Lesson 9b -Homogenous Systems ^ Particular Solution

3 23092013 example1

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Unformatted text preview: solution c. This means that Ap=b and Ac=b, this means that Ap=Ac which gives Ap‐Ac=0. Factoring A gives A(p‐c)=0. This means that p‐c is a solution to the Homogenous system Ax=0. This means we can p c t1v1 t 2 v2 ... t k vk for some t1 , t 2 ,..., t k R find , and isolating for c gives: c p t1v1 t 2 v2 ... t k vk for some t1 , t 2 ,..., t k R which is of the form set in S. 3 23/09/2013 Example 1 Determine a particular solution to the following system 1 0 2 0 1 0 1 1 0 1 0 0 0 0 0 The easiest way (when the solution is in RREF like the one above) to find a particular solution is to choose all of our free variables as 0, and to choose each basic variable as what is on the right hand side (everything else will be 0, so the equations will be easier to solve. This means our particular solution will be: (note you could do the matrix multiplication Ax to check) 1 1 p 0 0 Example 2 Find the solution set to the following system. State the particular solution, and the solution set for the homogenous system. 3 1 8 12 1 1 4 6 2 0 4 6 3 2t | t R S 3 2t t 3 2 S 3 t 2 | t R 0 1 1 0 2 3 0 1 2 3 0 0 0 0 3 p 3 0 2 H t 2 | t R 1 4...
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