By Wilfred Kaplan
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Additional resources for Advanced Calculus
Gauss procedure. 67) where A = (ai,) and k = col (k, , . . , k,). One forms the "augmented matrix" formed from A by adjoining k as (n + 1)-st column. One operates on the matrix B by I. adding multiples of one row to another row, 11. interchanging two rows. 69) obtained by augmntation from an upper triangular matrix D-that is, a matrix with 0's below the principal diagonal. 44) by new ones with the same solutions. Since D is upper triangular, the new equations have the form . and are easily solved for x,,, x , - ~ .
__f -2- 3. [Hint: Write the left side as + - f 4. Prove thc Law of Cosines in En: If P I P2 and P I P3 are nonzero vectors, then '+ where 8 is the angle between PI Pr and PI P3. 5. a) In E 5 , find the sides and angles of the triangle with vertices (1,2,3,4,5), (5,4,2,3,1), and (2,2,2,2,2). b) Prove: In E n , the sum of the angles of a triangle is n. [Hint: For a triangle with sides a , b, c in En, construct the triangle with the same sides in the plane E2; why is this possible? ] 6. Prove the rules ( I .
X, in turn. Thus we have found the unique solution. 71) where C is the number of row interchanges. Thus, for nonsingular A , det A and det D . . dnn # are both nonzero. 5, det D = d l 0, . . 71) one can at once obtain det A. Gauss-Jordan procedure. In addition to steps I and 11, we use a step: 111. multiplying a row by a nonzero scalar. By applications of these three steps we can obtain from the matrix B a new matrix E=[l PI, and the new equations (still having the same solutions) have the form we have again obtained the unique solution.