Change of basis, am I doing this right? The basis B in space $\Bbb R^3$, 
where 
$$b_1 = (1, 1, -2)$$ 
$$  
b_2 = (0, 1, -1) $$ 
$$ 
b_3 = (-1, -1, 3) $$
Give the change of basis  $: E\rightarrow B$. E is the standard basis in $\Bbb R^3$. 
So the solution I got is:
$$A=\begin{pmatrix}2&1&1 \\ -1&1&0 \\ 1&1&1  \end{pmatrix} $$.
But my friend seem to have different solution. Am I doing something wrong? 
 A: Let be $w$ any vector given in the canonical basis and indicate with: $b_1=(1,1,-2)$, $b_2=(0,1,-1)$ and $b_3=(-1,-1,3)$ the three vectors of the basis $\mathcal{B}$, also given with respect to the canonical basis.
We are loking for the coefficient $x_1$, $x_2$ and $x_3$ such that:
$$w=x_1\cdot b_1+x_2\cdot b_2+x_3\cdot b_3$$
or in matrix form:
$$w=B\cdot x$$
$$B=\begin{bmatrix}
    1  & 0 & -1\\
    1  & 1 & -1\\
    -2 &-1 & 3
    \end{bmatrix}$$
Note that $B$ contains as column the vectors of the new basis with respect to the canonical basis. It is important to note that $B$ represent the matrix of change of basis from $\mathcal{B}$ to the canonical.
Thus, the components of any vector $w$ with respect to the new basis are given by:
$$x=B^{-1}\cdot w=A\cdot w$$
$$A=B^{-1}=\begin{bmatrix}
    2  & 1 & 1\\
    -1 & 1 & 0\\
    1  & 1 & 1
    \end{bmatrix}$$
Thus $A=B^{-1}$ represent the matrix of change of basis from the canonical to the new basis $\mathcal{B}$.
So I think you have done well. 

Keep in mind that you can always verify the result by the product $AB$.

