$T$ is a linear transformation
$A = \begin{bmatrix} 2\\4\\8\\ \end{bmatrix}$
How can you define a vector $ x \in \mathbb R^4 $ which satisfies $T(x) = A$
this is just a made up example. Forgive me if it's not valid.
this is homework, so a point in the right direction would be preferred.
Thanks.
In a matrixe word, $A*x=b$ imply $x=(A^{-1})*b$ if $A$ is inversible, and the inverse is $A^{-1}$, but you need to make intention on the dimension of $A$, in your example If you considere $T(x)=A$ with your hypothesis $T \in \mathbb{M}_{(3,4)}(\mathbb{R}^4)$ :
\begin{matrix}a_1x_1 + b_1x_2 + c_1x_3 = d_1\\ a_2x_1 + b_2x_2 + c_2x_3 = d_2\\ a_3x_1 + b_3x_2 + c_3x_3 = d_3\end{matrix}
If : $A = \begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}$ with $X= \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$ et $b = \begin{pmatrix} d_1\\ d_2\\ d_3 \end{pmatrix}$. The system admet one unique solution if an only if $det(A)\neq 0$ : $x_1 = \frac{\det(A_1)}{\det(A)} = \frac{\begin{vmatrix}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{vmatrix}}{\det(A)}$ $x_2 = \frac{\det(A_2)}{\det(A)} = \frac{\begin{vmatrix}a_1&d_1&c_1\\a_2&d_2&c_2\\a_3&d_3&c_3\end{vmatrix}}{\det(A)}$ and $x_3 = \frac{\det(A_3)}{\det(A)} = \frac{\begin{vmatrix}a_1&b_1&d_1\\a_2&b_2&d_2\\a_3&b_3&d_3\end{vmatrix}}{\det(A)}$
Or simply : $X=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \frac1{\det(A)} \cdot \begin{pmatrix} \det(A_1)\\ \det(A_2)\\ \det(A_3)\end{pmatrix}.$
There is no solution if and only if : $\det(A) = 0$ and $\Big( \det(A_1) \ne 0\quad\text{ou}\quad\det(A_2) \ne 0\quad\text{ou}\quad\det(A_3) \ne 0 \Big)\,$
In the case $\det(A) = \det(A_1) = \det(A_2) = \det(A_3) = 0$, we can have an infinity of solution or no solution .
In your example, first the matrix is not a scare matrix, so it is already a singe that you can have infinity of solution, under the fact that the number of equation that you have is not the same as the number of variable in your model. This is due to the fact that one variable is consider as parameter.