ee231_HW6 - Ee231_B1_Vorobyov Damir_Iraliyev_1246540...

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Unformatted text preview: Ee231_B1_Vorobyov Damir_Iraliyev_1246540 Assignment 6 9.5 Given equations: .- =- . 1 0 5x1 x2 9 5 .- =- . 2 1 02x1 2x2 18 8 (a) Solve graphically. (b) Compute the determinant. (c) On the basis of (a) and (b), what would you expect regarding the systems conditions? (d) Solve by the elimination of unknowns (e) Solve again, but with a11 modified slightly to 0.52. Interpret your result. Solution: (a) For the Graph, please refer to the back of this report. The solution seems to be at = = . x1 10 and x2 14 5 (b) = . - .- = . - -- . = . det0 5 11 02 2 0 5 2 11 02 0 02 (c) On the basis of (a) and (b), regarding systems conditions, I would expect to be dealing with ill- conditioned system; (a) there is no clear point of intersection, (b) the is small. (d) .- =- . = . + . 1 0 5x1 x2 9 5 i x2 0 5x1 9 5 (e) . .- . + . =- . sub iinto 2 1 02x1 20 5x1 9 5 18 8 (f) . = . 0 02x1 0 2 (g) = x1 10 (h) = . x2 14 5 (i) (e) .- =- . = . + . ( ) 1 0 52x1 x2 9 5 x2 0 52x1 9 5 i (j) .- =- . 2 1 02x1 2x2 18 8 (k) . : sub iinto 2 and we get (l) .- . + . =- . 1 02x1 20 52x1 9 5 18 8 (m) - . = . 0 02x1 0 2 (n) =- & = . x1 10 x2 4 3 (o) Slight change in a11 caused a dramatic results changes, coming from 0.2-> -0.2 for xcaused a dramatic results changes, coming from 0....
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This note was uploaded on 12/20/2010 for the course E E 231 taught by Professor Vorobyov during the Spring '10 term at University of Alberta.

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ee231_HW6 - Ee231_B1_Vorobyov Damir_Iraliyev_1246540...

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