The set $B=\{−4−3x^2, −16−3x−12x^2, −35−9x−27x^2\}$ is a basis for $P_3$ (or $\Bbb R_3[x]$, if you wish.) Find the coordinates of $p(x)=−160−39x−123x^2$ relative to the basis $B$.
What I did to try to solve this problem was:
I let $x = 1$ making the set $B$ into a matrix:
\begin{bmatrix}-4&0&-3\\-16&-3&-12\\-35&-9&-27\end{bmatrix}
Then I transposed it:
\begin{bmatrix}-4&-16&-3\\0&-3&-12\\-3&-12&-27\end{bmatrix}
Then I made a matrix out of $p(x)$:
\begin{bmatrix}-160\\-39\\-123\end{bmatrix}
Then I multiplied the two matrices and received:
\begin{bmatrix}5569\\1224\\4269\end{bmatrix}
But this answer is wrong!
What am I doing wrong?
Best Answer
You should inverse the matrix, not transpose it.
But it is more efficient to solve the linear system.
$$\begin{bmatrix}-4&0&-3\\-16&-3&-12\\-35&-9&-27\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}-160\\-39\\-123\end{pmatrix} $$