1. In exercise 3.3 #19, the area of the parallelogram is 8, because that is the determinant of A=Matrix([[5,6],[2,4]]). Can we find a rectangle that is row equivalent to A with the same area?
a) Impossible with the conditions given
b) It is possible but I am stuck on how to do so
c) Yes and I can give it to you now

2. What is {s*Vector([1,1,1]) + t*Vector([2,2,2]}
a) a plane
b) a vector space
c) a line
d) a) and b)
e) b) and c)

3. If a matrix is square and the reduction shows that the last row is [0 0 .... 1], what can you conclude?
a) The n column vectors span Rnumber of rows.
b) The n column vectors do not span Rnumber of rows.
c) Further work must be done.

4. If a matrix is square and its determinant is zero, what can you conclude?
a) The column vectors are linearly independent.
b) The column vectors are not linearly independent.
c) Further work must be done.

5. Which of the following describes a basis for a subspace V ?
a) A basis is a linearly independent spanning set for V .
b) A basis is a minimal spanning set for V .
c) A basis is a largest possible set of linearly independent vectors in V.
d) All of the above
e) Some but not all of the above

6. Is x+y+z=1 a subspace in R3?
If so, then provide a basis and specify what geometric object this is.
If not, first demonstrate why using the subspace axioms.

a) I have created a basis for x+y+z=1
b) x+y+z=1 is not a subspace - one of the axioms fails
c) x+y+z=1 is not a subspace - two of the axioms fail
d) x+y+z=1 is not a subspace - all three of the axioms fail
e) x+y+z=1 is not a subspace - but I am unsure about which axioms fail

7. Next, can you create a subspace that is similar to the original set, by just changing one component of x+y+z=1 (ie the coefficients or the =)? If so, provide a basis for the subspace.

a) impossible
b) possible but I am stuck
c) possible and I showed how

8. Look at the vectors (1,2,3), (4,5,6), (5,7,10), and (-3,-3,-3).
Part A Are these vectors linearly independent in R3?

a) Yes
b) No
c) Unsure

Part B Do these vectors (1,2,3), (4,5,6), (5,7,10), and (-3,-3,-3) span R3? If not, what is a basis for the column space of the matrix A with these vectors as columns.

Part C Use the system of equations that is used in the definition of linearly independent and solve them for solutions. Do the solutions (the Null Space) form a vector space? Why or why not? If so, what is a basis?
Answers to 5 and 6: Not it is not a subspace. It violates axiom 1 and 6: For axiom 6, choose c=0 and u=(1,0,0) which satisfies the equation. Note that c*u=(0,0,0) which no longer satisfies the equation. However, x+y+z=0 is a vector space - a plane. To fine a basis for this plane, set z=t, y=s, and then x=-s-t. Factor out the s and t to see that a basis is (-1,1,0) and (-1,0,1).

Answers to 7: The vectors are not linearly independent since there are too many of them (4 - more than the dimension). To check span, we set up the corresponding span equations and use Gaussian on the augmented matrix. We see that the spanning system can never be inconsistent, so these do span. For the last part, set up the li equation, form the augmented matrix and reduce. Since this is a homeogeous system, the solutions are a vector space - in this case they are a line through the origin in R4. Then c4=t, c3=0, c2=t, and c1=-t. Factor out the t, and (-1,1,0,1) is a basis.