Study Guide 3 Solution Spring 2008 on Differential Equations with Linear Algebra 1 - 2r2 r3 r3 r2 r4 r4 2 1 2 0 M341 Study Guide for E 3(S Zhang 1 Find

Study Guide 3 Solution Spring 2008 on Differential Equations with Linear Algebra 1

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M341 Study Guide for E 3 (S. Zhang) . 1. Find the row echelon form (okay to have non-one diagonals, 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 so- lution would be (choosing 1 for one of x 3 , x 4 , x 5 , and 0 for the other two, then solving for x 1 and x 2 ) 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. Find A - 1 . 1 2 1 - 2 - 3 - 1 3 10 6 ans: We do row operations to transform the matrix [ A | I ] [ I | A - 1 ] 1 2 1 | 1 - 2 - 3 - 1 | 1 3 10 6 | 1 2 r 1 + r 2 1 2 1 | 1 1 1 | 2 1 3 10 6 | 1 - 3 r 1 + r 3 1 2 1 | 1 1 1 | 2 1 4 3 | - 3 1 - 4 r 2 + r 3 1 2 1 | 1 1 1 | 2 1 - 1 | - 11 - 4 1 ( - 1) r 3 1 2 1 | 1 1 1 | 2 1 1 | 11 4 - 1 ( - 1) r 3 + r 2 1 2 1 | 1 1 | - 9 - 3 1 1 | 11 4 - 1 ( - 1) r 3 + r 1 1 2 | - 10 - 4 1 1 | - 9 - 3 1 1 | 11 4 - 1 ( - 2) r 2 + r 1 1 | 8 2 - 1 1 | - 9 - 3 1 1 | 11 4 - 1 A - 1 = 8 2 - 1 - 9 - 3 1 11 4 - 1 L=[1 0 0;-2 1 0; 3 4 1] U=[1 2 1;0 1 1; 0 0 -1] A=L*U, det(A) inv(A) 4. (1) Use the Gauss elimination to solve the system.

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