18.06 Linear Algebra, Final Exam Solution - MIT …

Note: As stated there is no solution (my apologies!). All solutions to AT Av = 0 are multiples of (1, 1, 1, 1) which rules out v1 = 1 and v4 = 0. Intended problem: I meant to solve the reduced equations using KCL only at nodes 2 and 3. In fact symmetry gives v2 = v3 = 1 2. Then the currents are w1 = w2 = w5 = w6 = 2 1 around the sides and w 3 ...

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GetATeacher2 - Vanderbilt University

Title: GetATeacher2.eps Author: g4 Created Date: 6/20/2005 5:17:29 PM

Solving Inequalities Date Period

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