- #1

jeff1evesque

- 312

- 0

**Theorem 4.3:**The determinant of an nxn matrix is a linear function of each row when the remaining rows are held fixed. That is, for 1<=r<=n, we have

det( a1, ..., a_(r-1), u+kv, a_(r+1), ..., a_n ) = det( a1, ..., a_(r-1), u, a_(r+1), ..., a_n )

+ det( a1, ..., a_(r-1), v, a_(r+1), ..., a_n )

whenever k is a scalar and u, v, and each a_i are row vectors of F^n.

**Proof:**The proof is by mathematical induction on n. The result is immediate if n = 1. Assume that for some integer n >= 2 the determinant of an (n-1)x(n-1) matrix is a linear function of each row when the remaining rows are held fixed. Let A be an nxn matrix with rows a1, a2, ..., an, respectively, and suppose that for some r (1<=r<=n), we have a_r = u+kv for some u,v in F^n and some scalar k. Let u = (b1, b2, ..., bn) and v = (c1, c2, ..., cn), and let B and C be the matrices obtained from A by replacing row r of A by u and v, respectively. We must prove that det(A) = det(B) + det(C). We leave the proof of this fact to the reader for the case r = 1.

*For r > 1 and 1 <= j <= n, the rows of A_(1,j), B_(1,j), C_(1,j) - where these are matrices from taking the determinant coefficient from the first row- are the same except for row r-1. Moreover, row r-1 of A_(1,j) is:*

(b_1 + kC1, ..., b_j-1 + kc_(j-1), b_(j+1) + kc_(j+1), ..., b_n + kc_n), which is the sum of the row r-1 of B_(1,j) and k times row r-1 of C_(1,j).

(b_1 + kC1, ..., b_j-1 + kc_(j-1), b_(j+1) + kc_(j+1), ..., b_n + kc_n), which is the sum of the row r-1 of B_(1,j) and k times row r-1 of C_(1,j).

**Question:**Above in italized, I have no idea what they mean by the row are the same (of the matrices left after taking the first determinant to the first row of A) except for row r-1???

Last edited: