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Math 235 Section 1 Solution of Midterm 2 Fall 2000

  1. (18 points) You are given below the matrix A together with its row reduced echelon form B

    tex2html_wrap_inline275 tex2html_wrap_inline277

    a) (10 points) Find a spanning set for (i.e., a set of vectors which spans) the null space Null(A) of A.

    Answer: We use the row reduced echelon form to write, in parametric form, the general solution of tex2html_wrap_inline283 (i.e., the general vector in Null(A)). The free variables are: tex2html_wrap_inline287 , tex2html_wrap_inline289 , and tex2html_wrap_inline291 . Hence, the general solution is


    and the three column vectors above span Null(A).

    b) (8 points) Are the column spaces tex2html_wrap_inline295 and tex2html_wrap_inline297 equal? Justify your answer carefully! (either explain why they are equal, or find a vector in one that does not belong to the other).

    Answer: No, the two column spaces are different! For example, the last column of A has a non-zero fourth entry, while any vector in Col(B) is a linear combination of the columns of B, and hence, its fourth entry is zero.

  2. (16 points) Let tex2html_wrap_inline305 be the tex2html_wrap_inline307 identity matrix and tex2html_wrap_inline309 .

    a) (5 points) Show that the matrix A-5I is invertible. Note that A-5I is simply tex2html_wrap_inline315 .

    Answer tex2html_wrap_inline317 . Since tex2html_wrap_inline319 , the matrix A-5I is invertible.

    b) (5 points) Express the determinant of the tex2html_wrap_inline307 matrix A-tI in terms of the scalar parameter t.

    Answer tex2html_wrap_inline329 .

    c) (6 points) Show that there are precisely two values of t for which the matrix A-tI is not invertible. Find these values of t.

    Answer A-tI is not invertible if and only if its determinant vanishes. We find those values of t buy solving the quadratic equation tex2html_wrap_inline341 . The two solutions are t=6 and t=-1.

  3. (18 points) Determine if the following set in tex2html_wrap_inline347 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).

    1. tex2html_wrap_inline355

      Answer This subset of tex2html_wrap_inline357 does not contain the zero vector. Hence, it is not a subspace.

    2. tex2html_wrap_inline359


      We can write the general element as a linear combination

      tex2html_wrap_inline361 Hence, this subset is a subspace, which is equal to the column space of the matrix tex2html_wrap_inline363

    3. tex2html_wrap_inline365

    Answer: This subset of tex2html_wrap_inline367 is the general solution of the system of homogeneous linear equations tex2html_wrap_inline369 It is hence a subspace which is equal to the Null space of tex2html_wrap_inline371

  4. (16 points) a) (8 points) Compute the volume of the parallelepiped in tex2html_wrap_inline357 with vertices

    tex2html_wrap_inline375 , tex2html_wrap_inline377 , tex2html_wrap_inline379 , tex2html_wrap_inline381 , tex2html_wrap_inline383 , tex2html_wrap_inline385 , tex2html_wrap_inline387 , tex2html_wrap_inline389 where


    Answer: The volume is the absolute value of the determinant of the tex2html_wrap_inline391 matrix M with columns tex2html_wrap_inline395 and tex2html_wrap_inline381 .


    b) (8 points) Let T be the linear transformation from tex2html_wrap_inline357 to tex2html_wrap_inline357 sending a vector tex2html_wrap_inline407 to tex2html_wrap_inline409 , where A is the matrix tex2html_wrap_inline413 Compute the volume of the parallelepiped obtained as the image of the one in part (a) under the transformation T. Express your answer in terms of a, b and c. Note: The image is the parallelepiped with vertices tex2html_wrap_inline375 , tex2html_wrap_inline425 , tex2html_wrap_inline427 , tex2html_wrap_inline429 , tex2html_wrap_inline431 , tex2html_wrap_inline433 , tex2html_wrap_inline435 , tex2html_wrap_inline437 .


    First Method: tex2html_wrap_inline439 where M is the matrix in part (a).

    Second Method: Calculate directly the determinant.

    tex2html_wrap_inline443 . Hence the volume is 3|c|.

  5. a) (6 points) Let A, B, and C be tex2html_wrap_inline391 matrices satisfying the equation


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

    Answer: Multiply both sides by tex2html_wrap_inline463 on the left to get tex2html_wrap_inline465 . Then multiply both sides by A on the right to get tex2html_wrap_inline469 .

    b) (10 points) Let tex2html_wrap_inline471 Compute its inverse tex2html_wrap_inline463 . (Check that tex2html_wrap_inline475 .)

    Answer: Augment A by the tex2html_wrap_inline391 identity matrix and row reduce:


    Hence, tex2html_wrap_inline481

    c) (6 points) Compute the (2,3) entry of tex2html_wrap_inline485 if A is given in part (b) and tex2html_wrap_inline489

    Answer: Calculate the product of the second row of A times B. Then take the product of the resulting row vector with the third column of tex2html_wrap_inline463 .


  6. (10 points) Let tex2html_wrap_inline497 be the vector space of polynomials of degree tex2html_wrap_inline499 . Recall that a vector in tex2html_wrap_inline497 is a polynomial p(t) of the form tex2html_wrap_inline505 where the coefficients tex2html_wrap_inline507 are arbitrary real numbers.

    a) (6 points) Show that the subset H of tex2html_wrap_inline497 of polynomials p(t) of degree tex2html_wrap_inline499 which in addition satisfy


    is a subspace of tex2html_wrap_inline497 . (The straightforward answer would include the definition of a subspace and a verification that H satisfies all the properties.)

    Answer: We check the three conditions for a subset of the vector space tex2html_wrap_inline497 to be a subspace:

    i) The zero vector tex2html_wrap_inline523 has value 0 when t=0 and when t=1. Hence, the zero vector is in H.

    ii) (H is closed under addition) If p and q are in H, then they satisfy equation (1). Consequently, p(0)=0, p(1)=0, q(0)=0, and q(1)=0. We need to check that their sum satisfied equation (1) as well. Indeed:

    (p+q)(0)=p(0)+q(0)=0+0=0 and (p+q)(1)=p(1)+q(1)=0+0=0.

    iii) (H is closed under scalar multiplication) We need to check that if p is in H and c is a scalar, then cp satisfy equation (1) as well. Indeed:

    tex2html_wrap_inline563 and tex2html_wrap_inline565 .

    b) (4 points) Find a spanning set for H. Explain why the set you found spans H.

    Answer: We can rewrite equation (1) explicitly in the form


    We find the general solution, regarding tex2html_wrap_inline571 , tex2html_wrap_inline573 and tex2html_wrap_inline575 as unknowns: tex2html_wrap_inline575 is free and


    If we choose tex2html_wrap_inline579 , we get the polynomial tex2html_wrap_inline581 . Any other polynomial in H is a scalar multiple of this one. Hence,


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Eyal Markman
Mon Nov 20 09:20:19 EST 2000