By N. Biggs
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Книга Graphs and Digraphs, 3rd variation Graphs and Digraphs, 3rd variation Книги Математика Автор: Gary Chartrand, L. Lesniak Год издания: 1996 Формат: pdf Издат. :Chapman & Hall/CRC Страниц: 432 Размер: 22,5 ISBN: 041298721X Язык: Английский0 (голосов: zero) Оценка:This is the 3rd variation of the preferred textual content on graph concept.
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22 Linear Algebra Proof: We prove part (d) and leave the others as exercises (see Problems 21 through 23). Let A ¼ [aij] and B ¼ [bij] have orders n Â m and m Â p, so that the product AB is defined. Then ÀÂ ÃÂ ÃÁT ðABÞT ¼ aij bij " m X ¼ aik bkj #T definition of matrix multiplication k¼1 " m X ¼ ajk bki # definition of the transpose k¼1 " m X ¼ aTkj bTik # definition of the transpose k¼1 " m X ¼ bTik aTkj # k¼1 h ih i ¼ bTij aTij definition of matrix multiplication ¼ BT A T Observation: The transpose of a product of matrices is not the product of the transposes but rather the commuted product of the transposes.
47 48 Linear Algebra Determinants are defined in terms of permutations on positive integers. The theory is arduous and, once completed, gives way to simpler methods for calculating determinants. Because we make such limited use of determinants, we will not develop its theory here, restricting ourselves instead to the standard computational techniques. Determinants are defined only for square matrices. Given a square matrix A, we use det(A) or |A| to designate the determinant of A. If a matrix can be exhibited, we designate its determinant by replacing the brackets with vertical straight lines.
GAUSSIAN ELIMINATION Step 1. Construct an augmented matrix for the given system of equations. Step 2. Use elementary row operations to transform the augmented matrix into an augmented matrix in row-reduced form. Step 3. Write the equations associated with the resulting augmented matrix. Step 4. Solve the new set of equations by back substitution. The new set of equations resulting from Step 3 is called the derived set, and it is solved easily by back-substitution. Each equation in the derived set is solved for the first unknown that appears in that equation with a nonzero coefficient, beginning with the last equation and sequentially moving through the system until we reach the first equation.