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Unformatted text preview: Lecture 23 Andrei Antonenko April 2, 2003 1 Properties of determinants2 Now we’ll give a first motivation of the determinant. Theorem 1.1 (Criteria of invertibility). A square matrix A is invertible if and only if det A 6 = 0 . Proof. Let’s use elementary row operations to transform a matrix A to its triangular (row echelon) form. Let’s note, that if the determinant was not equal to 0, then it will not be equal to 0 after elementary row operations, and if it was equal to 0, it will be equal to 0 after elementary row operations. The matrix A is invertible, if it’s REF doesn’t have a row of zeros, i.e. the determinant of its REF is not equal to 0. So, the determinant of the initial matrix is not equal to 0. Moreover, A is not invertible if we have a row of zeros in its REF, so the determinant of REF equals to 0, and so, the determinant of the initial matrix A is equal to 0. Example 1.2. We computed the following determinant: fl fl fl fl fl fl fl 1 2 3 4 5 6 7 8 9 fl fl fl fl fl fl fl = 0 . So, this matrix is not invertible. This theorem gives us a criteria, when the matrix is invertible. But if we know that the determinant is not equal to 0, and so the matrix is invertible, the theorem doesn’t give us a method of computing the inverse. Now let’s continue with properties of determinants. Theorem 1.3 (The determinant of the transpose). det A > = det A . Example 1.4. Since fl fl fl fl fl fl fl 1 2 3 4 5 6 7 8 9 fl fl fl fl fl fl fl = 0 , we have that fl fl fl fl fl fl fl 1 4 7 2 5 8 3 6 9 fl fl fl fl fl fl fl = 0 . 1 Proof. The determinant of A > is equal to the sum of all possible products of matrix elements, taken 1 from each column and 1 from each row, as well as the determinant of A . So we have to check that the products are included with the same signs. To figure out the sign before a 1 k 1 a 2 k 2 ...a 3 k 3 in the expression for the determinant of A > , we have to reorder multiplicands by the second subscript. So, we can interchange multiplicands, and transpositions will occur simultaneously in the first subscripts and in the second subscripts. So, the sign of the final permutation will be the same, i.e. if at the end we’ll get a l 1 1 a l 2 2 ...a l n n , then sgn( k 1 ,k 2 ,...,k n ) = sgn( l 1 ,l 2 ,...,l n ) , and so, det A = det A > ....
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.
 Spring '03
 Andant

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