Cfrgtkky

$$B=left{begin{bmatrix}
1 \
2 \
3
end{bmatrix},
begin{bmatrix}
3 \
2 \
1
end{bmatrix},
begin{bmatrix}
1 \
1 \
0
end{bmatrix}right}
quad C=left{begin{bmatrix}
4 \
5 \
6
end{bmatrix},
begin{bmatrix}
2 \
4 \
7
end{bmatrix},
begin{bmatrix}
6 \
7 \
8
end{bmatrix}right} $$

I need to find such a matrix
$A=begin{bmatrix}
a & b & c \
d & e & f \
g & h & i \
end{bmatrix}$ that transforms vectors from $B$ into vectors from $C$, so for example:

$$begin{bmatrix}
a & b & c \
d & e & f \
g & h & i \
end{bmatrix} begin{bmatrix}
1 \
2 \
3 \
end{bmatrix}=begin{bmatrix}
4 \
5 \
6 \
end{bmatrix}text{ etc...}$$
I want to understand the general idea of solving such tasks, so please don't give away the answer!

Consider the following so called commutative diagram:
$$require{AMScd}begin{CD}
mathbb R^3 @>A>> mathbb R^3\
@A(b_1 b_2 b_3)AA @AA(c_1 c_2 c_3)A \
B @>>I> C end{CD}$$
If you have a vector with respect to basis $B$, such as (1,0,0), this vector is actually $mathbf b_1$, the first vector in the basis of $B$. More generally, (x,y,z) with respect to $B$ is $xmathbf b_1+ymathbf b_2+zmathbf b_3$. It means that we can convert a vector with respect to $B$ to the standard basis of $mathbb R^3$ by simply putting the basis vectors of $B$ as columns in a matrix.

In your case, you're interested in the conversion from $B$ to $C$, which is effectively the identity matrix $I$. That is, for instance (1,0,0) with respect to $B$, which is $mathbf b_1$, must be mapped to (1,0,0) with respect to $C$, which is $mathbf c_1$.

To find the corresponding matrix $A$ with respect to the standard basis, we need to follow the arrows, and in reverse where necessary:
$$A = begin{pmatrix}mathbf c_1&mathbf c_2&mathbf c_3end{pmatrix}
cdot I cdotbegin{pmatrix}mathbf b_1&mathbf b_2&mathbf b_3end{pmatrix}^{-1}
$$
Reading from right to left, we start with a vector in the standard basis, convert it to $B$ by the corresponding inverse, multiply by $I$, and convert from $C$ to the standard basis.

Here's a good way to start:
let
$$
E = Bigg{begin{bmatrix}
1 \
0 \
0
end{bmatrix},
begin{bmatrix}
0 \
1 \
0
end{bmatrix},
begin{bmatrix}
0 \
0 \
1
end{bmatrix}Bigg}
$$
and ask "Can I find a transformation taking the $E$ vectors to the $B$ vectors?" The answer is "sure, just put the $B$ vectors into the columns of a 3x3 matrix."

You should do this and convince yourself that this matrix, which I'll call $M_{Eto B}$does in fact take the $E$ vectors to the corresponding $B$ vectors.

Now...what matrix would do the opposite, would take the $B$ vectors to the $E$ vectors?

It's $M_{Eto B}^{-1}$, of course! So we now know that
$$
M_{B to E} = M_{E to B}^{-1}.
$$

Now you can do the same thing and find a matrix that takes the $E$ vectors to the $C$ vectors, right?

Now what happens if you multiply the $E$ vectors by the matrix
$$
Q = M_{Eto B}M_{C to E}?
$$

If you take it one step at a time, you'll see that $c_1$ gets send to $e_1$, which then (by the first matrix) gets sent to $b_1$.

Try this out on a $2 times 2$ example to convince yourself it works --- it's a lot easier to invert a $2 times 2$ matrix!