University:
Massachusetts Institute of Technology
Course:
18.06SC | Linear Algebra
Academic year:

2022
Views:

320

Pages:

2
Author:

Roderick Lutz

Exercises on properties of determinants Problem 18.1: (5.1 #10. Introduction to Linear Algebra: Strang) If the en tries in every row of a square matrix A add to zero, solve Ax = 0 to prove that det A = 0. If those entries add to one, show that det( A − I ) = 0. Does this mean that det A = 1? Solution: If the entries of every row of A sum to zero, then Ax = 0 when x = (1, . . . , 1) since each component of Ax is the sum of the entries in a row of A. Since A has a non-zero nullspace, it is not invertible and det A = 0. If the entries of every row of A sum to one, then the entries in every row of A − I sum to zero. Hence A − I has a non-zero nullspace and det( A − I ) = 0. If det( A − I ) � = 0 it is� not necessarily true that det A = 1. For example, the rows of A = 0 1 1 0 sum to one but det A = −1. Problem 18.2: (5.1 #18.) Use row operations and the properties of the determinant to calculate the three by three “Vandermonde determinant”: ⎡ ⎤ 1 a a2 det ⎣ 1 b b2 ⎦ = (b − a)(c − a)(c − b). 1 c c2 Solution: have: Using row operations and properties of the determinant, we ⎤ ⎡ ⎤ 1 a a2 1 a a2 det ⎣ 1 b b2 ⎦ = det ⎣ 0 b − a b2 − a2 ⎦ 1 c c2 1 c c2 ⎡ ⎤ 1 a a2 = det ⎣ 0 b − a b2 − a2 ⎦ 0 c − a c2 − a2 ⎡ ⎤ 1 a a2 1 b+a ⎦ = (b − a) det ⎣ 0 1 c − a c2 − a2 ⎡ ⎤ 1 a a2 ⎦ = (b − a) det ⎣ 0 1 b+a 0 0 (c − a)(c − b) ⎡ ⎤ 1 a a2 = (b − a)(c − a)(c − b) det ⎣ 0 1 b + a ⎦ 0 0 1 ⎡ ⎡ ⎤ 1 0 0 = (b − a)(c − a)(c − b) det ⎣ 0 1 0 ⎦ 0 0 1 = (b − a)(c − a)(c − b).�