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Let $M$ be the real vector space of $2\times 3$ matrices with real entries. Let $T:M \rightarrow M$ be defined by $$T(\begin{bmatrix} x_{1} & x_{2} & x_{3} \\ x_{4} & x_{5} & x_{6} \\ \end{bmatrix})=\begin{bmatrix} x_{6} & x_{4} & x_{1} \\ x_{3} & x_{5} & x_{2} \\ \end{bmatrix}$$ Then the determinant of $T$ is

$A. 1.$

$B. 2.$

$C.-1.$

$D.0.$

Now one method is as usual firstly find the matrix of the transformation $T$ with respect to usual basis of $W$ and then find the determinant of the matrix thus obtained which come as $-1.$ But i want some short trick that gives the determinant in less time. One more thing is that matrix of $T$ with respect to usual basis is an Orthogonal matrix. Please give some suggestion. Thank you.

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Hint: The map performs a permutation of the canonical basis. Its determinant is the signature of this permutation. I get $\det T= +1$.

More detailed: Let $e_i$ be the 2 by 3 matrix where $x_i=1$ and the others zero (in the matrix on the left hand side in your definition). Sort of a 'canonical' basis when you think of the matrix as a vector with 6 components. Then $ T : e_1\mapsto e_3, e_2 \mapsto e_6$, etc... Thus $T e_j = \sum_i e_i t_{ij}$ where $t_{ \sigma(j) j}=1$ (all others zero) with $\sigma$ being the permutation taking $(1,2,...)$ to $(3,6,...)$. Then $\det T = \det t = {\rm sign} \sigma = +1$.

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  • $\begingroup$ But my simple calculation det comes as -1.... $\endgroup$ – neelkanth Aug 20 '16 at 11:50
  • $\begingroup$ where is mistake? $\endgroup$ – neelkanth Aug 20 '16 at 11:51
  • $\begingroup$ Justed checked it by writing down the matrix. Still got $+1$ $\endgroup$ – H. H. Rugh Aug 20 '16 at 11:51
  • $\begingroup$ Would you care to elaborate? I'm not getting it? What specifically can be the canonical basis $B$ and its image under the transformation? $\endgroup$ – Vim Aug 20 '16 at 12:06
  • $\begingroup$ @H.H.Rugh can you tell something about canonical basis and relation with determinant for more clear $\endgroup$ – neelkanth Aug 20 '16 at 12:17
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I would start by decomposing the permutation into disjoint cycles. It is easy to find out the sign from the decomposition. In this case, the decomposition gives one cycle of length 5 and one trivial cycle ($x_5$ doesn't move). A cycle of odd length is even, so it seems like my method gives 1 as the answer...

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