[Math] Number of non singular matrices over a finite field of order 2

Question

I have to find out the number of $3×3$ non singular matrices over a field of order $2$.
I tried in the following way.
First to find out a non singular matrix $A,$ clearly any row of $A$ can't be full of $0$s.
So the first row (say) can be filled up by $(8-1)$ ways.
Once the row is filled up,the next row can't be the same and also can't be full of zeros,so we can fill the next row by $
(8-2)$ ways.
And at last the third row also can't be full of zeros,same as the first row,and same as the second row also.So we have
$(8-3)$ choices.
Hence the number of non singular matrices seems to be $7×6×5=210$.
Am I right? Or there are more non singular matrices ? May be less also.
Please correct me if I am wrong.
Thank you.

Answer 1

Here field of order 2 is $\mathbb{F}_2$ = $\mathbb{Z}_2$ = $\{0,1\}$. The result is actually the number of elements in the General Linear Group $GL_3(\mathbb{F}_2)$ or $GL_3(\mathbb{Z}_2)$.

To count the number of non-singular matrices of order 3 with 0 and 1 as its elements only, we have to make sure that all the rows are linearly independent and non-zero.

For the first row we have $(2^3 – 1)$ choices.

For the second row we have $(2^3 – 1) – 1 $ = $(2^3 – 2)$ choices. Because we cannot count the vector already has been used in first row.

For the third row we have $(2^3 – 1) – 2 – 1 $ = $(2^3 – 2^2)$ choices. Because we have to omit 2 vectors from the count that already have been used in first and second row. And we have to omit one more vector that can be the linear combination of the first and second rows.

So, in total we can have $(2^3 – 1)(2^3 – 2)(2^3 – 2^2)$ = $7\times6\times4$ = $168$ matrices.