M314 Practice Exam 1

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.
• (a) Which variables are the free variables? Answer: the free variables are the ones corresponding to columns without a leading 1; i.e., x2 and x5.
• (b) How many solutions does the system of equations have? Answer: it is a consistent since the augmented column does not have a leading 1, so there is at least one solution, and it has free variables, so there are infinitely many solutions if there are any (which there are because it is consistent).
• (c) Write down the general solution of the system of equations in vector form.
```[x1]    [-2]      [6]     [-3]
[x2]    [ 0]      [1]     [ 0]
[x3] =  [ 7] +  x2[0] + x5[-4]
[x4]    [ 8]      [0]     [-5]
[x5]    [ 0]      [0]     [ 1]
```
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).
• (a) If A is the coefficient matrix of a linear system with infinitely many solutions, then the number of variables is more than the rank of A. Answer: True. (There must be a free variable, so not every variable can be a leading variable; i.e., there are more variables than leading variables. But the number of leading variables is the same as the number of leading 1's in the reduced row echelon form of A, which is the rank of A.)
• (b) A linear system with fewer variables than equations never has infinitely many solutions. Answer: False. (For example, x+y=0, 2x+2y=0 and 3x+3y=0 has 2 variables, 3 equations and infinitely many solutions.)
• (c) A homogeneous linear system with a unique solution must have at least as many equations as variables. Answer: True. (It can't have a free variable if it has only one solution, so each variable must be a leading variable, so the rank of the coefficient matrix A equals the number of variables. But the number of rows of A (which is also the number of equations) is always at least equal to the rank. So there are at least as many equations as variables.)
• (d) A linear system with fewer equations than variables is never inconsistent. Answer: False. (For example, x+y+z=0, 2x+2y+2z=1 is inconsistent.)
• (e) Suppose a matrix has n columns, and that the matrix is taller than it is wide. Then the rows span Rn. Answer: False. (The matrix could, for example, be the 0 matrix.)
• (f) Suppose a matrix is taller than it is wide. Then the rows of the matrix are linearly dependent. Answer: True. (This is what Theorem 2.8 says.)
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.
• (a) Find the cosine of the acute angle between the planes defined by x+y+z=2 and x+2y+z=0. Answer: Just find the absolute value of the cosine of the angle between the normal vectors for the planes: (1,1,1).(1,2,3)/[||(1,1,1)||*||(1,2,3)||] = 6/(3*14)1/2.
• (b) Find an equation (in vector form) for the line L where the two planes intersect. Answer: Find any two solutions P and Q to the system of equations given by the equations of the two planes. For example, P = (1, -2, 3) and Q=(4, -2, 0). Then the vector d = (3, 0, -3) from P to Q is a direction vector for L, so x = P + td gives an equation for the line L.
• (c) Find (in normal form) an equation for the plane passing through the point R=(7,3,2) and perpendicular to L. Answer: The vector d is a normal for the desired plane, so the normal form is d.(x-R)=0, where d and R are as above.