Let $A=\{(1,0,5) (4, 5,5)(1,1,4)\}; B= \{(1,3,2)(-2,-1,1)(1,2,3) \}$. To find change of basis matrix.
$\boxed{ M_{A\to B} = M_{A\to e}M_{e\to B} } ---(1)\\= \begin{pmatrix} 1 & 4 & 1 \\ 0 & 5 & 1 \\ 5 & 5 & 4 \end{pmatrix}^{-1} \begin {pmatrix} 1 & -2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end {pmatrix} = \boxed {\begin {pmatrix} -2 & -2 & -1 \\ 0 & -1 & 0 \\ 3 & 4 & 2 \end {pmatrix} } -----(2)$
Is equation (1) correct??? or it is $M_{A \to B} = M_{e \to B} M_{A \to e}$ as given in solution part of LINK:- How to construct change of basis matrix (I am confused here pls correct me if i am wrong...)
To reconfirm i did it in following way:-
to find coordinates of B interms of A
$(1,3,2)= a(1,0,5)+b(4,5,5)+c(1,1,4)=(a+4b+c,5b+c,5a+5b+4c)=[-2,0,3]_B$
Similarly $(-2,-1,4) \sim [-2,-1,4]_B \\ (1,2,3) \sim [-1,0,2]_B$
THis boils down to same above matrix by (2)
We need
$$M_{A\to B} = M_{e\to B}M_{A\to e}$$
that is
$$\begin {pmatrix} 1 & -2 & 1 \\ 3 & -1 & 2 \\ 2 & 1 & 3 \end {pmatrix}^{-1}\begin{pmatrix} 1 & 4 & 1 \\ 0 & 5 & 1 \\ 5 & 5 & 4 \end{pmatrix} $$
indeed the first matrix takes a vector in basis $A$ to the standard basis and the second takes a vector in the standard basis to the basis $B$, therefore the product matrix takes a vector from basis $A$ to basis $B$.