2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

$2^3-2^0=7$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Shailesh Oct 7 '15 at 10:31
  • $\begingroup$ @Shailesh There cannot be first row + first row, since field has only two elements I think $\endgroup$ – luka5z Oct 7 '15 at 10:32
  • $\begingroup$ @luka5z. Yes.Absolutely. $\endgroup$ – Shailesh Oct 7 '15 at 10:36
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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' $\endgroup$ – arya_stark Apr 18 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.