M114_Ast4_Soln - MATH 114 Assignment 4 Solutions Due Tuesday October 14th From PROBLEMS 2.2 Homework Problems(p 89-90 B1 Find the RREF and rank of the

M114_Ast4_Soln - MATH 114 Assignment 4 Solutions Due...

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MATH 114 - Assignment 4 Solutions Due Tuesday October 14th From PROBLEMS 2.2 - Homework Problems (p. 89-90): B1 Find the RREF and rank of the following matrices: (a) The RREF of 2 3 3 - 1 3 2 is 1 0 0 1 0 0 ; rank = 2. (c) The RREF of 2 4 8 1 1 3 1 - 1 1 is 1 0 2 0 1 1 0 0 0 ; rank = 2. B2 Let the given matrix be the coefficient matrix for a homogeneous system , already in RREF. Determine the number of parameters in the general solution and write the general solution in standard form. (a) RREF = 1 3 0 - 1 0 0 1 2 0 0 0 0 # of parameters = # variables - rank of coefficient matrix = 4 - 2 = 2. The 2nd and 4th columns do not contain leading ones, so we choose x 2 and x 4 as our parameters. Let x 2 = s R and let x 4 = t R . Then the general solution is given by x 1 x 2 x 3 x 4 = - 3 s + t s - 2 t t = s - 3 1 0 0 + t 1 0 - 2 1 , s, t R (b) RREF = 1 1 0 - 2 0 0 0 1 1 0 0 0 0 0 1 # of parameters = # variables - rank of coefficient matrix = 5 - 3 = 2. The 2nd and 4th columns do not contain leading ones, so we choose x 2 and x 4 as our parameters. Let x 2 = s R and let x 4 = t R . Then the general solution is given by x 1 x 2 x 3 x 4 x 5 = - s + 2 t s - t t 0 = s - 1 1 0 0 0 + t 2 0 - 1 1 0 , s, t R 1
(c) RREF = 1 2 0 0 0 - 3 0 0 1 - 5 0 4 0 0 0 0 1 1 0 0 0 0 0 0 # of parameters = # variables - rank of coefficient matrix = 6 - 3 = 3. The 2nd, 4th and 6th columns do not contain leading ones, so we choose x 2 , x 4 and x 6 as our parameters. Let x 2 = r R , x 4 = s R and x 6 = t R . Then the general solution is given by x 1 x 2 x 3 x 4 x 5 x 6 = - 2 r + 3 t r 5 s - 4 t s - t t = r - 2 1 0 0 0 0 + s 0 0 5 1 0 0 + t 3 0 - 4 0 - 1 1 , r, s, t R B3 (b) Consider the system x 1 + 4 x 2 - 2 x 3 = 0 2 x 1 + - 3 x 3 = 0 4 x 1 + 8 x 2 - 7 x 3 = 0 The coefficient matrix is 1 4 - 2 2 0 - 3 4 8 - 7 The RREF of the coefficient matrix is 1 0 - 3 / 2 0 1 - 1 / 8 0 0 0 The rank of the coefficient matrix is 2. The # of parameters is 3 - 2 = 1. Let x 3 = t R . Then the general solution to the homogeneous system is x 1 x 2 x 3 = (3 / 2) t (1 / 8) t t = t 3 / 2 1 / 8 1 , t R From PROBLEMS 2.2 - Conceptual Problems (p. 90) D3 What can you say about the consistency of a system of m linear equations in n variables and the number of parameters in the general solution if: (a) m = 5, n = 7 and the rank of the coefficient matrix, A , is 4?

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