Finding a coordinate vector for specific basis

Question

I am having trouble with a specific problem concerning coordinate vectors. I have a basis $B' = \{t^2 + 2, t + 3, t^2 + t + 1\} $and a polynomial $p(t) = 4t^2 - t$. so I believed the way to go about finding a coordinate vector with respect to a specific basis was to do the following: $[4t^2 -t]_{B'} = a(t^2 + 2) + b(t + 3) + c(t^2 + t + 1)$ Now I tried solving this by throwing it into the matrix $$ \left(\!\! \begin{array}{ccc|c} 1 & 0 & 2 & 4 \\ 0 & 1 & 3 & -1 \\ 1 & 1 & 1 & 0 \end{array} \!\!\right) $$ so i got a wild solution consisting of fractions and it didn't add up. I got $$ \frac{1}{4} \begin{pmatrix} 10 \\ -13 \\ 3 \end{pmatrix} $$ (I tried RREF) so I'm left a little confused how to approach this problem and any help would be appreciated. I've never had to deal with a more complex basis then the regular standard basis. Thank you.

Answer 1

I don't think it needs a matrix version. According to what you noted, you need to solve $$(4t^2-t)=a(t^2+2)+b(t+3)+c(t^2+t+1)$$ for some $a,b,c\in\mathbb R$. So we have $$(a+c)t^2=4t^2\\(b+c)t=-t\\(2a+3b+c)=0$$ or $$(a+c)=4\\(b+c)=-1\\(2a+3b+c)=0$$

Answer 2

The equations obtained in Babak S.'s answer shows that you had the wrong matrix. The equation is $$ \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 2 & 3 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix}. $$ For some reason you got the transpose of the actual matrix. So you need to do RREF on the following matrix $$ \left(\!\! \begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & 1 & -1 \\ 2 & 3 & 1 & 0 \end{array} \!\!\right) $$ and you should get the right answer.