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Math 130 Second Test Answers. Nov 2009. Graderanges: 84100 A, 7383 B, 5672 C. Median 80.

Problem 1. On cofactor expansion to evaluate determi-nants [15]

Use cofactor expansion to evaluate the following determi-nant. (Your choice of which row or column to use.) Showyour work.

1 2 0 02 0 3 10 1 3 04 0 2 3

You can use any row or column. Here are the computa-tions if you expand along the first row which leads to thecomputations of two 3 3 determinants.

1

0 3 1

1 3 00 2 3

2

2 3 10 3 04 2 3

+ 0 0

Now, the first 3 3 determinant equals 11 while the secondone equals 30, so the original determinant equals 11 60 =49.

Problem 2. On abstract vector spaces [25].Prove that in any vector space, if the product of a scalar

and a vector is the zero vector, then either the scalar is 0 orthe vector is 0 (or both). That is,

cv = 0 implies c = 0 or v = 0.

(Hint. Try showing that ifc is not 0, then v is 0.) Work outyour proof on the back of one of the test pages, then copy ithere when youre satisfied you have it correct.

Heres one proof. Suppose that cv = 0 but c = 0.Multiply the equation by c1 (which exists since c = 0)to get c1cv = c10. But c1cv = 1v = v (since 1times any vector equals that vector) while c10 = 0 (sinceany scalar times the zero vector equals the zero vector), sov = 0. q.e.d.

Note that proving that both c = 0 implies cv = 0 andv = 0 implies cv = 0 is not relevant. Thats a proof of the

converse of the statement, and the converse of a statementis not logically equivalent to the statement.

An argument that uses coordinates is only valid when V=Rn. Other vector spaces do not have coordinates.

Problem 3. On lines in R3. [14;7 points each part]

a. Give a parametric equation for the line through thepoints (2,3, 4) and (2, 3, 5).

In general, a parametric equation for the line through aand b is the equation x = a + (b a)t. With a = (2,3, 4)and b = (2, 3, 5), that gives

x = (2,3, 4) + (0, 6, 1)t

which you can also write as three equations:

x = 2y = 3 + 6tz = 4 + t

b. Are the points (4, 6, 2), (3, 5, 5), and (4, 2, 4) on thesame line? Explain why or why not.

Heres one way to do it. Look at two displacement vec-tors between any two pairs of points, say u = (3, 5, 5) (4, 6, 2 ) = ( 7,1, 3), and v = (4, 2, 4) (4, 6, 2) =(8,4, 2). These two vectors u and v are not parallel (sincetheyre not scalar multiples of each other), so the points they

connect do not lie on the same straight line.There are a couple of alternative methods. For one, you

could show that the area of the parallelogram with sides uand v is not zero, so u and v cant be parallel.

For another method, you could find the parametric equa-tion for a line through two of the points and show the thirdpoint doesnt satisfy that equation.

Finally, you could show the 33 determinant of the matrixwhose rows are the three vectors is not zero, since that showsthat the parallelopiped with those three vectors as edges hasa nonzero volume, hence they cant be collinear. (But if ithappens to be zero, all you would know is that they lie in

the same plane.)

Problem 4. On vector operations. [20; 4 points each part]Which of the following expressions makes sense? For each,write valid or invalid. No explanation is necessary. Thevectors u, v, and w belong to R3, while a and b are scalarsin R. (For example, the expression uv is invalid since youcant multiply two vectors, but both u v and uv are validsince you can take the dot product and the cross product oftwo vectors.)

a. a u + v.

Invalid. You cant take a cross product of the scalar a

with a vector.

b. bw u.

Valid. bw is a vector, and you can take the dot productof it with another vector.

c. a + u v.

Invalid. You cant add the scalar a to a vector.

d. (w u) v.

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Valid. The cross product w u is a vector, and youcan take the dot product of it with another vector. (Youllrecognize this as the triple product [w,u,v].)

e. (w u) v.

Invalid. The dot product w u is a scalar, and you cant

take the cross product of a scalar with a vector.

Problem 5. On cross products [14; 7 points each part].

a. If the cross product v w of two vectors in R3 equals(14, 3, 2), then what does the cross product w v equal?

Since w v = v w, therefore v w = (14,3,2).

b. If the cross product vw of two vectors in R3 equals0, then what can you say about the two vectors?

Since they are two sides of a parallelogram with 0 area,theyre parallel, that is to say, one is a multiple of the other.

Problem 6. On linear transformations. [12] Consider thelinear transformation L of R3 such that L(i) = 2k, L(j) =i + k, and L(k) = 3i 4j + 5k. Find the standard matrix Awhich represents L. (Recall that A represents L ifL(v) =Av for every vector v R3.)

In general, the columns ofA are the images of the standard

matrix vectors. Since L

100

=

002

, L

010

=

101

,

and L

001

=

34

5

, therefore

A =

0 1 30 0 4

2 1 5

.

Note that the transpose of A describes a different lineartransformation: ATi = j + 3k, ATj = 4k, and ATk =2i +j + 5k.

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