Week_5

# Week_5 - Week 5 3.2 Determinants and Matrix Inverses...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Week 5 3.2 Determinants and Matrix Inverses Theorem 1: Suppose A is an n × n matrix and k is a scalar. Then det( kA ) = k n det( A ) since there are n rows and we can pull a common factor of k out of each row. Theorem 2: If A and B are square matrices of the same size, then det( AB ) = det( A ) det( B ) Theorem 3: A square matrix A is invertible iff det( A ) 6 = 0. Proof: Let R be the RREF of A . Then there is a sequence of EROs that reduce A to R . Hence, there exist elementary matrices E 1 , E 2 , .. . E k such that E k ··· E 2 E 1 A = R . ( ⇒ ) Assume det( A ) 6 = 0. Taking the determinant of both sides of the equation above, we have det( E k ··· E 2 E 1 A ) = det( R ) det( E k ) ··· det( E 2 ) det( E 1 ) det( A ) = det( R ) Since the determinant of an elementary matrix is either 1, -1 or k 6 = 0, then the left side of the equation is non-zero. Hence the right side of the equation is non-zero and so det( R ) 6 = 0. Therefore, R does not contain a row of zeros. Thus, R = I and A is invertible. ( ⇐ ) Assume A is invertible. Then R = I . Then there is a sequence of EROs that reduce A to I . Hence, there exist elementary matrices E 1 , E 2 , .. . E k such that E k ··· E 2 E 1 A = I . Taking the determinant of both sides of the equation above, we have det( E k ··· E 2 E 1 A ) = det( I ) det( E k ) ··· det( E 2 ) det( E 1 ) det( A ) = I Thus, det( A ) 6 = 0. 1 Theorem 4: If A is invertible, then det( A- 1 ) = 1 det( A ) . Proof: Since AA- 1 = I , then det( AA- 1 ) = det( I ) or det( A ) det( A- 1 ) = 1, and since det( A ) 6 = 0, we have det( A- 1 ) = 1 det( A ) . Examples: 1. If det( A ) = 2 and det( B ) = 5, find det( A 3 B- 1 A T B 2 ). 2. If A is a 2 × 2 matrix where | A | =- 4, find det(3 A ), det(3 A 2 ), and det[(3 A ) 2 ]. 3. Is A invertible? Definition: The adjugate of an n × n matrix is the transpose of the matrix of cofactors, or adj ( A ) = [ c ij ( A )] where [ c ij ( A )] is the matrix whose ( i, j ) entry is the ( i, j ) cofactor of A . 2 Example 4: If A = 1 3- 2 1 5- 2- 6 7 , find adj ( A ). Theorem 5: If A is an n × n matrix, then A ( adj ( A )) = det( A ) I . Furthermore, if A is invertible, then A- 1 = 1 det( A ) ( adj ( A )). Proof: Example 5 Calculate the determinant of A , where A is the matrix in the Example 4. 3 Another use of determinants is in solving systems of equations. So far, we have seen two methods: augmented matrices and coefficient matrices (if A is invertible). Cramer’s Rule If A is an invertible n × n matrix, then the solution to A~x = ~ b is x 1 = | A 1 | | A | , x 2 = | A 2 | | A | , .. . ,x n = | A n | | A | where A j is the matrix obtained by replacing the entries in the j th column of A by the entries in the matrix ~ b = b 1 b 2 ....
View Full Document

## This note was uploaded on 08/05/2008 for the course MATH 115 taught by Professor Dunbar during the Fall '07 term at Waterloo.

### Page1 / 11

Week_5 - Week 5 3.2 Determinants and Matrix Inverses...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online