$T(x_1, x_2, x_3) = (x_1 -2x_2 +5x_3, 3x_1 -4x_3)$
What I have so far is:
$$T\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix}x_1 -2x_2 +5x_3 \\ 3x_1 -4x_3 \end{bmatrix}$$
$$T\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = x_1 \begin{bmatrix}1 \\ 3 \\ \end{bmatrix} + x_2\begin{bmatrix}-2 \\ 0 \\ \end{bmatrix} + x_3 \begin{bmatrix}5 \\ -4 \\ \end{bmatrix}$$
If $T(x) = Ax$:
$$A(x_1, x_2, x_3) = x_1 \begin{bmatrix}1 \\ 3 \\ \end{bmatrix} + x_2\begin{bmatrix}-2 \\ 0 \\ \end{bmatrix} + x_3 \begin{bmatrix}5 \\ -4 \\ \end{bmatrix}$$
$$A = \begin{bmatrix}1 &-2 & 5 \\ 3 & 0& -4 \end{bmatrix}$$
The domain of T would be $\mathbb{R^3}$ and the codomain would be $\mathbb{R^2}$.
Best Answer
Yes, domain is correct and codomain also, since vectors $$\begin{bmatrix}1 \\ 3 \\ \end{bmatrix}$$ and $$\begin{bmatrix}-2 \\ 0 \\ \end{bmatrix}$$ are lineary independent.