find the associated matrix with respect to change in basis under identity transformation 
In each following cases, find $M_{B'}^B(id)$. The vector space in each case is $\Bbb R^3$
  1) $B = \{ ( 1,1,0), (-1,1,1 ), (0,1,2 )\}$
     $\;\;\;B' = \{ (2,1,1),(0,1,1), (-1,1,1)\}$
  2) $B = \{ ( 3,2,1), (0,-2,5), (1,1,2)\}$
     $\;\;\;B' = \{ (1,1,0),(-1,2,4), (2,-1,1)\}$  

Here $id$ is identity mapping. Since there was no example of this problem (this is from Linear Algebra - Serge Lang (Chapter VI section 3), I am confused how to do it. A solution of either of one or an example or links to it would suffice. 
 A: The procedure I described in my comment is actually not the most efficient one, since it makes you invert the same matrix three times...
Let us suppose we work with bases of $\mathbb{R}^n$, $B=\{x_1,\ldots,x_n\}$ and $B'=\{y_1,\ldots,y_n\}$. 
Now let us introduce the canonical basis of $\mathbb{R}^n$, $B_0=\{e_1,\ldots,e_n\}$.
By Theorem 3.4 in Lang, we have
$$
M_{B'}^B(id)=M_{B'}^{B_0}(id)M_{B_0}^B(id).
$$
And by Corollary 3.5, $M_{B'}^{B_0}(id)=M_{B_0}^{B'}(id)^{-1}$, so 
$$
M_{B'}^B(id)=M_{B_0}^{B'}(id)^{-1}M_{B_0}^B(id).
$$
Now observe that $M_{B_0}^B(id)$ (respectively $M_{B_0}^{B'}(id)$) is simply the matrix whose columns are the vectors $x_1,\ldots,x_n$ (resp. $y_1,\ldots,y_n$) with coordinates as they are given, namely in the canonical basis.
So all you have to is to invert once and for all $M_{B_0}^{B'}(id)$, and then multiply the two matrices.
A: What you should do was noted by @julien. I am writing some notation just to clear the point. Let $\{e_1,e_2,...,e_n\}$ is a basis for a vector space $V$, and $\{f_1,f_2,...,f_n\}$ be anothe basis for the vector space. Moreover assume that we can write each $f_i$ as a linear combination of $\{e_i\}_{i=1..n}$. I mean: $$f_1=a_{11}e_1+...+a_{1n}e_n\\f_2=a_{21}e_1+...+a_{2n}e_n\\.\\.\\.\\\\f_n=a_{n1}e_1+...+a_{nn}e_n$$ wherein $a_{ij}\in K$ our field. Therefore the transpose of the matrix: $$P=\left( \begin{array}{ccc}
a_{11} & a_{12} & ...& a_{1n} \\
a_{21} & a_{22} & ...& a_{2n} \\.\\.\\
a_{n1} & a_{n2} & ...& a_{nn} \end{array} \right)$$ is called the transition matrix from the basis $\{e_i\}_{i=1..n}$ to a new basis $\{f_i\}_{i=1..n}$. Note that since $\{f_i\}_{i=1..n}$ are independent so $P^{-1}$ is invertible.
