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

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