determinant of a a linear operator 
I understand that matrix of $T$ will be $6\times 6$ and will consists of $1,0$ but not able to find out what will be the determimant.
 A: If we didn't have the -1 coefficient in the first entry (look carefully at graphic), $T$ would be a permutation of entries, and the determinant of $T$ would be $\pm 1$ accordingly as $T$ is an even or odd permutation (as defined by the number of transpositions needed to write the permutation in a combinatorial/group theory sense).  Throw in the -1 on that row of the matrix representation, and the determinant will still be $\pm 1$, but opposite in sign to what the permutation by itself would give.
A: Here the space of matrices is viewed as a 6-dimensional space.  The 2 by 3 matrix structure is irrelevant.  Therefore it is preferable to view this map as $T:\mathbb{R}^6\to \mathbb{R}^6$ given by $(x_1,x_2,x_3,x_4,x_5,x_6)\mapsto(x_6,x_4,x_1,x_3,x_5,x_2)$.  Then one immediately recongnizes a permutation matrix.  Hence its determinant is either $1$ or $-1$.  Now all you have to do is determine whether the permutation is even or odd, for example by decomposing it into cycles.
I just noticed the minus sign on $x_6$, so this is not exactly a permutation matrix.  That changes the sign of the determinant to its opposite.
A: $T$ is a permutation of the elements of the base, so its matrix has only one $1$ in every row and every column. The determinant of such matrix equals $\pm 1$.
A: Is the permutation $(1,6,2,4,3)\in S_6$ even or odd? The determinant of the linear operator equals the negative of the sign of this permutation (the sign of the permutation is $1$ if it's even and $-1$ if it's odd). Why?
I hope this helps! (The details are minimal so you might have to think a little bit to understand my answer.)
A: I think that for the transformation to be defined, $T$ must be a $2x2$ matrix.After which you will have a system of equations. Then you may solve for the entries of $T$.
