WhyRowOps - ROW OPERATIONS AND DETERMINANTS TERRY A LORING The fast way to calculate determinents is usually to use row operations Row operations

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

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

Unformatted text preview: ROW OPERATIONS AND DETERMINANTS TERRY A. LORING The fast way to calculate determinents is usually to use row operations. Row operations do change determinents, but do so predictably. The main technical lemma needed is to show that exchanging the top two rows in a matrix will alter just the sign of the determinent. 1. D ELETING TWO ROWS AND COLUMNS AT ONCE If A is an n-by-n matrix, with n at least 2, define A \ {i; k } to be the (n − 1)-by-(n − 1) matrix the comes from deleting row i and column k in A. If A is an n-by-n matrix, with n at least 3, define A \ {i, j ; k, l} to be the (n − 2)-by-(n − 2) matrix the comes from deleting rows i and j and columns k and k in A. The row numbering is as in the original matrix. For example, if 1234 5 6 7 8 A= a b c d efgh 568 A \ {1; 3} = a b d efh A \ {1, 2; 3, 2} = The determinant is defined, for A = [aij ], 1 then and ad eh . 2 TERRY A. LORING as n det(A) = − j =1 (−1)j a1,j det (A \ {1; j }) . Applying this to the smaller determinants is a little tricky, since the indexing in the smaller matrices is no the same as the indexing of the aij in A. Let us define 1 if j < k 0 σ (j, k ) = −1 j > k to tell us is two numbers are “in order,” equal or “out of order.” As to the big A and its determinant: • If j < k the scalar a2k will show up in A \ {1; j } in row 1 and column k − 1. • If j = k the scalar a2k will have been deleted when making A \ {1; j }. • If j > k the scalar a2k will show up in A \ {1; j } in row 1 and column k. So, in terms of the original aij , the definition of determinant tells us n det(A \ {1; j }) = − k =1 (−1)k σ (k, j )a2,j det (A \ {1, 2; j, k }) Notice the zeroed out term at k = j, so there isn’t really one to many terms here. Notice also the order in σ (k, j ) is as needed to put in the extra minus sign we need when j < k. So, in one formula, we find n n det(A) = − n (−1) a1,j j =1 n j − k =1 (−1)k σ (k, j )a2,k det (A \ {1, 2; j, k }) = j =1 k =1 σ (k, j )(−1)j +k a1,j a2,k det (A \ {1, 2; j, k }) . Another perspective: det(A) = j <k σ (k, j )(−1)j +k a1,j a2,k + σ (j, k )(−1)k+j a1,k a2,j det (A \ {1, 2; j, k }) (−1)j +k+1 det j <k = a1,j a1,k a2,j a2,k det (A \ {1, 2; j, k }) ROW OPERATIONS AND DETERMINANTS 3 So if we let A ∩ {r, s; t, w } be the result of tossing all rows except rows r and s and tossing all columns except columns t and w we have the formula det(A) = j <k (−1)j +k+1 det (A ∩ {1, 2; j, k }) det (A \ {1, 2; j, k }) Now it should be clear we can do row operations in the first two rows and the last n − 2 rows (induction). The formula n det(A) = j =1 (−1)j +1 a1,j det (A \ {1; j }) involve the last n − 1 rows. shows we can do row operations that That’s enough! Back to the example 123 5 6 7 A= a b c efg Then det(A) is 4 8 d h 567 568 578 678 1 b c d −2 a c d +3 a b d −4 a b c efg efh egh fgh or, by the other formula, 12 56 + + 23 67 34 78 13 cd − 57 gh ad 24 − eh 68 ab . ef 14 bd + 58 fd ac eg bc fg U NIVERSITY OF N EW M EXICO URL: www.math.unm.edu/~loring ...
View Full Document

This note was uploaded on 12/11/2010 for the course MATH 311 taught by Professor Coluos during the Spring '10 term at UNMSM.

Ask a homework question - tutors are online