http://www.maths.nuigalway.ie/~rquin...section2-5.pdf

has everything you could want to know about row operations and determinants

Results 1 to 3 of 3

- January 25th 2014, 03:27 AM #1

- Joined
- Jun 2013
- From
- Israel
- Posts
- 158

## det(A)=-det(B)

We have matrix A in the order of (n x n) , and we have matrix B that can be derived from A by adding to the row j the multiplication of row i with a scalar.

Then det(B)=-det(A)

I tried to prove it with the multilinear property but I am stuck... please help. thank you

- January 25th 2014, 04:14 AM #2

- Joined
- Nov 2013
- From
- California
- Posts
- 3,439
- Thanks
- 1397

## Re: det(A)=-det(B)

http://www.maths.nuigalway.ie/~rquin...section2-5.pdf

has everything you could want to know about row operations and determinants

- January 25th 2014, 11:39 AM #3

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,530
- Thanks
- 835

## Re: det(A)=-det(B)

Again, what you are trying to prove IS NOT TRUE. What IS true is that det(A) = det(B). For convenience we will assume j < i (if i < j, apply the matrix that switches rows i and j first, see note below).

Let P be the matrix:

then PA has the effect of adding r times row i to row j.

It follows that det(B) = det(PA) = det(P)det(A). Expanding by successive minors along the first column shows that det(P) = 1, so:

det(B) = det(A).

Or, regarding det as a multilinear function of the rows of a matrix, we have:

det(B) = det(B_{1},...,B_{j},...,B_{i},...,B_{n}) = det(A_{1},...,A_{j},...rA_{j}+A_{i},...,A_{n})

= det(A_{1},...,A_{j},...rA_{j},...,A_{n}) + det(A_{1},...,A_{j},...,A_{i},...,A_{n})

= r(det(A_{1},...,A_{j},...A_{j},...,A_{n}) + det(A)

= r(0) + det(A) = det(A) (since the first term is a determinant of a matrix with 2 equal rows).

Note (from above):

If i < j, then switching the rows multiplies the determinant by -1 (the determinant is an ALTERNATING multilinear function). Then adding r times row j (which *was* row i) to row i ( which *was* row j, before) does not change the determinant, as above.

Now we have our original row i times r plus our original row j currently in the i-th row, and our original row i in the j-th position. Switch them again, which again multiplies the determinant times -1.

This gives us our original row i times r plus our original row j in the j-th position, as desired, and puts our original row i back in the i-th position. Thus:

det(B) = (-1)(det(A))(-1) = det(A).

(Or: you could simply use the matrix:

instead of P).