Cfrgtkky

To be specific :
How many $4times4$ matrices with entries from $0$, $1$ have odd determinant?

P.S : Please do post comment/answer by fully reading it first and which satisfies what I asked for which I cleared at my best possible.

Approach :
I didn't go for combinatorial approach as it was a multiple choice question .
So what i thought of is :

Consider Half of them are even and remaining are odd.
Total is $2^{4times 4}$ so half would be $65536/2$ hence 2 of the options i have eliminated.
Now, One option was "20160" and other was "32767" but reason i choose 1st option because there are some matrices with "0" determinant so it wont be "32767" , actually less than that so i go for "20160" .

But though I got correct answer , i wasn't satisfied by own intuition so tried it for $2times 2$ matrix.
What I found is "10" zero matrices , "6" odd and "0" even matrices.
( I didn't considered "0" in the category of even just to separate it from odd/even)

With the $2times 2$ matrix it didn't satisfy my above intuition of dividing into half.
So my question is :

1. Whether I got the correct answer by luck?




2. Where my intuition gone wrong?




3. How to proceed with generalisation?

Note that in $mathbb{F}_2$ the only possible odd determinant has value $1$ and the only even determinant has value $0$, implying the matrix is not invertible and thus the columns are not linearly independent.

Now this is just a counting problem that asks: How many matrices are there in $text{M}_{4 times
4}left(mathbb{F}_2right)$ with linearly independent columns?

For the first column, there are $2^4-1$ options, all columns except the $0$ column. For the second column, the only stipulation is that it cannot be the same exact column and it cannot be $0$, and it is linearly independent. So there are $2^4-2$ possibilities. For the third, it cannot be either of those columns or the sum or $0$, so there are $2^4-4$ possibilities. And for the remaining column, any possible candidates would have to be not the sum of any combination of the three columns, for which there are $2^3$ sums (including $0$, the sum of none of them).

Thus, $(2^4-1)(2^4-2)(2^4-4)(2^4-2^3) = 20,160$.

I think your intuition would work reasonably well for fields with more elements, where the determinant would act more randomly.

"Half of them are even and remaining are odd." is not correct.

According to Consider the set of all $ntimes n$ matrices, how many of them are invertible modulo $p$., the number of $ntimes n$ invertible matrices modulo $p$ is
$$prod_{i=0}^{n-1} (p^n- p^i).$$
In your case $p=2$ and a matrices modulo $2$ is invertible if and only if its determinant is non-zero modulo $2$, i.e. it is odd. For $n=4$, the above formula yields
$$(2^4-1)(2^4-2)(2^4-4)(2^4-8)=20160.$$
On the other hand, the number of binary matrices with even determinant for $n=4$ is the cardinality of the complement:
$$2^{16}-20160=45376.$$

P.S. The number of binary matrices whose determinant is exactly zero is harder to find. For $n=4$ they are $42976$ which is less than the even ones $45376$. As a reference see http://oeis.org/A046747.