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$T:\Bbb{P}_3 \to \Bbb{P}_3$ is a linear transformation such that:

$$\begin{align} T\left(-2 x^2\right) &= 3 x^2 + 3 x \\ T(0.5 x + 4) &= -2 x^2 - 2 x - 3 \\ T\left(2 x^2 - 1\right) &= -3 x + 2 \end{align}$$

With respect to these three input vectors, I got $$\begin{bmatrix}0\\3\\3\end{bmatrix} = \frac{45}2\begin{bmatrix}0\\0\\-2\end{bmatrix}+6\begin{bmatrix}4\\\frac12\\0\end{bmatrix}+24\begin{bmatrix}-1\\0\\2\end{bmatrix}$$

I'm pretty sure this means the first column of my transformation matrix is

$$\begin{bmatrix}\frac{45}2&\cdot&\cdot\\6&\cdot&\cdot\\24&\cdot&\cdot\end{bmatrix}$$

Following this algorithm, I got a final matrix:

$$\begin{bmatrix}\frac{45}2&-12&-26\\6&-4&-6\\24&-13&-26\end{bmatrix}$$

Is this correct so far?

The question wants me to find $T(1)$.

I did this:

$$T\left(\begin{bmatrix}1\\0\\0\end{bmatrix}\right)=\begin{bmatrix}\frac{45}2&-12&-26\\6&-4&-6\\24&-13&-26\end{bmatrix}\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}\frac{45}2\\6\\24\end{bmatrix}$$

Apparently, I'm wrong. Why?

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You've expressed your matrix in terms of a specific basis, namely the three input vectors. In order to evaluate $T(1)$, you need to express $1$ as a linear combination of these basis vectors. In this case, it is $-1$ times the first plus $-1$ times the third. Apply your matrix to $(-1,0,-1)$ and see if you get the right answer.

More importantly, matrices are overkill here. You know the function is linear, so just use that $T(x+y)=T(x)+T(y)$ and $T(ax)=aT(x)$

$$\begin{align} T(1)=T(-(-2x^2)-(2x^2-1))\\ =-T(-2x^2)-T(2x^2-1)\\ =-(3x^2+3x)-(-3x+2)\\ =-3x^2-2 \end{align} $$

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Note that the matrix you found is, properly saying, the matrix representation of $T$ with respect to the basis $\beta = \{-2x^2, 0.5x+4, 2x^2-1\}$, that is \begin{equation} [T]_\beta^\beta \end{equation} You need to be specific in which basis you are considering, as if you use different basis you would obtain a different matrix. Please also note that in general the domain space $V$ and the codomain space $W$ of $T:V\to W$ may not be the same, and even they are the same, we need not consider the same basis. The general notation for a matrix representation of $T:V\to W$ with respect to a basis $\beta$ for $V$ and $\gamma$ for $W$ is \begin{equation} [T]_\beta^\gamma \end{equation}

Up to this point I see no mistake in your calculation, if this is the matrix you want. But the matrix representation is not used in a correct way. In order to calculate $T(1)$, you have to

  1. Write $1 \in \mathbb{P}_3$ as a linear combination of $\beta$, i.e. $1 = \alpha_1(-2x^2)+\alpha_2(0.5x+4)+\alpha_3(2x^2-1)$.
  2. Write the coordinates into a column vector, i.e. $[1]_\beta = (\alpha_1, \alpha_2, \alpha_3)^T$.
  3. Multiply the matrix to the column vector of coordinates yields the coordinates of $T(1)$ with respect to $\gamma = \beta$, i.e. $[T(1)]_\beta = [T]_\beta^\beta[1]_\beta = (\lambda_1,\lambda_2,\lambda_3)^T$.
  4. Write $T(1)$ as an element in $\mathbb{P}_3$ according to its coordinates with respect to $\beta$, i.e. $T(1) = \lambda_1(-2x^2) + \lambda_2(0.5x+4) + \lambda_3(2x^2-1)$.
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