How is the Determinant of a Matrix Affected by Row Operations?

[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!

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

 

[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.

 

[epkid08]

$$B_i = a_kA_i + A_j$$



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

then use the multilinearity of the determinant over the rows of the matrix.

 

[slider142]

$$B_i = a_kA_i + A_j$$



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

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

 

[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?

 

[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.