788 CHAPTER 11 Systems of Equations and Inequalities Finding Minors of a 3 by 3 Determinant For the determinant = − − − A 2 1 3 2 5 1 0 6 9 , find: (a) M12 (b) M23 Solution EXAMPLE 3 (a) M12 is the determinant that results from removing the first row and the second column from A. ( )( ) = − − − = − − = − − − ⋅ = A M 2 1 3 2 5 1 0 6 9 2 1 0 9 2 9 0 1 18 12 (b) M23 is the determinant that results from removing the second row and the third column from A. ( ) = − − − = − = ⋅ − − = A M 2 1 3 2 5 1 0 6 9 2 1 0 6 2 6 0 1 12 23 Referring to formula (9), note that each element aij in the first row of the determinant is multiplied by its minor, but sometimes this term is added and other times subtracted. To determine whether to add or subtract a term, we must consider the cofactor . DEFINITION Cofactor For an n by n determinant A, the cofactor of entry a , ij denoted by A , ij is given by ( ) = − + A M 1 ij i j ij where Mij is the minor of entry a . ij The exponent of ( ) − + 1 i j is the sum of the row and column of the entry a , ij so if +i j is even, ( ) − = + 1 1, i j and if +i j is odd, ( ) − = − + 1 1. i j To find the value of a determinant, multiply each entry in any row or column by its cofactor and sum the results. This process is referred to as expanding across a row or column . For example, the value of the 3 by 3 determinant in formula (9) was found by expanding across row 1. Expanding down column 2 gives ( ) ( ) ( ) ( ) = − + − + − =− ⋅ + ⋅ + − ⋅ + + + a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a 1 1 1 1 1 1 11 12 13 21 22 23 31 32 33 1 2 12 21 23 31 33 2 2 22 11 13 31 33 3 2 32 11 13 21 23 12 21 23 31 33 22 11 13 31 33 32 11 13 21 23 ↑ Expand down column 2. Expanding across row 3 gives ( ) ( ) ( ) ( ) = − + − + − = ⋅ + − ⋅ + ⋅ + + + a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a 1 1 1 1 1 1 11 12 13 21 22 23 31 32 33 3 1 31 12 13 22 23 3 2 32 11 13 21 23 3 3 33 11 12 21 22 31 12 13 22 23 32 11 13 21 23 33 11 12 21 22 ↑ Expand across row 3.
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