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

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

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

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