C3092110 - i th row and j th column of A So M ij is a k 1 × k 1 matrix with two identical rows By our assumption we have det M ij = 0 for all j =

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
10. (Statement of the problem.) Proof. Step One : Suppose that A is a 2 × 2 matrix with two identical rows. Then we must have A = ± a b a b . and therefore det A = ab - ab = 0. So the statement is true for n = 1. Step Two : Suppose that the statement is true for n = k . That is, for any ( k + 1) × ( k + 1) matrix M with two identical rows, det M = 0. ( We set up the induction hypothesis here. ) Let A be a ( k + 2) × ( k + 2) matrix with two identical rows. ( We want to prove the statement for n = k + 1 . This is the point we begin with. ) Choose a row that is not one of the two identical rows, and call this row the i th row. Let M ij be the matrix obtained by deleting the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: i th row and j th column of A . So M ij is a ( k + 1) × ( k + 1) matrix with two identical rows. By our assumption, we have det M ij = 0 for all j = 1 , ··· ,k + 2 By the definition of A ij , we have A ij = (-1) i + j det M ij = 0. Hence by the cofactor expansion theorem, det A = a i 1 A i 1 + a i 2 A i 2 + ··· + a i,k +2 A i,k +2 = 0 . So the statement is true for n = k + 1. ( This is the end of step two. ) By induction, det A = 0 for any ( n +1) × ( n +1) matrix with two identical rows for any n ≥ 1. / 1...
View Full Document

This note was uploaded on 04/01/2008 for the course MATH 309 taught by Professor Samiabdul during the Spring '08 term at Michigan State University.

Ask a homework question - tutors are online