Math 3A Winter 2016 Midterm

Transcription

1 Math 3A Winter 016 Midterm Name Signature UCI ID # address There are 7 problems for a total of 115 points. Present your work as clearly as possible. Partial credit will be awarded, and you must show your work to be eligible for full credit. If you find yourself stuck somewhere, move on and come back to the problem later. You are not permitted the use of a calculator, phone, or other electronic aid. Academic dishonesty in any form will result in a score of zero.

2 { [ [ 3 } 1. Is the set,, 4 linearly independent? [15 points 8 1 We know this set is independent iff every column of is pivot. By swapping the first and third rows we can quickly put this matrix into an echelon form, namely , from which it follows that every column is indeed pivot, so the given set is linearly independent.

3 . Consider the system x 1 + x x 3 = 1 x 1 + x + x 3 = 3 (a) Write the corresponding augmented matrix. [ points [ (b) Put the augmented matrix in reduced echelon form. [10 points We replace row by row 1 minus row to get echelon form , 0 0 then replace row by 1 times row and finally replace row 1 by row 1 plus row to obtain the reduced echelon form [ (c) Write the solution set in parametric form. [8 points Evidently x 1 and x 3 are basic, while x is free. In fact row of the reduced echelon [ form tells us that x 3 = 1 and row 1 tells us that x 1 + x =, so x 1 = x. Thus x1 x x 3 is a solution iff x 1 x = 0 + t 1 1 x for some scalar t.

4 3. Express [ 1 0 as a linear combination of [ 1 and [ 1. [15 points We need to find weights x 1 and x so that 1 x 1 + x 1 = so we write the corresponding augmented matrix [ 1, and row reduce, first replacing row by row 1 minus twice row to get 1 1, then multiplying row by 1 3 to get [ , 3 next replacing row 1 by row 1 minus row to get [ and finally multiplying row 1 by 1 to yield the (apparently unique) solution x 1 = and x 3 = 1. Thus 3 1 = [ [ 1. 3,

5 4. Let T : R R be a linear map and suppose T [ 1 0 = [ 0 0 and T [ 0 1 = [ 3 3. [4 points each (a) Evaluate T [ 5 4. [ ( [ [ T = T ) [ [ 1 0 = 5T + 4T 1 [ [ 0 3 = = 1 (b) Write the matrix A representing T (so T v = A v for every v in R ). We know that A is matrix with first column the image under T of [ 1 0 and second column the image under T of [ 0 1, so 0 3 A =. 0 3 (c) Is T one-to-one? How do you know? [ [ 1 0 Every multiple of is mapped to by T, so T is not one-to-one. (Equivalently you can observe that the columns of A are not linearly independent, or you can put A in an echelon form and observe that not every column is pivot.) (d) Is T onto? How do you know? We can see that the image of T is just the line x 1 = x, so T does not map onto R. (Equivalently you can observe that the columns of A do not span R, or you can put A in an echelon form and observe that not every row has a pivot.)

6 5. Suppose that A is an m n matrix and that the equation A x = 0 has a nontrivial solution. Given these assumptions, determine if each of the following statements is true or false. Include a brief explanation. [4 points each (a) A x = 0 has infinitely many solutions. True. If a homogeneous linear system has a nontrivial solution, then in fact it has infinitely many solutions. (b) The columns of A are linearly dependent. True. In fact A x = 0 has a nontrivial solution iff the columns of A are linearly dependent. (c) If the equation A x = b is consistent, then it has a unique solution x. False, since solutions to the associated homogeneous system are not unique. (d) Every column of A is pivot. False. In fact every column of A is pivot iff A x = 0 has only the trivial solution. (e) One column of A is in the span of the others. columns.) (Assume that A has at least two True. In a linearly dependent set there must be at least one vector which is in the span of the others.

7 6. Let A = (a) What is the size of A? 7 3 [ points (b) In the usual way we can use A to define a linear transformation by the rule T v = A v. What are the domain and target (codomain) of T? Answer by filling in the boxes: T : R 3 R 7. [4 points (c) Is T onto? How do you know? You can (and should) answer the question without doing any computation. [4 points No, T is not onto, because A has more rows than columns and therefore cannot have a pivot in every row. (d) Is in the image (range) of T? If not, how do you know? If so, find a vector in the domain that T maps to the given vector. You can answer these questions without doing any computation. [4 points Yes, in fact we can see that this is the middle column of A, so [ 01 0 has the desired image under T. (The map T is not one-to-one; there are other vectors with the same image, which of course also received full credit.)

8 7. Suppose C is 3 3 matrix and C 1 = (a) Find C. [10 points Since the problem gives us C 1, it must be the case that C is invertible. We also know that C = (C 1 ) 1, so we will find C by finding the inverse of C 1. We do this by row reduction; since C 1 is already in an echelon form, we just have to carry out the backward phase : We conclude that R R+R3 R1 R1+R3 R3 1 R3 C = ( C 1) 1 = (b) Without doing any more computation, determine whether 11 1 is in the span of the 14 columns of C. How do you know? [4 points (c) Is Yes. In fact we can answer this without even doing (a) first. The wording of the problem tells us that C is invertible. This tells us that the columns of C must span R 3 (every row contains a pivot), and therefore every vector in R 3, including the given one, belongs to the span of the columns of C. 0 in the span of the columns of C? [1 point 14 No. The given vector belongs to R, while the columns of A live in R 3, so their span does too..

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