Question

_{Theorem 4.3. The determinant of an n × n matrix is a linear function of each row when the remaining rows are held fixed. That is, for 1 Sr S n, we have ar-1 ar-1 ar-1 ar+1 ar+1 ar+1 an an rt whenever k is a scalar and u, v, and each a are row vectors in F". Proof. The proof is by mathematical induction on n. The result is imme- diate if n-1. Assume that for some integer n 2 2 the determinant of any (n-1) x (n - 1) matrix is a linear function of each row when the remaining 213 Sec. 4.2 Determinants of Order n rows are held fixed. Let A be an n × n matrix with rows al, aa-, . , an, respec- tively, and suppose that for some r (1< r < n), we have ar-u+kv for some u,v F" and some scalar k. Let u- (bi,b2,...,bn) and v- (ci,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) +kdet(C). We leave the proof of this fact to the reader for the case r -1. For r > 1 and 1 Sj s n, the rows of Ay, Bij, and Ciy are the same except for row r - 1. Moreover, row r-1 of Aij which is the sum of row r- 1 of Bıj and k times row r -1 ofCij. Since Bj and Ciy are (n - 1) x (n - 1) matrices, we have by the induction hypothesis. Thus since A,-B,-Gj, we have = Σ(-1)1+)Alj" |det(By) + k det@g) = det(B) + k det(C) This shows that the theorem is true for n × n matrices, and so the theorem is true for all square matrices by mathematical induction.}

_{QUESTION: PROVE THE FOLLOWING 4.3 THEOREM IN THE CASE r=1(no induction required, just use the definition of the determinants)}

Answer 1

Consider an n x n matrix of real entries given by A - [aijl. By exspanding by first row we get signed minor oof aij . Consider the nx n matrix E whose first row is (c.a1,1 + bi,1 , c.a1.2 + b1,2 , С.a1,3 + b1,2 , cal,n + b1,n) but all other rows are same as that of (c.an+bin).M Thus T is linear as function of the first row.