Linear Transformation finding parameters

Question

Let

$f_t\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ with

$f_t\left(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\right)=\begin{pmatrix} -1 \\ 2 \\ -4-t \end{pmatrix}, \, \, f_t\left(\begin{pmatrix} 0 \\ 1 \\ t \end{pmatrix}\right)=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}, \, \, f_t\left(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right)=\begin{pmatrix} 7 \\ 2 \\ 2t+7 \end{pmatrix}$

For which $t \in \mathbb{R}$ is $f_t$ a linear transformation?

My idea is take 3 base elements and wirte it below as linear combination. But I need a fourth vector or not?

Answer 1

Hints:

1. The vectors are from $\Bbb R^3$, which has $3$ dimensions, so the image of $3$ linearly independent vectors determine a unique linear mapping, since they form a basis of $\Bbb R^3$.
2. Clearly, if $t=1$, the function $f_1$ is not well defined.
3. For $t\ne1$, we can epress $(0,0,1)^T$ by a linear combination of the last 2 vectors of the $3$ where $f_t$ is defined. Then we can also express $(0,1,0)^T$, and using the first one also $(1,0,0)^T$.