Determinant Relationship: det(A) and det(-A) for n x n Matrices

[johndoe3344]

For an n x n matrix A, what is the relationship between det(A) and det(-A)?

I tried it with a 1x1 matrix, and det (-A) = - det (A)
I tried it with a 2x2 matrix, and det(A) = det(-A)
I tried it with a 3x3 matrix, and the results were the same as that with a 1x1.

This leads me to believe that for all odd n's, det(-A) = - det(A) and that for all even n's the two are the same.

Is this the case? And if so, how would I show that in a more mathematical manner that just intuition?

 

[radou]

Let A be an nxn matrix. What happens if we multiply a row/column of A with a scalar $\lambda$? What happens if we do this to all the rows/columns? How does it affect the first relation?

 

[johndoe3344]

Why are we multiplying each individual row/column by a scalar?

When we multiply the matrix by a scalar, doesn't this automatically distribute the scalar to every single entry of the matrix?

I guess what I'm confused at is this:

Suppose we let the square matrix B be obtained from matrix A by multiplying the matrix A by the scalar k. Then det(B) = k*det(A). Then if we let k = -1, as in my original question, then won't det(-A) = det(A) for all n?

But obviously this isn't the case. What am I doing wrong?

 

[Hurkyl]

Suppose we let the square matrix B be obtained from matrix A by multiplying the matrix A by the scalar k. Then det(B) = k*det(A).

Not so. You're thinking of the case when B is obtained by multiplying a single row of A by k.

 

[johndoe3344]

So since multiplying each row of a scalar by the constant gives us det(B) = k*det(A)...

If we multiply each row of a n x n matrix by the same constant, i.e. n times, then the formula would be for the case of k = -1:

det(B) = [(-1)^n]*det(A)

Which would explain the difference for odd n, and the same for even n.

Is this correct?

 

[morphism]

Yes, that's correct.

There's an another way to see this if you know that det(A*B) = det(A)*det(B): If we let I be the nxn identity matrix, then det(-1*I) = (-1)^n, so that det(-A) = det(-1*I * A) = det(-1*I) det(A) = (-1)^n det(A). Of course this can be generalized to any k.