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Homework Help: Column space

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Does b = [ 2 15 ]T lie in the column of the matrix A

    [1 -3]
    [2 5]


    2. Relevant equations

    a basis of CS(U) forms a basis for the corresponding columns for CS(A)

    3. The attempt at a solution
    I reduced the given matrix into row echelon form and determined its column space. But did not figure out if [2 15]T lies in the column space of A.
     
  2. jcsd
  3. Nov 22, 2008 #2
    Your matrix A reduces to the identity matrix in reduced row echelon form; so then the column space is made up of all the columns of the original matrix;

    [tex]
    \text{Col}(A)=\left\{
    \begin{pmatrix}
    1 \\
    -3
    \end{pmatrix}
    ,
    \begin{pmatrix}
    2 \\
    5
    \end{pmatrix} \right\}
    [/tex]

    So does the vector they're asking lie in that space? In other words is it a linear combination of those vectors in the space?
     
  4. Nov 22, 2008 #3
    Is the

    [tex]
    \text{Col}(A)=\left\{
    \begin{pmatrix}
    1 \\
    2
    \end{pmatrix}
    ,
    \begin{pmatrix}
    -3 \\
    5
    \end{pmatrix} \right\}
    [/tex]

    or

    [tex]
    \text{Col}(A)=\left\{
    \begin{pmatrix}
    1 \\
    -3
    \end{pmatrix}
    ,
    \begin{pmatrix}
    2 \\
    5
    \end{pmatrix} \right\}
    [/tex]
    ?
    The given matrix is

    [tex]

    \begin{pmatrix}
    1; -3\\
    2; 5
    \end{pmatrix}
    [/tex]
     
  5. Nov 22, 2008 #4
    thanks, i resolved it!
     
  6. Nov 22, 2008 #5
    Oh yea sorry I read your matrix backwards accidentally. Glad you got it.
     
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