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Permutations in Matrix!

Question-Let \(A\) be the set of all \(3\)x\(3\) symmetric matrices whose entries are \(1, 1, 1, 0, 0, 0, -1, -1, -1\). \(B\) is one of the matrix in set \(A\).

Number of such matrices \(B\) in set \(A\) is \(k\). Then, what is the value of \(k\)?

Note by Advitiya Brijesh
4 years, 1 month ago

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A \( 3 \times 3 \) symmetric matrix follows this pattern: \[ \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \end{pmatrix} \] So we have to choose which numbers are \( a \), \(b\), \(c\), \(d\), \(e\) and \(f\). Firstly, we'll focus on the principal diagonal: \[ \begin{pmatrix} a & & \\ & b & \\ & & c \end{pmatrix} \] If we had \( a = b \neq c \), the third entry of \(a\) wouldn't have a symmetric entry. For the same reason, we can't have \( a \neq b = c \) nor \( a = c \neq b\). If we had \( a = b = c \), then we couldn't build a symmetric matrix: we could choose a number for \(d\) and \(e\), but there couldn't be symmetry in the remaning places. Hence \(a\), \(b\) and \(c\) are all different numbers, and we have \( 3! \) possibilities. Now, we'll focus on the other numbers, outside the principal diagonal: \[ \begin{pmatrix} & d & e \\ d & & f \\ e & f &
\end{pmatrix} \] Since \(a\), \(b\) and \(c\) are all different numbers, we have 2 entries of \(-1\), \(0\) and \(1\) left. It's obvious that \(d\), \(e\) and \(f\) must also be all different numbers, and we have \(3!\) possibilities for them too. So the cardinality of the set \(A\) is \( { \left( 3! \right) }^2 = 36 \). I'm quite sure this is the right answer, but I'm open to corrections :) Battista Lonardi · 4 years, 1 month ago

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@Battista Lonardi yup! Thanks! :) Advitiya Brijesh · 4 years, 1 month ago

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Here's a better phrasing of the question: How many symmetric 3x3 matrices exist, with entries consisting of three 1's, three 0's, and three -1's.

Also, I'm going to put this answer in code in case any users don't want to see it: (5^2 - 13)*3 Bob Krueger · 4 years, 1 month ago

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How would you define a symmetric matrix? Tim Vermeulen · 4 years, 1 month ago

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@Tim Vermeulen Think about the diagonal that goes from the upper left to the bottom right corners. A symmetric matrix has symmetry about this line. This really only works for square matrices. Bob Krueger · 4 years, 1 month ago

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@Tim Vermeulen If \(A\) is a matrix of some order. Then, \(A\)=\(A^T\). Where, \(A^T\) denotes the transpose of matrix \(A\). For Transpose of matrix, see here, http://en.wikipedia.org/wiki/Transpose Advitiya Brijesh · 4 years, 1 month ago

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