Results 1 to 7 of 7

Math Help - Elementary linear algebra college work

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    8

    Elementary linear algebra college work

    test tomorrow need the instructions to how to do this problem.


    (1 1)
    Question: A=( 0 2) find a matrix B such that C(A)=N(B)
    (1 0)

    any kind of help please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    148
    What are C(A) and N(B)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    8
    C(A) is a matrix and N(B) is the north space of the matrix
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2008
    Posts
    148
    C(A) may be a matrix but how does C operate on A? North space, I am not familiar with I am sorry, how is it defined?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,938
    Thanks
    338
    Awards
    1
    Quote Originally Posted by fatboy View Post
    C(A) is a matrix and N(B) is the north space of the matrix
    I think the question was, "C(A) is a matrix with what relationship to A?"

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2008
    Posts
    8
    figured it out C(A) means the column space and N(A) is the nullspace hope that helps
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2008
    Posts
    148
    Ok so the column space of A, C(A) is the set of all linear combinations of the columns of A.

    That is all vectors of the following form:

    c_1 \left( \begin{array}{c}<br />
1  \\<br />
0  \\<br />
1  \end{array} \right) + c_2 \left( \begin{array}{c}<br />
1  \\<br />
2  \\<br />
0  \end{array} \right) = \left( \begin{array}{c}<br />
c_1+c_2  \\<br />
2c_2  \\<br />
c_1  \end{array} \right) where c_1, c_2 are scalars.

    We could think about these vectors as points in 3-space with certain restrictions: x = c_1+c_2, y=2c_2, z = c_1 which leads to:

    z = x-\frac{y}{2}

    We want B = \left( \begin{array}{ccc}<br />
a & b & c  \\<br />
d & e & f  \end{array} \right)

    such that

    B \left( \begin{array}{cc}<br />
c_1+c_2  \\<br />
2c_2 \\<br />
c_1  \end{array} \right) = \left( \begin{array}{c}<br />
0 \\<br />
0 \\<br />
\end{array} \right)

    B = \left( \begin{array}{ccc}<br />
2 & -1 & -2  \\<br />
2 & -1 & -2  \end{array} \right) works I think.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: November 14th 2011, 09:06 AM
  2. Replies: 7
    Last Post: August 30th 2009, 10:03 AM
  3. Help with elementary Linear Algebra proof?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 7th 2009, 01:53 PM
  4. college algebra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 18th 2009, 05:39 PM
  5. College Algebra
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 27th 2006, 07:54 AM

Search Tags


/mathhelpforum @mathhelpforum