$T$ is a map which sends this $2 \times 3$ matrix
\begin{bmatrix}x_1&x_2&x_3\\x_4&x_5&x_6\end{bmatrix}
to
\begin{bmatrix}-x_6&x_4&x_1\\x_3&x_5&x_2\end{bmatrix}
I want to calculate the determinant of $T$.
My approach: I think this question is wrong because if $x_1 = 1$ and rest all $0$, then we cannot construct any $2 \times 2$ matrix $T$ because by rule of matrix multiplication, if I had such a $T = \begin{bmatrix}a&b\\c&d\end{bmatrix} $, then $ax_3 + bx_4 = a.0 + b.0 = x_1 = 1$ which is non-sense.
Since this question was asked in one of the previous exams, I tried answering it by trial and error (and non-mathematical reasoning). I first argues that $T$ which is a $2 \times 2$ matrix can only only have element as $0$ and $1$ otherwise there would there would be some scaling in the image which is not there in our case. So this narows down search to either $1$ or $-1$. I choose $-1$ because of the negative sign in $x_6$ and suprisingly it matched the answer.
Can anyone comment/justify the answer to be $-1$ and also improve upon my reasing given in 2nd half on my question.
$e_1, e_2,\cdots, e_6 = \begin{bmatrix} 1&0&0\\0&0&0\end{bmatrix},\begin{bmatrix} 0&1&0\\0&0&0\end{bmatrix},\cdots, \begin{bmatrix} 0&0&0\\0&0&1\end{bmatrix}$
Is a basis for the set of $2\times 3$ matrices
The matrix $T$ in this basis:
$T = \begin {bmatrix} 0&0&0&0&0&-1\\ 0&0&0&1&0&0\\ 1&0&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&1&0\\ 0&1&0&0&0&0\end{bmatrix}$
To calculate $\det (T)$.... Sum the indexes where there is a 1, and if the result is even flip the sign and if it is odd leave it alone. And multiply them all together.
Or, you could say that for each transposition you get one more sign flip.
I see a cycle $(13426)$ and that is even. And then multiply buy $(-1)$ becuase $x_6$ changes sign.
Either way, I get $-1$