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Unformatted text preview: Solutions to problems from section 1.5 6. Write the solution set of the given homogeneous system in parametric vector form. x 1 + 3 x 2 5 x 3 = 0 x 1 + 4 x 2 8 x 3 = 0 3 x 1 7 x 2 + 9 x 3 = 0 Solution: First we find the general solution: 1 3 5 0 1 4 8 0 3 7 9 0 NR 2 = R 2 R 1→ 1 3 5 0 1 3 0 3 7 9 0 NR 3 = R 3 +3 R 1→ 1 3 5 0 0 1 3 0 0 2 6 0 NR 3 = R 3 2 R 2→ 1 3 5 0 0 1 3 0 0 0 0 0 NR 1 = R 1 3 R 2→ 1 0 4 0 0 1 3 0 0 0 0 0 Thus the general solution is x 1 = 4 x 3 x 2 = 3 x 3 x 3 is free. We write the solution set in parametric vector form as follows: x = x 1 x 2 x 3 =  4 x 3 3 x 3 x 3 = x 3  4 3 1 so that x = t  4 3 1 ( t ∈ R ) 12. Describe all solutions of A x = in parametric vector form where A is row equivalent to the matrix 1 5 2 6 9 0 0 1 7 4 8 0 0 0 0 0 1 0 0 0 0 0 Solution: Let’s reduce the above matrix to RREF: 1 5 2 6 9 0 0 1 7 4 8 0 0 0 0 0 1 0 0 0 0 0 NR 2 = R 2 +8 R 3→ 1 5 2 6 9 0 0 0 1 7 4 0 0 0 0 0 0 1 0 0 0 0 0 0 NR 1 = R 1 2 R 2→ 1 5 0 8 1 0 0 0 1 7 4 0 0 0 0 0 0 1 0 0 0 0 0 0 Looking at the above matrix we see the general solution to the equation A x = is x 1 = 5 x 2 8 x 4 x 5 x 2 is free x 3 = 7 x 4 4 x 5 x 4 is free x 5 is free x 6 = 0 We write the solution set in parametric vector form as follows: x = x 1 x 2 x 3 x 4 x 5 x 6 =  5 x 2 8 x 4 x 5 x 2 7 x 4 4 x 5 x 4 x 5...
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This note was uploaded on 01/28/2010 for the course MATH 307 taught by Professor Axenovich during the Fall '08 term at Iowa State.
 Fall '08
 AXENOVICH

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