M314 Practice Exam 1

Instructions: Show all of your work; your work must justify your answers.

Note: These are mostly questions I've given on exams when I've taught M314 before, so this should give you an idea what my exams are like. And although it should help you to review, this is not a review sheet in the usual sense. There may be things you need to know that aren't on it, so be sure to review the book, the homework and the quizzes. On the other hand, problem 5 will be on the exam Wednesday (verbatim!), so be sure you know how to do problem 5.

Problem 1 (18 points): For each matrix A below, find its reduced row echelon form RA and determine its rank. Also, in each case, determine if the rows of the given matrix span Rn, where n in each case is the number of columns of the matrix, and determine if the rows of the given matrix are linearly independent. Show your work and explain your answers.
By the addition to Theorem 2.7 (p. 98) given in class,
the rows of matrix A with m rows and n columns span Rn
if and only if rank(A) = n. So we can do the part about the spans
just knowing the rank. Moreover Theorem 2.7 says the rows are linearly dependent
if and only if m > rank(A), so we can do the part about
linear independence also just knowing the rank. To save myself some typing,
I won't show my work row reducing the matrices, but you should show it!

(a) A=  [1   4   0   4  -1] 
        [0   0   1   4  -1]
        [1   4   1   8  -2]
We immediately see that the rows do not span
R5, since the rank of a matrix with m rows and n columns
is at most the minimum of m and n, so here rank(A) <= 3.
To check independence of the rows we need to row reduce.
RA =  [1   4   0   4  -1] 
      [0   0   1   4  -1]
      [0   0   0   0   0]
The rank is 2, so the rows are linearly dependent
(in fact, the bottom row is the sum of the top two rows).

(b) A= [1   1    1] 
       [0   1   -1]
       [0   1   -1] 
RA =  [1   0   0] 
      [0   1   0]
      [0   0   0]
The rank is again 2, so the rows are
linearly dependent but do not span R3.

(c) A= [1   2   3]
       [1   0   4]
       [0   2   0]
RA =  [1   0   0] 
      [0   1   0]
      [0   0   1]
The rank is 3, so the rows are linearly independent
and span R3.

Problem 2 (15 points): Suppose
[1  -6  0  0  3  | -2]
[0   0  1  0  4  |  7]
[0   0  0  1  5  |  8]
[0   0  0  0  0  |  0]
is the reduced echelon form of the augmented matrix of a system of 4 equations in the variables x1,..., x5. Problem 3 (12 points): For each of the following statements, determine whether it is True or False; if it is false, give a specific example which shows it is not true (it is not enough to just say in general terms why it is false). Problem 4 (15 points): Problem 21 p. 100. (On the actual exam, I will type out every problem.) Answer: We proved parts (a) and (b) just before the quiz in class on Wed Sept 14. To do part (c), let v1, v2 and v3 be the vectors whose span we're asked to show is R3. We already know that v1, v2 and v3 are in the span of e1, e2 and e3 (since span(e1, e2, e3)=R3, so everything is in the span of e1, e2 and e3). On the other hand, e1, e2 and e3 are in the span of v1, v2 and v3: e1=v1, e2=v2-v1, and e3=v3-v2. Thus part (b) says that span(e1, e2, e3) = span(v1, v2, v3), hence R3=span(v1, v2, v3). Problem 5 (15 points) Do Problem 58 on p. 27. Use complete sentences and good grammar in your answer! Answer: (a) First, ||u+v||=||u-v|| if and only if ||u+v||2=||u-v||2. Using the definition of || . ||, this is just u.u + 2u.v + v.v = (u+v).(u+v) = (u-v).(u-v) = u.u - 2u.v + v.v. Thus ||u+v||=||u-v|| if and only if u.u + 2u.v + v.v = (u+v).(u+v) = (u-v).(u-v) = u.u - 2u.v + v.v, hence by canceling, if and only if 2u.v = - 2u.v, which is true if and only if 4u.v = 0; i.e., if and only if u.v = 0, which is he same thing as saying if and only if u and v are orthogonal.
(b) Think of u and v as adjacent sides of a parallelogram as shown in the figure. Then u+v and u-v are the diagonals. Thus (a) says that the diagonals of a parallelogram have the same length if and only if the adjacent sides are perpendicular (i.e., if and only if the parallelogram is a rectangle). Problem 6 (10 points) In Problem 14 on p. 14, write the vector from C to E as a linear combination of a and b. Answer: Starting from C, a - 2b takes you to E, so the vector from C to E is a - 2b. Problem 7 (15 points) Consider the planes defined by x+y+z=2 and x+2y+z=0.