1.5soln

# 1.5soln - Solutions to problems from section 1.5 6 Write...

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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.

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1.5soln - Solutions to problems from section 1.5 6 Write...

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