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How many non-singular matrices of order 3x3 we can form in which 4 entries are 1 and other entries are zero.. Any short​trick??

Praj
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    Did you try anything? – Ralff Apr 04 '17 at 02:53
  • 100 010 001 fixing this and adding one 1 at any one of the zero's place det remains non zero E.g 101 010 001 and so on.. also matrices of the form 110 100 001 , 100 011 010... So Atleast 8 such we can have.. – Praj Apr 04 '17 at 03:07
  • Please use MathJax. Specifically, Matrices. – Em. Apr 04 '17 at 03:13
  • Think about determinants. You know that there can't be any zero row or column, so you can restrict your options for the first 3 ones you place. The fourth one can then go anywhere you want. – AspiringMathematician Apr 04 '17 at 03:19
  • Since permuting rows doesn't change being singular vs. non-singular, I would start by breaking down cases in which the rows are ordered by the number of ones in each. With only four entries $1$, there are not a huge number of possibilities. – hardmath Apr 04 '17 at 03:22

2 Answers2

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We need at least one triple of $1$s appearing in ${\rm det}(A)$. If such a triple has been chosen, e.g., the diagonal of $A$, then the fourth $1$ cannot create a second such triple. Since there are six such triples in all, and for each of them there are six spots to place the last $1$ there are $36$ regular matrices of this kind.

Christian Blatter
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Hints.

  • If the matrix is nonsingular, it must possess a circulant diagonal/anti-diagonal $\{a_{i\sigma(i)}:i=1,2,3,\ \sigma\in S_3\}$ of ones. How many such diagonals/anti-diagonals are there?
  • Having fixed the diagonal of ones, how many ways are there to put the remaining one?
user1551
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