This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Linear Algebra Explore Assignment 1 I. Exercise on Linear Systems (a). A = [1 1 1 1  0 ] [ 1 2 2 2   4] [0 3 2 1  10] [2 0 0 1  4 ] (b). B = [1 0 0 0  4 ] [ 0 1 0 0   6 ] [ 0 0 1 0   10] [0 0 0 1  12 ] (c). Yes, x4 =12, x3 = 10, x2 = 6, and x1 = 4. (d). A = [1 1 1 1  0 ] [ 2 2 2 2   4] [0 3 2 1  10] [2 0 0 1  4 ] (e). ). C = [1 0 0 .5  0] [0 1 0 .4  0] [0 0 1 1.1  0] [0 0 0 0  2] (f). No, because the last row is all zeros and has inconsistency equal 2. (g). B is in the reduced echelon form and C isn’t. C has rows 1 through 3 equaling to zero while the last row is 2. This means that B has a unique solution while C does not have a solution. I I. Finding Polynomials – Cardiac Hill (a). A = [0 0 1  5] [1 1 1  3] [9 3 1  2] B = [1 0 0  .5 ] [0 1 0  2.5] [0 0 1  5 ] a = .5 b = 2.5 c = 5 (b). (c). (0,5), (1,3), (3,2), and (2,10). (0^3)a + (0^2)b + (0)c + 1d = 5 (1^3)a + (1^2)b + (1)c + 1d = 3 (3^3)a + (3^2)b + (3)c + 1d = 2 (2^3)a + (2^2)b + (2)c + 1d = 10 A = [0 0 0 1  5 ] [1 1 1 1  3 ] [27...
View
Full Document
 Fall '09
 Dwang
 Algebra, Harry Connick, Jr., F3, f4, downward sloping line

Click to edit the document details