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M 340L - M 340L Fall 2010(55430 Midterm 1 Answer the...

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M 340L Fall 2010 (55430), Midterm 1 Answer the questions on the scratch paper provided. Box your answers. Correct answers will always get full credit even if no work is shown; if you want partial credit, show your work. All scratch paper will be collected at the end. Name: Question Points Score 1 20 2 15 3 10 4 20 5 15 6 30 7 5 Total: 115

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Problem 1 (20 points). Consider the matrix A = - 3 - 4 6 1 2 0 0 2 6 . (a) (8 points) Find all solutions of A x = - 1 3 8 . If the solution set is nonempty, describe it in parametric form. (Recall that “parametric form” means giving a description of the solution set of the form x = t v + p , or x = t 1 v 1 + t 2 v 2 + p .) Solution: Build the augmented matrix - 3 - 4 6 - 1 1 2 0 3 0 2 6 8 . Reduce it to re- duced row echelon form, 1 0 - 6 - 5 0 1 3 4 0 0 0 0 . This amounts to reducing the equa- tions to x 1 = - 5 + 6 x 3 , x 2 = 4 - 3 x 3 , with x 3 a free variable. In parametric form, writing x 3 as t , this is x = t 6 - 3 1 + - 5 4 0 . (b) (8 points) Find all solutions of A x = 1 0 0 . If the solution set is nonempty, describe it in parametric form. Solution: Build the augmented matrix - 3 - 4 6 1 1 2 0 0 0 2 6 0 . Reducing it to row echelon form you see it has a pivot in the last column. (The reduced row echelon form is 1 0 - 6 0 0 1 3 0 0 0 0 1 , although you don’t have to go all the way to the reduced row echelon form to see where the pivots are.) So the system is inconsistent: there are no solutions. (c) (4 points) Do the columns of A span R 3 ? If not, name a vector in R 3 that is not in the span of the columns of A . Solution: A vector b is in the span of the columns of A if and only if the equation A x = b has a solution. In part b we saw that this equation does not have a solution when b = 1 0 0 . So the columns of A do not span R 3 , and the Page 2
vector b = 1 0 0 is not in the span of the columns of A . Problem 2 (15 points). Suppose v 1 = 1 2 3 , v 2 = 4 5 6 , v 3 = 7 8 9 . (a) (5 points) Is the set { v 1 , v 2 } linearly independent? If it is not, give a nontrivial solution of the equation x 1 v 1 + x 2 v 2 = 0 . Solution: Yes, this set is linearly independent: it is a set of two vectors, neither of which is a scalar multiple of the other.

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