Exam 3 Study Guide Solution Fall 2007 on Differential Equations with Linear Algebra 1 - 2r2 r3 r3 r2 r4 r4 2 1 1 2 1 3 0 3 M341 Study Guide 3(S Zhang 1

Exam 3 Study Guide Solution Fall 2007 on Differential Equations with Linear Algebra 1

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M341 Study Guide 3 (S. Zhang) . 1. Find the row echelon form (okay to have non-one diago- nals, Gauss elimination) and solve the system. Find the reduced row echelon form (Gauss-Jordan elimination) and solve the system. 2 x 1 + x 2 + x 3 = 3 - 2 x 1 + x 2 + 0 x 3 = - 3 4 x 1 - 2 x 2 + 3 x 3 = 12 0 x 1 - 2 x 2 + 2 x 3 = 6 ans: Gauss elimination: 2 1 1 | 3 - 2 1 0 | - 3 4 - 2 3 | 12 0 - 2 2 | 6 - 2 r 1 + r 3 r 3 , r 1 + r 2 r 2 , 2 1 1 | 3 2 1 | 0 - 4 1 | 6 0 - 2 2 | 6 2 r 2 + r 3 r 3 , r 2 + r 4 r 4 , 2 1 1 | 3 2 1 | 0 3 | 6 0 3 | 6 - r 3 + r 4 r 4 , 2 1 1 | 3 2 1 | 0 3 | 6 0 0 | 0 Backward substitution, x = 1 - 1 2 Gauss-Jordan elimination: 2 1 1 | 3 - 2 1 0 | - 3 4 - 2 3 | 12 0 - 2 2 | 6 - 2 r 1 + r 3 r 3 , r 1 + r 2 r 2 , 2 1 1 | 3 2 1 | 0 - 4 1 | 6 0 - 2 2 | 6 2 r 2 + r 3 r 3 , r 2 + r 4 r 4 , 2 1 1 | 3 2 1 | 0 3 | 6 0 3 | 6 - r 3 + r 4 r 4 , 2 1 1 | 3 2 1 | 0 3 | 6 0 0 | 0 1 2 r 1 r 1 , 1 2 r 2 r 2 , 1 3 r 3 r 3 , 1 1 / 2 1 / 2 | 3 / 2 1 1 / 2 | 0 1 | 2 0 0 | 0 - 1 2 r 3 + r 1 r 1 , 1 3 r 3 + r 2 r 2 , 1 1 / 2 | 1 / 2 1 | - 1 1 | 2 0 0 | 0 - 1 2 r 2 + r 1 r 1 1 | 1 1 | - 1 1 | 2 0 0 | 0 It shows the solution, x = 1 - 1 2 2. Solve the system Ax = b by the Gaussian elimination. A = 1 1 - 1 2 1 2 1 - 1 2 2 0 - 1 1 - 2 0 , b = 1 3 1 ans: We do row operations on the matrix. 1 1 - 1 2 1 | 1 2 1 - 1 2 2 | 3 0 - 1 1 - 2 0 | 1 - 2 r 1 + r 2 1 1 - 1 2 1 | 1 - 1 1 - 2 0 | 1 0 - 1 1 - 2 0 | 1 - r 2 + r 3 1 1 - 1 2 1 | 1 - 1 1 - 2 0 | 1 0 0 0 0 0 | 0 1
Therefore, we can let x 3 , x 4 , x 5 be free variables. The solution would be x = 2 - 1 0 0 0 + C 1 0 1 1 0 0 + C 2 0 - 2 0 1 0 + C 3 - 1 0 0 0 1 a=[1 1 -1 2 1] b=[1 0 0 0 1] A=[a; b+a; b-a] x=[1 0 1 0 1]’; A*x 3. (1) Use the Gauss elimination to solve the system. (2) Find the LU decomposition of A , and use it to solve the system. (2) Find the A - 1 , and use it to solve the system. 2 x 1 + x 2 + x 3 = 3 - 2 x 1 + x 2 + 0 x 3 = - 3 4 x 1 - 2 x 2 + 3 x 3 = 12 ans: (1) Gauss elimination: 2 1 1 | 3 - 2 1 0 | - 3 4 - 2 3 | 12 - 2 r 1 + r 3 r 3 , r 1 + r 2 r 2 , 2 1 1 | 3 2 1 | 0 - 4 1 | 6 2 r 2 + r 3 r 3 , 2 1 1 | 3 2 1 | 0 3 | 6 Backward substitution, x = 1 - 1 2 (2) LU decomposition: We repreat GE steps, but record the negative row operations: 2 1 1 - 2 1 0 4 - 2 3 - 2 r 1 + r 3 r 3 , r 1 + r 2 r 2 , 2 1 1 - 1 2 1 2 - 4 1 2 r 2 + r 3 r 3 , 2 1 1 - 1 2 1 2 - 2 3 So we can read out L = 1 - 1 1 2 - 2 1 , U = 2 1 1 2 1 3 We can check the product of LU is really A itself.

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