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Let M be real vector space of order $2\times3$ matrices with the real entries. Let $T:M\longrightarrow M$ be defined by

$T\Bigg( \begin{pmatrix} x_{1} & x_{2} & x_{3} \\ x_{4} & x_{5} & x_{6}\end{pmatrix}\Bigg)$=$\begin{pmatrix}-x_{6}& x_{4}& x_{1}\\ x_{3}& x_{5}& x_{2}\end{pmatrix}$

how to calculate the determinant of T?

Jearson Narvaez Rojas
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amit
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  • What are $x_i$? Real? you want the determinant written with those paramenter? – Exodd Sep 27 '14 at 08:41
  • yeah they are real..it will be written with zeros and one.it will 6X6 matrix.calculate the determinant of this matrix. – amit Sep 27 '14 at 08:42

1 Answers1

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We can write $T = RS$, where $R$ is the map the multiplies the $(1, 1)$ entry by $-1$ and fixes all of the other entries and $S$ is the permutation matrix for the permutation $(16243)$ of six elements.

The matrix representation of $R$ is diagonal with entries $(-1, 1, \ldots 1)$, so $\det R = (-1)(1) \cdots (1) = -1$. The above permutation is even, so $\det S = 1$ and thus $$\det T = \det(RS) = \det R \det S = (-1)(1) = -1.$$

Travis Willse
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  • Shouldn't the permutation be $(641352)$? – Jonas Dahlbæk Sep 27 '14 at 08:49
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    There are two notations for permutations, I'm using the one that decomposes the permutation as a product of cycles. – Travis Willse Sep 27 '14 at 08:50
  • may i know how did u write detR=-1...can you explain this line ...R is the map the multiplies the (1,1) entry by −1 ? – amit Sep 27 '14 at 13:20
  • We can identify $M \leftrightarrow \mathbb{R}^6$ just be ordering the entries of a $2 \times 3$ matrix in a $6 \times 1$ matrix, say, sending the $(1, 1)$ entry to the top entry. Then, $R$ is the diagonal map that sends that first entry to its negative, and fixes the remaining entries. – Travis Willse Sep 27 '14 at 13:44