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Unformatted text preview: ECE 130C HW-7 Solutions June 3, 2008 Problem 4.2.12
1 a a2 1 b b2 1 c c2 1 a a2 0 b - a b2 - a2 0 c - a c2 - a2 1 b+a 1 c+a = = (b - a)(b - c) = (b - a)(c - a)(c - b) Problem 4.2.16
(a) det(A) = 0 since R1 + R3 = 2R2 (b) det(A) = (1 - t2 )3 after LU factorization. Problem 4.3.26
1 -1 |B4 | = 2 -1 2 -1 -1 2 Pivots are all Ones. 1 -1 -1 2 -1 -1 + = 2|B3 | - |B2 | Problem 4.3.34
(a) Gaussian elimination leads to simultaneous triangularization of both the blocks A and D. NOTE: A, B,C and D are matrices. So the given matrix is as such not triangular. If A and D are made triangular, then the overall matrix itself is triangular and so the determinant of the matrix is the product of the determinants of A and D. 1 (b),(c) Choose A=[1,0;0 0], B=[0 0;1 0], C=[0 1;0 0] and D=[0 0;0 1] Problem 4.4.38
Partial derivatives and evaluation of the determinant gives the result. Do not forget the dV. This is the volume increment that is used for volume evaluation. Problem 4.4.42
x y z 1 1 0 1 2 1 =0x-y+z =0 This is the volume of the bounding box enclosed by the the planes through these points. This is just a plane as a plane passes through these points and so the volume, which is the determinant is ZERO. Problem 4.4.44
By Cramer's rule, the components of x = A-1 b are the ratios |Bk /|A||. If b = e1 , then |Bk | = Cofactor C1k . Therefore x is the first row of the cofactor matrix divided by det(A), which is the first column of inv(A). 2 ...
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