Matematické Fórum

stuart clark · 14. 07. 2012 06:05

If $A$ is a $3\times 3$ matrices with entries are from the set $\{-1,1\}$ . Then total number of Non- Singular matrices $A$ is

vanok · 14. 07. 2012 10:39 — Editoval vanok (14. 07. 2012 21:48)

Hi ↑ stuart clark:,

1) The total number of all possible matrices is $2^9=512$ .

2) By considering the linear dependence of the columns of a matrix, we obtain the total number of all singular matrices is $2^3\cdot( 2\cdot 2^3+ 2^3 \cdot 2- 2\cdot 2 +2\cdot 6 )=320$

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3) The total number of all regular matrices is $512 -320= 192$

Edit: A correction of an oversight and addition of details.

anes · 14. 07. 2012 20:47 — Editoval anes (14. 07. 2012 20:48)

↑ vanok:

I understand that in step 2 you consider all matrices with 2nd row being multiple of the first (and vice versa, since there's no 0 involved), $+$ matrices with 3rd and 1st row being multiples, $-$ matrices, where all rows are multiples of each other, that you've counted twice in previous cases. I suppose at this point you forgot to count the matrices "sharing" only the 2nd and 3rd row. That should add another $2\cdot2^3-4$ to the brackets.

Let's take a look from the point of constructing a non-singular matrix.

We have $2^3$ picks for the first row (there's no worry about a row of zeroes).
Having picked first row in any way, for the second row we have $2^3-2$ options. The $-2$ corresponds to $\pm 1$ multiples of the first row, which are foridden.
For the third row we have $2^3 - 4$ options. The 4 corresponds to $\pm 1$ multiples of first and second row. It's easy to check that there is no other combination of the first two rows that we'd need to worry about.

So altogether we should have $2^3 \cdot (2^3 - 2) \cdot (2^3-4) = 8\cdot 6\cdot 4 = 8\cdot 24 = 192$ different non-singular matrices.