# Determinant 2

1. Nov 8, 2009

### epkid08

If I have a matrix $$A$$, and I use $$n$$ different row operations of this form: $$a_kR_i + R_j \rightarrow R_i$$ to construct a new matrix $$B$$, what is the determinant of $$A$$ in terms of $$B$$?

Solved!

$$|A|=|B|\prod^n_{k=1}\frac{1}{a_k}$$

Last edited: Nov 9, 2009
2. Nov 8, 2009

### slider142

Not entirely. Write B out in terms of the rows of A, then use the multilinearity of the determinant over the rows of the matrix.

3. Nov 8, 2009

### epkid08

$$B_i = a_kA_i + A_j$$

I'm not sure what you mean when you say:

4. Nov 8, 2009

### slider142

If we write A as a list of rows: A1,...,Am, where the Ai is the ith row of the matrix A, we know that det(A1, ..., r*Ai+Aj, ..., Am) = r*det(A1, ..., Ai, ..., Am) + det(A1, ..., Aj, ..., Am) for all scalars r and each Ak. That is, the determinant is a linear operator with respect to each row; it is multilinear.

5. Nov 9, 2009

### epkid08

Wow, after a week of looking for it, I found what I was doing wrong, and it turns out it was just a stupid mistake.

The actual formula to the problem in my first post should be:

$$|A|=|B|\prod^n_{k=1}\frac{1}{a_k}$$

I assume that's what you were trying to hint at slider142?

6. Nov 9, 2009

### slider142

Yep. Each scalar can be pulled out as a factor due to linearity, while the second determinant in the sum vanishes, so you end up with the product of each scalar multiplied by the original determinant.