Thread: 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

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

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

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:

$\displaystyle P = \begin{bmatrix}1&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\ddots&\vdots&{}&\vdots&{}&\vdots\\0&\cdots &1&\cdots&r&\cdots&0\\ \vdots&{}&\vdots&\ddots&\vdots&{}&\vdots\\0&\cdots &0&\cdots&1&\cdots&0\\ \vdots&{}&\vdots&{}&\vdots&\ddots&\vdots\\0&\cdots &0&\cdots&0&\cdots&1 \end{bmatrix}$

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(B₁,...,B_j,...,B_i,...,B_n) = det(A₁,...,A_j,...rA_j+A_i,...,A_n)

= det(A₁,...,A_j,...rA_j,...,A_n) + det(A₁,...,A_j,...,A_i,...,A_n)

= r(det(A₁,...,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:

$\displaystyle Q = \begin{bmatrix}1&\cdots&0&\cdots&0&\cdots&0\\ \vdots&\ddots&\vdots&{}&\vdots&{}&\vdots\\0&\cdots &1&\cdots&0&\cdots&0\\ \vdots&{}&\vdots&\ddots&\vdots&{}&\vdots\\0&\cdots &r&\cdots&1&\cdots&0\\ \vdots&{}&\vdots&{}&\vdots&\ddots&\vdots\\0&\cdots &0&\cdots&0&\cdots&1 \end{bmatrix}$

instead of P).