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 through the origin
    b) a basis for a subspace of R3
    c) a line through the origin
    d) a) and b)
    e) b) and c)

  3. Which of the following are true about Matrix([[1,4],[2,5],[3,6]])?
    a) The column space is the plane b1-2b2+b3=0 in R3
    b) The column space is the plane sVector([1,2,3]) + tVector([4,5,6]) in R3
    c) The nullspace is the 0 vector in R2
    d) more than one of the above, but not all of them
    e) all of a, b, c

  4. If an matrix is not square, then
    a) the column space is a subspace of Rnumber of rows
    b) the column space is a subspace of Rnumber of columns
    c) Further work must be done to tell.

  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?

    a) It is a subspace
    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 =)?

    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 the column space?

    Part C What is the nullspace of the matrix with those vectors as columns?
    Answers to 6 and 7: It is not a subspace. It violates all the axioms: For example, for axiom 3, 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 parametrization and 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 8: 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 R3, and the pivots are in the 1st, 2nd and 3rd spots (the last vector is a linear combination of the first 2). The column space 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.