I was given the following question (I am taking this course on my own which is why I am lacking a lot of clarity):
Let $A=${$a_1$,$a_2$,$a_3$} and $B=${$b_1,b_2,b_3$} be bases for a vector space $V$ and suppose $b_1=6a_1-6a_3,b_2=-a_1+4a_2,b_3=a_1+a_2+5a_3.$ Find the change of coordinates from $B$ to $A$. Find $[x]_A$ for $x=b_1-4b_2+4b_3$.
I am trying to understand the question so I can solve it and others like it and am looking for some guidance. These are my thoughts: The given $x$ at the end of the question is $x$ in the $B$-basis. So in the $B$-basis it can be written as $\left[\begin{matrix}1\\-4\\4\\\end{matrix}\right]$. We want to convert it to the $A$-basis. The relationship between $A$ and $B$ is given - each of the vectors of $B$ is a linear-combination of the $A$ vectors.
Here is where I'm unsure of what my next step should be. I can set up a system of equations using $[x]_B$ as my solution set - like this: $$6a_1-6a_3=1\\-a_1+4a_2=-4\\a_1+a_2+5a_3=4$$ I can set this up as a matrix: $\left[\begin{matrix}6&0&-6&1\\-1&4&0&-4\\1&1&5&4\\\end{matrix}\right]$ and solve to get my solution. This would make my solution: $[x]_A=\left[\begin{matrix}\frac{14}{15}\\\frac{-23}{20}\\\frac{23}{30}\\\end{matrix}\right]$.
I don't believe this answer is right, and I think I may be attempting this problem in entirely the wrong way. Can anyone help me out?
The approach looks correct to me though i didnt check if the calculation is correct.
I would anyway try a different aproach. From the equations $$ b_1=6a_1−6a_3,b_2=−a_1+4a_2,b_3=a_1+a_2+5a_3 $$
we can create a Matrix $M$ which changes the basis from $A$ to $B$. So that we have that $M[x]_A = [x]_B$ This Matrix is: $$ M = \left(\begin{matrix}6&0&-6\\-1&4&0\\1&1&5 \end{matrix}\right) $$ We can invert that Matrix and get $M^{-1}$ and then $M^{-1}[x]_B$ changes $[x]_B$ to $[x]_A$.