Find value of $T$ - linear transformation of matrix (incorrect solution)

Question

So I have a screenshot of a question below. I keep getting 2. ii) wrong below and I'm not sure what I'm doing incorrectly in my steps. I began by placing the values of the tuples in a $4\times5$ matrix consisting of:

\begin{bmatrix}1&0&-2&2&4\\2&1&-7&3&3\\1&1&-5&2&0\\-3&-2&12&-5&-3\end{bmatrix}

Using an RREF calculator, I got it to:

\begin{bmatrix}1&0&-2&0&2\\0&1&-3&0&-4\\0&0&0&0&1\\0&0&0&1&0\end{bmatrix}

I'm getting stumped because I'm not exactly sure what I'm supposed to do with column 3 since it's a free variable. I assumed I would just ignore the 3rd column as well as the 3rd tuple given in the question. I did the following by taking each value of the last column in the augmented matrix and ended up ignoring the 3rd value and 3rd column:

\begin{equation} \begin{split} 2(0 + 2x + 2x^2 + 1x^3) - 4(2 - x + 2x^2 + 2x^3) + 0(-1 + 0x + (-1)x^2 + 1x^3) &=0 + 4x + 4x^2 + 2x^3 - 8 + 4x - 8x^2 - 8x^3\\ &=-8 + 8x - 4x^2 - 6x^3\\ \end{split} \end{equation}

but it keeps getting marked as wrong. I tried to just count the free variable/third column and tuple in anyways to see what happens but the new answer I got is also marked wrong. Are there steps that I'm missing or doing wrong completely? I would appreciate some help. I redid the RREF and substitutions multiple times but I keep getting the original answer that I got. This is the method I used to solve a bunch of other questions like this so I'm not sure why I'm getting a wrong answer for this question in particular. I would appreciate some help.

Answer 1

That third row shows that there is no solution to this system, and hence you cannot find the transformation with the information given, as the given vector cannot be expressed as a linear combination of the four vectors provided

Answer 2

Your RREF is incorrect. It should be $$\begin{pmatrix} 1 & 0 & -2 & 0 & 2 \\ 0 & 1 & -3 & 0 & -4 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ https://rrefcalculator.com/ gives the equivalent but not technically RREF $$\begin{pmatrix} 1 & 0 & -2 & 0 & 2 \\ 0 & 1 & -3 & 0 & -4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ \end{pmatrix}$$ In either case, the $1$'s in the last two columns should be in the same row. From this, it's easy to see (labeling the $i$'th column of the original matrix as $a_i$) that $a_5 = 2a_1 - 4a_2 + a_4$, from which it follows that $Ta_5 = 2Ta_1 - 4Ta_2 + Ta_4$. Since you are given $Ta_1$, $Ta_2$, and $Ta_4$, it's now a routine calculation to obtain $Ta_5$.