Let $M$ be the real vector space of $2\times 3$ matrices with real entries.
Let $T:M\rightarrow M$ be defined by $$T\left(\begin{bmatrix}x_1 &x_2&x_3\\x_4&x_5&x_6\end{bmatrix}\right)=\begin{bmatrix}-x_6&x_4&x_1\\x_3&x_5&x_2\end{bmatrix}$$
Find the determinant of $T$?
My efforts
$M$ is 6 dimensional vector space. Consider the basis $\beta=\{A_{ij}\}_{1\leq i\leq2,1\leq j\leq3}$
where $A_{ij}$ has $1$ in $ij^{th}$ place and $0$ elsewhere.
Let us put an order of basis to be $\{A_{11}, A_{12},A_{13},A_{23},A_{22},A_{21}\}$
Now if we observe closely the matrix of linear transformation would be a permutation matrix baring a negative sign which we can take out.
So instead of writing that big $6\times 6$ matrix, we write everything in the permutation notation(the notation we use to write element of symmetric group)
So we get something like $$(16243)$$
Since this cycle is even and we have taken out the negative sign out of the matrix of linear transformation, we conclude that determinant is $-1$.
Edit:
Is there a flaw in my working? Please have a look at the argument.