \\ -a_2+2a_3+2a_4=0 \end{align} \right. $$ Solving this, we have $a_1=-2/3$, $a_2=2/3$, $a_3=0$ and $a_4=1/3$. Hence $$[v_2]_\mathcal{C}=\left(-\frac{2}{3},\frac{2}{3},0,\frac{1}{3}\right)^t$$ and this is what would go in the second column of your matrix. Repeat the same process for $v_3$ and $v_4$ and it's done. $\textsf{Edit:}$ Just to make sure, I add the rest: $$[v_3]_\mathcal{C}=\left(-\frac{17}{3},\frac{2}{3},1,-\frac{2}{3}\right)^t$$ $$[v_4]_\mathcal{C}=\left(\frac{4}{3},-\frac{1}{3},0,\frac{1}{3}\right)^t$$ Therefore, the change of basis matrix is the following $$\textsf{M}_{\mathcal C}^{\mathcal B}=\begin{pmatrix} 1&-\tfrac{2}{3}&-\tfrac{17}{3}&\tfrac{4}{3} \\ 0&\tfrac{2}{3}&\tfrac{2}{3}&-\tfrac{1}{3} \\ 0&0&1&0 \\ 0&\tfrac{1}{3}&-\tfrac{2}{3}&\tfrac{1}{3} \end{pmatrix}$$ $\textsf{Another way to do so are:}$ Note that the vectors of $\mathcal{C}$ can be written easily as a linear combination of the elements in $\mathcal B$, like that $$v_1=1v_1+0v_2+0v_3+0v_4$$ $$2v_1+v_2-v_4=2v_1+1v_2+0v_3+(-1)v_4$$ $$3v_1+v_3+2v_4=3v_1+0v_2+1v_3+2v_4$$ $$-2v_1+v_2+2v_4=(-2)v_1+1v_2+0v_3+2v_4$$ Hence $$\textsf{M}_{\mathcal B}^{\mathcal C}=\begin{pmatrix} 1&2&3&-2 \\ 0&1&0&1 \\ 0&0&1&0 \\ 0&-1&2&2 \end{pmatrix}$$ And then $$\textsf{M}_{\mathcal C}^{\mathcal B}=\big(\textsf{M}_{\mathcal B}^{\mathcal C}\big)^{-1}=\begin{pmatrix} 1&2&3&-2 \\ 0&1&0&1 \\ 0&0&1&0 \\ 0&-1&2&2 \end{pmatrix}^{-1}=\begin{pmatrix} 1&-\tfrac{2}{3}&-\tfrac{17}{3}&\tfrac{4}{3} \\ 0&\tfrac{2}{3}&\tfrac{2}{3}&-\tfrac{1}{3} \\ 0&0&1&0 \\ 0&\tfrac{1}{3}&-\tfrac{2}{3}&\tfrac{1}{3} \end{pmatrix}$$ It may be easier, maybe not.