Given the $R^2 \rightarrow R^3$ transformation how can I determine without RREF is the vector (1,3,5) exists in the $R^3$ co-domain of the transformation?
A = $\begin{bmatrix} 1 & 3 \\\\ 4 & 5 \\\\ 6 & 1 \end{bmatrix}$
I know the basis of the subspace is the vectors (1,4,6) and (3,5,1), but I'm still unclear how to quickly determine? I am also aware the range of the subspace is all reals (x,y) in $R^2$
Just try to verify if there is a vector $[x , y ]$ such that $$\begin{bmatrix} 1 & 3 \\ 4 & 5 \\ 6 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\3 \\5 \end{bmatrix}$$ If it exists, then $\begin{bmatrix} 1 \\3 \\5 \end{bmatrix}$ is in the co-domain.
So in our case, let's pick any $2 \times 2$ subsystem to solve (let's pick the first two)
$$\begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\3 \end{bmatrix}$$ which will give \begin{align} x^* &= 0.5714 \\ y^* &= 0.1429 \end{align} Now verify if the solution solves the third equation, that is $$6x + y =5$$ We get $$6x^* + y^* =3.5714 \neq 5 $$