Change of basis for a vector

[burt]

I was given the following problem:

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 $a_1=3b_1-b_2,a_2=-b_1+5b_2+b_3,a_3=b_2-6b_3$. Find the change of coordinates from $A$ to $B$. Find $[x]_B$ for $x=4a_1+5a_2+a_3$.

It seems to me the matrix that changes the vector from $A$ to $B$ is $\left(\begin{matrix}3&-1&0\\-1&5&1\\0&1&-6\\\end{matrix}\right)$. I can take the inverse of that matrix and multiply it by $\left(\begin{matrix}4&5&1\\\end{matrix}\right)$. The only problem is that doing this gives me $\left(\begin{matrix}\frac{155}{87}&\frac{39}{29}&\frac{5}{87}\\\end{matrix}\right)$- a different answer than given in the answer key. I will attach a picture of the answer in the answer key.

 

[HallsofIvy]

I wouldn't worry about a "change of basis matrix" at all!

You are given that $\displaystyle a_1= 3b_1- b_2$, $\displaystyle a_2= -b_1+ 5b_2+ b_3$, and $\displaystyle a_3= b_2- 6b_3$.

I had a misprint before.

So $\displaystyle x= 4a_1+ 5a_2+ a_3= 4(3b_1- b_2)+ 5(-b_1+ 5b_2+ b_3)+ b_2- 6b_3= (12- 6)b_1- (4- 25- 1)b_2+ (5+ 6)b_3= 6b_1+ 22b_2+ 11b_3$.

 

[burt]

I wouldn't worry about a "change of basis matrix" at all!

You are given that $\displaystyle a_1= 3b_1- b_2$, $\displaystyle a_2= -b_1+ 5b_2+ b_3$, and $\displaystyle a_3= b_2- 6b_3$.

I had a misprint before.

So $\displaystyle x= 4a_1+ 5a_2+ a_3= 4(3b_1- b_2)+ 5(-b_1+ 5b_2+ b_3)+ b_2- 6b_3= (12- 6)b_1- (4- 25- 1)b_2+ (5+ 6)b_3= 6b_1+ 22b_2+ 11b_3$.

That is so much simpler! Thanks!

 

[burt]

I wouldn't worry about a "change of basis matrix" at all!

You are given that $\displaystyle a_1= 3b_1- b_2$, $\displaystyle a_2= -b_1+ 5b_2+ b_3$, and $\displaystyle a_3= b_2- 6b_3$.

I had a misprint before.

So $\displaystyle x= 4a_1+ 5a_2+ a_3= 4(3b_1- b_2)+ 5(-b_1+ 5b_2+ b_3)+ b_2- 6b_3= (12- 6)b_1- (4- 25- 1)b_2+ (5+ 6)b_3= 6b_1+ 22b_2+ 11b_3$.

Do you know why my work before didn't work?

 

[Steven G]

Sure.
Let that matrix you wrote be matrix A. All you had to do was compute A(4 5 1)^T and NOT A^-1(4 5 1)^T