Math 221: LINEAR ALGEBRA 3-1. The Cofactor Expansion Le Chen 1 - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA §3-1. The Cofactor Expansion Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

det Determinant of 2 × 2 matrix � a � b Recall that if A = , then the determinant of A is defined as c d det A = ad − bc , and that A is invertible if and only if det A � = 0 .

Determinant of 2 × 2 matrix � a � b Recall that if A = , then the determinant of A is defined as c d det A = ad − bc , and that A is invertible if and only if det A � = 0 . � a � � � b a b � � Notation. For det , we often write � , i.e., use vertical bars � � c d c d � instead of square brackets.

Determinant of 2 × 2 matrix � a � b Recall that if A = , then the determinant of A is defined as c d det A = ad − bc , and that A is invertible if and only if det A � = 0 . � a � � � b a b � � Notation. For det , we often write � , i.e., use vertical bars � � c d c d � instead of square brackets. Problem How to define determinant for a general n × n matrix?

2 × 2 − a 12 a 22 a 11 a 21 +

3 × 3 − − − a 13 a 23 a 33 a 13 a 23 a 12 a 22 a 32 a 12 a 22 a 11 a 21 a 31 a 11 a 21 + + +

4 × 4 a 14 a 24 a 34 a 44 a 14 a 24 a 34 a 13 a 23 a 33 a 43 a 13 a 23 a 33 a 12 a 22 a 32 a 42 a 12 a 22 a 32 a 11 a 21 a 31 a 41 a 11 a 21 a 31

5 × 5 · · ·

5 × 5 · · · The determinant of an n × n matrix is more effectively defined through recursion...

det det det Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix.

det det det Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix. – The sign of the ( i , j ) position is ( − 1) i + j . Thus the sign is 1 if ( i + j ) is even, and − 1 if ( i + j ) is odd.

det det det Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix. – The sign of the ( i , j ) position is ( − 1) i + j . Thus the sign is 1 if ( i + j ) is even, and − 1 if ( i + j ) is odd. Let A ij denote the ( n − 1) × ( n − 1) matrix obtained from A by deleting row i and column j.

det det Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix. – The sign of the ( i , j ) position is ( − 1) i + j . Thus the sign is 1 if ( i + j ) is even, and − 1 if ( i + j ) is odd. Let A ij denote the ( n − 1) × ( n − 1) matrix obtained from A by deleting row i and column j. – The ( i , j ) -cofactor of A is c ij ( A ) = ( − 1) i + j det ( A ij ) .

det det Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix. – The sign of the ( i , j ) position is ( − 1) i + j . Thus the sign is 1 if ( i + j ) is even, and − 1 if ( i + j ) is odd. Let A ij denote the ( n − 1) × ( n − 1) matrix obtained from A by deleting row i and column j. – The ( i , j ) -cofactor of A is c ij ( A ) = ( − 1) i + j det ( A ij ) . Finally,

Cofactor and cofactor expansion Definitions Let A = [ a ij ] be an n × n matrix. – The sign of the ( i , j ) position is ( − 1) i + j . Thus the sign is 1 if ( i + j ) is even, and − 1 if ( i + j ) is odd. Let A ij denote the ( n − 1) × ( n − 1) matrix obtained from A by deleting row i and column j. – The ( i , j ) -cofactor of A is c ij ( A ) = ( − 1) i + j det ( A ij ) . Finally, – det A = a 11 c 11 ( A ) + a 12 c 12 ( A ) + a 13 c 13 ( A ) + · · · + a 1 n c 1 n ( A ) , and is called the cofactor expansion of det A along row 1 .

Example   1 2 3 Let A = 4 5 6  . Find det A.  7 8 9 Using cofactor expansion along row 1, det A = 1 c 11 ( A ) + 2 c 12 ( A ) + 3 c 13 ( A ) � � � � � � 5 6 4 6 4 5 1( − 1) 2 � � + 2( − 1) 3 � � � + 3( − 1) 4 � � � = � � � � � � 8 9 7 9 7 8 � � � � = (45 − 48) − 2(36 − 42) + 3(32 − 35) = − 3 − 2( − 6) + 3( − 3) = − 3 + 12 − 9 = 0

det Example (continued)   1 2 3 A = 4 5 6   7 8 9 Now try cofactor expansion along column 2.

Example (continued)   1 2 3 A = 4 5 6   7 8 9 Now try cofactor expansion along column 2. det A = 2 c 12 ( A ) + 5 c 22 ( A ) + 8 c 32 ( A ) � � � � � � 4 6 1 3 1 3 2( − 1) 3 � � + 5( − 1) 4 � � � + 8( − 1) 5 � � � = � � � � � � 7 9 7 9 4 6 � � � � = − 2(36 − 42) + 5(9 − 21) − 8(6 − 12) = − 2( − 6) + 5( − 12) − 8( − 6) = 12 − 60 + 48 = 0 .

Example (continued)   1 2 3 A = 4 5 6   7 8 9 Now try cofactor expansion along column 2. det A = 2 c 12 ( A ) + 5 c 22 ( A ) + 8 c 32 ( A ) � � � � � � 4 6 1 3 1 3 2( − 1) 3 � � � + 5( − 1) 4 � � � + 8( − 1) 5 � � = � � � � � � 7 9 7 9 4 6 � � � � = − 2(36 − 42) + 5(9 − 21) − 8(6 − 12) = − 2( − 6) + 5( − 12) − 8( − 6) = 12 − 60 + 48 = 0 . We get the same answer!

Why is this signifjcant? Theorem (Cofactor Expansion Theorem) The determinant of an n × n matrix A can be computed using the cofactor expansion along any row or column of A. That is det A can be computed by multiplying each entry of the row or column by the corresponding cofactor and adding the results.

Why is this signifjcant? Theorem (Cofactor Expansion Theorem) The determinant of an n × n matrix A can be computed using the cofactor expansion along any row or column of A. That is det A can be computed by multiplying each entry of the row or column by the corresponding cofactor and adding the results.

det det Example   0 1 2 1 5 0 0 7   Let A =  . Find det A.   0 1 − 1 0  3 0 0 2

Example   0 1 2 1 5 0 0 7   Let A =  . Find det A.   0 1 − 1 0  3 0 0 2 Cofactor expansion along row 1 yields det A = 0 c 11 ( A ) + 1 c 12 ( A ) + 2 c 13 ( A ) + 1 c 14 ( A ) = 1 c 12 ( A ) + 2 c 13 ( A ) + c 14 ( A ) , whereas cofactor expansion along, row 3 yields det A = 0 c 31 ( A ) + 1 c 32 ( A ) + ( − 1) c 33 ( A ) + 0 c 34 ( A ) = 1 c 32 ( A ) + ( − 1) c 33 ( A ) , i.e., in the first case we have to compute three cofactors, but in the second we only have to compute two.

Example (continued) We can save ourselves some work by using cofactor expansion along row 3 rather than row 1.   0 1 2 1 5 0 0 7   A =   0 1 − 1 0   3 0 0 2 det A = 1 c 32 ( A ) + ( − 1) c 33 ( A ) � � � � 0 2 1 0 1 1 � � � � 1( − 1) 5 � � + ( − 1)( − 1) 6 � � = 5 0 7 5 0 7 � � � � � � � � 3 0 2 3 0 2 � � � � � � � � 5 7 5 7 ( − 1)2( − 1) 3 � � � + ( − 1)1( − 1) 3 � � = � � � � 3 2 3 2 � � � = 2(10 − 21) + 1(10 − 21) = 2( − 11) + ( − 11) = − 33 .

Example (continued)   0 1 2 1 5 0 0 7   Try computing det  using cofactor expansion along other   0 1 − 1 0  3 0 0 2 rows and columns, for instance column 2 or row 4. You will still get det A = − 33 .

det det Problem   − 8 1 0 − 4 5 7 0 − 7   Find det A for A =  .   12 − 3 0 8  − 3 11 0 2

det Problem   − 8 1 0 − 4 5 7 0 − 7   Find det A for A =  .   12 − 3 0 8  − 3 11 0 2 Solution Using cofactor expansion along column 3, det A = 0 .

Problem   − 8 1 0 − 4 5 7 0 − 7   Find det A for A =  .   12 − 3 0 8  − 3 11 0 2 Solution Using cofactor expansion along column 3, det A = 0 . Remark If A is an n × n matrix with a row or column of zeros, then det A = 0 .

det det det det Elementary Row Operations and Determinants Example  2 0 − 3  Let A = 0 4 0  . Then  1 0 − 2

det det det Elementary Row Operations and Determinants Example  2 0 − 3  Let A = 0 4 0  . Then  1 0 − 2 � � 2 − 3 det A = 4( − 1) 4 � � � = 4( − 1) = − 4 . � � 1 − 2 �

Elementary Row Operations and Determinants Example  2 0 − 3  Let A = 0 4 0  . Then  1 0 − 2 � � 2 − 3 det A = 4( − 1) 4 � � � = 4( − 1) = − 4 . � � 1 − 2 � Let B 1 , B 2 , and B 3 be obtained from A by performing a type 1, 2 and 3 elementary row operation, respectively, i.e.,       2 0 − 3 2 0 − 3 2 0 − 3 B 1 = 1 0 − 2  , B 2 = 0 4 0  , B 3 = 0 4 0  .    0 4 0 − 3 0 6 5 0 − 8 Compute det B 1 , det B 2 , and det B 3 .

det det det det Elementary Row Operations and Determinants Example (continued) � � 2 − 3 det B 1 = 4( − 1) 5 � � � = ( − 4)( − 1) = 4 = ( − 1) det A . � � 1 − 2 �