Determinant of 3x3 matrices: Is a matrix with all odd entries a multiple of 4?

[fox1]

I need to prove that a 3x3 matrix with all odd entries will have a determinant that is a multiple of 4.

This is how I set it up:

I let A = { {a, b, c}, {d, e, f}, {g, h, i} } with all odd entries

then I define B = { {a, b, c}, {d + na, e + nb, f + nc}, {g + ma, h + bm, i + cm} }
where I add the multiple of first row to second and third row. So only the first row will have odd integers entries while the second and third row will be even entries.

det(A) = det(B) since adding multiple of one row to another doesn't change the determinant

After this I was going to show that the each of three 2x2 matrix will have a even determinants. This is where I'm kind of stuck. I can show that the determinant of 3x3 will be even, but how can I show that it will be a multiple of 4?

 

[SteamKing]

IDK why you are using row operations and minors to calculate a 3x3 determinant. Certainly, the row operations complicate the calculation.

The Rule of Sarrus is a more straightforward method for calculating the determinant.

http://en.wikipedia.org/wiki/Rule_of_Sarrus

 

[PeroK]

Maybe think about what happens when most of the entries are even and what you can say about the product of two even numbers.

 

[D H]

then I define B = { {a, b, c}, {d + na, e + nb, f + nc}, {g + ma, h + bm, i + cm} }
where I add the multiple of first row to second and third row. So only the first row will have odd integers entries while the second and third row will be even entries.

You're overcomplicating things. Why the arbitrary n and m? What's wrong with n=m=1?

Your main problem with this problem is that you aren't representing that matrix as consisting of odd numbers only. Instead of a use 2a+1; for b use 2b+1, etc.

What can you say about the determinant of a 2x2 matrix that consists of only even numbers? (What is it a multiple of?)