MIT OpenCourseWare http ocw mit edu 18 085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use visit http ocw mit edu terms 18 085 Mathematical Methods for Engineers I Solutions Problem Set 2 Section 1 3 7 Suppose A is rectangular m by n and C is symmetric m by m matrix i AT CA T AT C T AT T since C C T symmetric AT CA AT CA is symmetry AT n m C m m A m n AT CA is n n a2 aT 1 ii Let A a1 AT A a T 2 aT n T a1 a1 a3 an a1 a2 an aT 2 a2 aT n an 2 Since aT i ai ai 0 we conclude that AT A has no negative numbers on its diagonal 1 3 A 3 2 1 0 1 3 3 1 0 7 0 1 3 1 0 1 0 1 3 1 0 7 L D LT 1 Prof Gilbert Strang A 1 b b c 1 0 b 1 1 b 0 c b2 1 0 1 0 1 b 0 1 b 1 0 c b2 L 2 1 A 1 2 0 1 1 1 2 0 1 1 2 0 LT D 0 1 2 0 0 1 0 2 3 1 0 0 1 0 2 3 1 L 2 1 0 0 3 2 1 2 0 0 4 3 2 0 0 0 3 2 0 0 0 4 3 D 1 1 2 0 0 1 2 3 0 0 1 LT Section 1 4 3 u x 1 3 x 2 3 General Fixed Fixed Solution 1 a x x a u 1 x a x a u x 1 3 1 1 3 x x 1 3 u 1 x 1 3 x 1 3 u x 2 3 1 2 3 x x 2 3 u 1 x 2 3 x 2 3 Combining two single load solutions 1 1 3 x 1 2 3 x x 1 3 u 1 x 1 3 1 2 3 x 1 3 x 2 3 1 x 1 3 1 x 2 3 x 2 3 x for x 1 3 1 3 for 1 3 x 2 3 1 x for x 2 3 2 Second Method Ax B u x Cx D Ex F x 1 3 1 3 x 2 3 x 2 3 u 0 0 A 0 B 0 B 0 1 0 A 1 3 B C 1 3 D A C 3D 2 3 A 1 C C 2 3 D E 2 3 F 4 2C 3D 2E 3F 5 C 1 E E 1 F 0 6 E F 2 3 C 3D 1 C 1 D 3 6 4 2C 3 1 3 2E 3 E 2C 1 E 7 7 2 5 1 3 3E E 1 C 1 E 1 1 C 0 E F F 1 A 1 C A 1 0 A 1 x u x 1 3 1 x x 1 3 1 3 x 2 3 x 2 3 3 5 Free Free condition u x R x a Cx D u 0 0 u 1 0 u 0 0 C 0 C 0 u 1 1 C 0 C 1 There are no solutions for C and D C cannot be 0 and 1 at the same time 7 f x x 1 3 x 2 3 u 1 0 u 0 0 Ax B u x Cx D Ex F x 1 3 1 3 x 2 3 x 2 3 u 0 A 0 0 B C 1 3 D A 1 3 3B C 3D A 1 C 2 C 1 C 2 3 D E 2 3 F 2C 3D 2E 3F C 1 E E 0 d Ex F 0 dx E 0 From 3 3 redundant less one equation 2C 3D 3F 3D 3F 2C 3F 2 D F 2 3 From 2 3B 1 3D 1 3F 2 3 F 1 3 B 4 F 1 3 x 1 3 u x x F 2 3 1 3 x 2 3 F x 2 3 F can take any value F R in nitely many solutions for u x 0 x 0 12 u x C x x3 x 0 6 Cubic spline C x is a particular solution for u u x C x Ax3 Bx2 Gx D Given that u 1 u 1 0 u 1 0 u 1 0 0 u 1 1 A B G D 0 6 A B G D 1 6 1 u 1 0 A B G D 0 2 A B G D 0 u 1 1 6A 1 2B 0 6A 2B 1 3 u 1 0 6A 2B 0 B 3A 1 1 B 4 12 1 1 G D 4 6 1 G D 0 4 A 1 12 1 12 G D 1 6 G D G 0 1 6 D u x C x 1 6 1 3 1 2 1 x x 12 4 6 5 Section 1 5 2 1 1 K 2 1 1 1 1 0 1 1 0 3 1 1 1 1 1 0 1 1 Q QT 0 3 1 1 2 1 1 2 1 1 3 1 1 1 3 1 1 2 1 4 2 4 2 2 2 1 2 1 1 Q 2 K 4 K toeplitz 2 1 zeros 1 3 Q E eig K DST Q diag 1 1 1 1 1 DST 0 2887 0 5000 0 5774 0 5000 0 5000 0 0000 0 5000 0 5000 0 5774 0 0000 0 5774 0 5000 0 5000 0 0000 0 2887 0 5000 0 5000 0 0000 0 2887 0 5774 0 5000 0 5000 0 5774 0 5000 0 2887 JK 1 5 1 5 sin JK pi 6 sqrt 3 ans 0 2887 0 5000 0 5774 0 5000 0 5000 0 5774 0 0000 0 5000 0 0000 0 5774 0 0000 0 5000 0 5000 0 0000 0 2887 0 5000 0 5000 0 5000 0 5774 0 5000 0 2887 0 5000 0 5774 0 5000 0 2887 6 DST sin JK pi 6 sqrt 3 DST ans 0 2887 0 5000 0 5774 0 5000 0 5000 0 0000 0 5000 0 2887 0 5000 0 5774 0 5000 0 2887 0 5000 0 5774 0 0000 0 5774 0 0000 0 5000 0 5000 0 0000 0 2887 0 5000 0 5000 0 5774 0 5000 inv DST ans 0 2887 0 5000 0 5774 0 5000 0 5000 0 0000 0 5000 0 5000 0 5774 0 0000 0 5774 0 0000 0 5000 0 5000 0 0000 0 2887 0 5000 7 2 1 0 5000 0 5774 0 5000 0 1 DST inv DST 0 2887 0 5000 0 5774 0 5000 0 2887 1 2 1 0 C4 0 1 2 1 …