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Math 235 Solution of Midterm 2 Spring 1999

  1. (24 points) You are given below the matrix A together with its row reduced echelon form B

    tex2html_wrap_inline243 tex2html_wrap_inline245

    a) (9 points) Find a basis for the null space Null(A) of A.

    Recall that Null(A) is the space of solutions of the linear system tex2html_wrap_inline253 . Since A and B are row equivalent, Null(A)=Null(B). We have three free variables: tex2html_wrap_inline261 , tex2html_wrap_inline263 , and tex2html_wrap_inline265 . Hence Null(B) is 3 dimensional. The equations


    and the three vectors tex2html_wrap_inline271 are a basis of Null(A).

    b) (9 points) Find a basis for the column space of A.

    Take the pivot columns of A : tex2html_wrap_inline279

    c) (6 points) Is the vector tex2html_wrap_inline281 in the column space of A? Justify your answer!

    No! The vector b is in Col(A) if and only if the system tex2html_wrap_inline289 is consistent. We could check that it is inconsistent by row reduction of the augmented tex2html_wrap_inline291 matrix. However, this would require a lot of work. Instead, use the fact that b is in Col(A) if and only if the vector b is a linear combination of the basis vectors in part (b). I.e., if and only if the augmented matrix tex2html_wrap_inline299 does not have a pivot in the fourth column. But it does (Check by row reduction).

  2. (10 points) Find a basis for the set of vectors in tex2html_wrap_inline301 in the plane x+2y+z=0.

    Regard the equation of the plane as a linear ``system'' of one equation. So y and z are free variables and the general solution has the form


  3. (24 points) Determine which of the following sets in tex2html_wrap_inline309 is a subspace. If it is not, find a property in the definition of a subspace which this set violates. If it is a subspace, find a matrix A such that this set is either Null(A) or Col(A). Note: part (a), (b), (c) are analogous to Homework problems number 7, 9 ,10, 15 in section 4.2

    (a) (9 points) tex2html_wrap_inline317

    This is a subspace.

    It is the set of vectors of the form tex2html_wrap_inline319 Hence, it is the subspace spanned by these four vectors. Equivalently, it is the column space of the tex2html_wrap_inline321 matrix having these 4 vectors as column vectors.

    (b) (9 points) tex2html_wrap_inline323

    This is not a subspace because the zero vector tex2html_wrap_inline325 is not in this set (the equation is not homogeneous).

    (c) (6 points) tex2html_wrap_inline327

    This is a subspace. If we rewrite the equations in the form tex2html_wrap_inline329 we get that this set is Null(A) where tex2html_wrap_inline333 .

  4. (16 points) a) (6 points) Compute the area of the parallelogram in tex2html_wrap_inline335 with vertices
    (1,1), (3,4), (6,2), (8,5). Caution: note that (0,0) is not a vertex.

    Translating by the vector (-1,-1) we get the parallelogram with vertices (0,0), (2,3), 5,1), (7,4) whose area is tex2html_wrap_inline357

    b) (6 points) Compute the volume of the parallelepiped in tex2html_wrap_inline301 with vertices

    tex2html_wrap_inline361 , tex2html_wrap_inline363 , tex2html_wrap_inline365 , tex2html_wrap_inline367 , tex2html_wrap_inline369 , tex2html_wrap_inline371 , tex2html_wrap_inline373 , tex2html_wrap_inline375 where


    The volume is the absolute value of the following determinants tex2html_wrap_inline377

    c) (4 points) Use your answer in part (b) and the algebraic properties of determinants to compute the volume of the parallelepiped obtained if tex2html_wrap_inline367 is replaced by
    tex2html_wrap_inline381 where a, b, c are real numbers. (Express your answer in terms of a, b, c).

    By the algebraic properties of determinants we know that


    The first and second equalities follow from the fact that adding a multiple of one column to another does not change the determinant. The third equality reflects the property that if a matrix B is obtained from A by multiplication of a column by a scalar c then tex2html_wrap_inline401 . The last equality is our computation in part (b). A more computational method is to express tex2html_wrap_inline403 in terms of a, b, c


  5. a) (4 points) Let A, B, and C be tex2html_wrap_inline417 matrices satisfying the equation


    with A invertible. Solve for C in terms of A and B.

    Answer: tex2html_wrap_inline427

    b) (8 points) The inverse of tex2html_wrap_inline429 is tex2html_wrap_inline431 .

    c) (4 points) Compute the (2,2) entry of C in part (a) if A is given in part (b) and tex2html_wrap_inline439

    tex2html_wrap_inline427 . The second row of tex2html_wrap_inline443 is (2, 1, 0). Multiplying by the second column of A we get tex2html_wrap_inline449 .

  6. (10 points) Let tex2html_wrap_inline451 be the vector space of polynomials of degree tex2html_wrap_inline453 . Recall that a vector in tex2html_wrap_inline451 is a polynomial p(t) of the form tex2html_wrap_inline459 where the coefficients tex2html_wrap_inline461 are arbitrary real numbers.

    (a) (6 points) Show that the subset H of tex2html_wrap_inline451 of polynomials p(t) of degree tex2html_wrap_inline453 which in addition satisfy


    is a subspace of tex2html_wrap_inline451 . (The straightforward answer would include the definition of a subspace and a verification that H satisfies all the properties.)

    Answer: A polynomial p(t) of the form tex2html_wrap_inline477 is in H if and only if


    So H is the subset of tex2html_wrap_inline451 of polynomials with the property that the sum of their coefficients is 0. To show that it is a subspace you can identify tex2html_wrap_inline451 with tex2html_wrap_inline301 by mapping the polynomial tex2html_wrap_inline477 to the vector tex2html_wrap_inline493 and argue that (1) is the equation of a plane in tex2html_wrap_inline301 through the origin. Alternatively, you can argue more abstractly using the definition of a subspace of a vector space:

    1) The zero polynomial tex2html_wrap_inline497 is in H.

    2) If tex2html_wrap_inline501 and tex2html_wrap_inline503 are in H, then p+q is in H because (p+q)(1)=p(1)+q(1)=0+0=0.

    3) If p is a polynomial in H and c is a scalar then cp is in H because tex2html_wrap_inline523 .

    (b) (4 points) Find a basis for H. Explain why the set you found is linearly independent and why it spans H.

    Answer: As a basis to the plane in tex2html_wrap_inline301 given by equation (1) we can take the two vectors tex2html_wrap_inline531 (use the same method as in problem 2). Thus, the two polynomials


    are a basis for H.

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Eyal Markman
Tue Apr 20 07:38:27 EDT 1999