How many invertible 3x3 matrices?

Question

How many invertible $ 3 \times 3 $ matrices exist over $2$-element field?

Obviously if some field has only $2$ elements, those elements must be $0$ and $1$. A matrix is invertible if and only if its determinant is non zero. I think it will be easier to find matrices which determinants are $0$ and subtract that number from $2^9$ (total number of matrices we can build over that field).

Is there any "clever" way of solving this problem?

Hint: it will be invertible precisely when the columns are linearly independent. So after picking the first column, how many can you pick for the next? And the next? — Tobias Kildetoft, Oct 07 '15 at 10:19
You can probably find several older related questions, for example: Number of invertible 0-1 matrices, Probability that a random binary matrix is invertible? or Consider the set of all $n\times n$ matrices, how many of them are invertible modulo $p$.. — Martin Sleziak, Apr 15 '17 at 08:22

score 4 · Accepted Answer · answered Oct 07 '15 at 10:25

4

$2^3-2^0=7$ choices for the first row as nozero vector. Then $2^3-2^1=6$ choices for the second row vector not in the span of the first. Then $2^3-2^2=4$ choices for the third row not in the span of the first two.

answered Oct 07 '15 at 10:25

Hagen von Eitzen

382,479

Wouldn't the second choice be only 5 ? Zero is exclude d, the first row is excluded, and the first row + first row is excluded. – Shailesh Oct 07 '15 at 10:31
@Shailesh There cannot be first row + first row, since field has only two elements I think – luka5z Oct 07 '15 at 10:32
@luka5z. Yes.Absolutely. – Shailesh Oct 07 '15 at 10:36

score 2 · Answer 2 · edited Apr 15 '17 at 08:46

2

Here we are finding invertible matrices:

None of columns can be entirely zero and the columns must linearly independent. We can construct the first in $2^3$ ways but $(0,0,0)$ is included so we are left with $8-1=7$ ways for the first column. Now again we can construct the second column in $2^3$ ways but $(0,0,0)$ and the arrangement of type of first column should be excluded for linear independence; thus for second column we are left with $8-2=6$ ways; similarly for third column we have $8-1{(0,0,0)}-1$(first column)$-1$ (second column)$-1$ (span of first two columns)$= 4$.

Thus, the total no. of ways = total number of invertible matrices = 7×6×4 = 168.

edited Apr 15 '17 at 08:46

Glorfindel

4,000

answered Apr 15 '17 at 08:07

Ashok Singh Bhandari

57

1

Why the 1, for the span of first two columns? Some more detail there would be appreciated, possibly explaining in a way without using 'span' – stoic-santiago Apr 18 '19 at 18:05
1

For third column excluded columns are 1) 0C1+0C2= 0. 2) 1C1+0C2= C1. 3) 0C1+1C2=Ç2. 4) 1C1+1C2= C1+ C2 – Ashok Singh Bhandari Aug 04 '19 at 18:58
@AshokSinghBhandari It can be proven that from 8 possible choices for a row, one and only one of them is in the set {(C1+C2), (C1−C2), (C2−C1)}. So the fourth linear composition you mentioned should be one the above compositions not just what you wrote. – therealak12 Nov 14 '19 at 15:08

How many invertible 3x3 matrices?

2 Answers2