Given the three dimension space $x + 2y + 3z + 4t = 0$, write the plane as the image of a 4x3 matrix and find a basis for this space.
We get $A=\begin{bmatrix} 1 & 2 & 3 & 4\end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix}$ = $x+2y+3z+4t$. Afterwards, we can find three unique solutions to the equation taht could suffice as our matrix: take {1, 1, -1, 0}, {0, 3, -2, 0}, {0, 0, 4, 3}. However, I don't really know how to move on from there. Any help?
The three vectors you find (with a correction in the third) $$\{1, 1, -1, 0\}, \{0, 3, -2, 0\}, \{0, 0, 4, -3\}$$ satisfy the plane equation and are linearly independent, since the plane has dimension 3 those vectors are a basis for the subspace/plane, thus the matrix we are looking for is
$$M=\begin{bmatrix} 1 & 0 & 0\\ 1 & 3 & 0\\ -1 & -2 & 4\\ 0 & 0 & -3\end{bmatrix}$$