Change of basis matrix to convert standard basis to another basis

Consider the basis $B=\left\{\begin{pmatrix} -1 \\ 1 \\0 \end{pmatrix}\begin{pmatrix} -1 \\ 0 \\1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\1 \end{pmatrix} \right\}$ for $\mathbb{R}^3$.

A) Find the change of basis matrix for converting from the standard basis to the basis B.

I have never done anything like this and the only examples I can find online basically tell me how to do the change of basis for "change-of-coordinates matrix from B to C".

B) Write the vector $\begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}$ in B-coordinates.

Obviously I can't do this if I can't complete part A.

Can someone either give me a hint, or preferably guide me towards an example of this type of problem?

The absolute only thing I can think to do is take an augmented matrix $[B E]$ (note - E in this case is the standard basis, because I don't know the correct notation) and row reduce until B is now the standard matrix. This is basically finding the inverse, so I doubt this is correct.

Denote $E$ the canonical basis of $\mathbb{R}^3$.

A) These three column vectors define a $3\times 3$ matrix $$P=\left(\matrix{-1&-1&1\\1&0&1\\0&1&1}\right)$$ which is the matrix of the linear map $$ Id:(\mathbb{R}^3,B)\longrightarrow (\mathbb{R}^3,E). $$ This means in particular that whenever you right multiply it by a column vector $(x_1,x_2,x_3)$ where $x_j$ are the coordinates of a vector $x=x_1B_1+x_2B_2+x_3B_3$ with the respect to the basis $B$, you obtain the coordinates of $x$ in the canonical basis $E$.

What you want is the matrix of $$ Id:(\mathbb{R}^3,E)\longrightarrow (\mathbb{R}^3,B). $$ That is $P^{-1}$, the inverse of the matrix above. This will transform, by right multiplication, the coordinates of a vector with respect to $E$ into its coordinates with respect to $B$. That's the change of basis matrix you need.

B) As explained above, you just have to right multiply the change of basis matrix $P^{-1}$ by this column vector.

Check your answer: you should find

$$P^{-1}=\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)$$ $$\left(\matrix{-1/3&2/3&-1/3\\-1/3&-1/3&2/3\\1/3&1/3&1/3} \right)\left(\matrix{1\\0\\0}\right)=\left(\matrix{-1/3\\-1/3\\1/3}\right).$$

By definition change of base matrix contains the coordinates of the new base in respect to old base as it's columns. So by definition $B$ is the change of base matrix. Key to solution is equation $v = Bv'$ where $v$ has coordinates in old basis and $v'$ has coordinates in the new basis (new basis is B-s cols) suppose we know that in old basis $v$ has coords $(1,0,0)$ (as a column) (which is by the way just an old base vector) and we want to know $v'$ (the old base vector coordinates in terms of new base) then from the above equation we get $$B^{-1}v = B^{-1}Bv' \Rightarrow B^{-1}v = v'$$

As a side-node, sometimes we want to ask how does that change of base matrix B act if we look at it as linear transformation, that is given vector v in old base $v=(v_1,...,v_n)$, what is the vector $Bv$? In general it is a vector whith i-th coordinate bi1*v1+...+bin*vn (dot product of i-th row of $B$ with $v$). But in particular if we consider v to be an old base vector having coordinates (0...1...0) (coordinates in respect the old base) where 1 is in the j-th position, then we get $Bv = (b_{1j},...,b_{nj})$ which is the j-th column of B, which is the j-th base vector of the new base. Thus we may say that B viewed as linear transformation takes old base to new base.