Nsol1 - Solutions to Review Problems One 1 Represent the following system of equations by an augmented matrix reduce this matrix to upper

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Solutions to Review Problems One 1. Represent the following system of equations by an augmented matrix, reduce this matrix to upper triangular form by Gaussian Elimination and then use back- substitution to find the solution. x + 2 y + 4 z = 0 3 x + 8 y + 8 z = 2 2 x + 6 y + 9 z = 7 (1) 1 2 4 | 0 3 8 8 | 2 2 6 9 | 7 may be reduced by the operations R 2 - 3 R 1 , R 3 - 2 R 1 , to upper triangular form 1 2 4 | 0 0 2 - 4 | 2 0 0 5 | 5 . Back-substitution then gives the solution z = 1, y = 4 z +2 2 = 3 and x = - 2 y - 4 z = - 10. 2. Put the matrix A = 1 2 0 0 3 2 4 1 3 0 2 4 2 3 1 into reduced row echelon form and then give the general solution to the equation A -→ x = -→ 0 and express in parametric form. (2) The operations R 2 - 2 R 1 , R 3 - 2 R 1 , R 3 - 2 R 2 , - 1 3 R 3 reduce A to row echelon form 1 2 0 0 3 0 0 1 3 - 6 0 0 0 1 - 7 3 . Finally the operation R 2 - 3 R 3 gives reduced row echelon form 1 2 0 0 3 0 0 1 0 1 0 0 0 1 - 7 3 . Then x 1 , x 3 and x 4 are the dependent variables, x 2 = s and x 5 = t are the independent variables and the solution is ( - 2 s - 3 t,s, - t, 7 3 t,t ). 3. Let A be an m × n matrix, where m < n . Explain why the homogeneous system of equation A -→ x = -→ 0 has a non-trivial solution. (3) There are three possible outcomes for any system of equations: no solution, a unique solution, or infinitely many solutions. A homogeneous equation always has solution -→ x = , -→ 0 , so the first outcome is impossible. Since m < n , the system A -→ x = 0 is underdetermined, which implies that the solution is not unique. Thus we must be in the third outcome where there are infinitely many solutions and therefore at least one non-trivial solution. 4. Let A be an m × n matrix with entries a ij , let B be an n × r matrix with entries b ij , and let C be an n × r matrix with entries c ij . Prove that A ( B + C ) = AB + AC . (4) B + C has entries b jk + c jk , so that A ( B + C ) has entries n j =1 a ij ( b jk + c jk ), which equals n j =1 ( a ij b jk + a i jc jk ), which equals n j =1 a ij b jk + n j =1 a ij c jk ), which equals the entries of AB + AC , since AB has entries n j =1 a ij b jk and AC has entries n j =1 a ij c jk ). 1
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2 5. Prove that if A is invertible, then A T is also invertible. (5) Suppose that A is invertible and let B be the inverse of A , so that AB = I = BA . Then we have B T A T = ( AB ) T = I T = I and A T B T = ( BA ) T = I T = I , so that A T has inverse B T . 6. Suppose that -→ u is a solution of A -→ x = -→ 0 and that -→ v is a solution of A -→ x = -→ c . Explain carefully why 2 -→ u + -→ v is a solution of A -→ x = -→ c . (6) We are given
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This note was uploaded on 12/27/2011 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.

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Nsol1 - Solutions to Review Problems One 1 Represent the following system of equations by an augmented matrix reduce this matrix to upper

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