Question

Let a={a1, a2, a3, a4}, b={b1,b2,b3,b4} and c ={c1, c2,c3,c4} be a basis for vector space V. Suppose we have
b1=3a1
b2=6a1+4a2
b3=5a1+4a2+7a3
b4=6a1+7a2+8a3+9a4
and
c1=5b4
c2=10b3+6b4
c3=7b2+5b3+4b4
c4=3b1+5b2+4b3+9b4

Find the transition matrix from basis a to the basis c

Answer 1

\[M_{a \rightarrow b} =\left[\begin{matrix}3 & 0&0&0 \\ 6&4&0&0\\5&4&7&0\\6&7&8&9\end{matrix}\right]\]\[M_{b \rightarrow c} = \left[\begin{matrix} 0&0&0&5 \\ 0&0&10&6\\0&7&5&4\\3&5&4&9\end{matrix}\right]\]

Answer 2

I get \[M_{b \rightarrow c}M_{a \rightarrow b} = \left[\begin{matrix} 30&35&40&45\\86&82&118&54 \\91&76&67&36\\113&99&100&81\end{matrix}\right]\]

Answer 3

What else do I need to do?

Answer 4

I have a stupid question. How did you get a 4by4 only given 3 basis for V?

Answer 5

I see you chose a=(1,0,0,0)
so b=(3,6,5,6)
so c=(30,86,91,113)
I'm having trouble getting the rest.

Answer 6

That's how the problem was set up.
For example, b1=3a1 so the matrix for that is (3 0 0 0) because a2, a3, a4 aren't included in it.
I'm not sure if that is what you were asking