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Matrices: Rows and Columns Meaning

  1. May 4, 2015 #1
    I'm trying to learn column space currently and I'm confused about the meaning of rows and columns.
    So I'm given this definition for column space:
    "The column space of matrix A is the set Col A of all linear combinations of the columns of A"

    Given the matrix A:
    [ 1 -3 -4 ]
    [ -4 6 -2 ]
    [ -3 7 6 ]

    b=
    [ 3 ]
    [ 3 ]
    [ -4 ]

    Determine if b is in the column space of A.

    My books solves by row reducing [ A b ].

    Has this always been what I was solving for whenever I row reduced an augmented matrix to obtain x for Ax = b?
    For example, when I'm given a system of linear equation such as:
    2x1 + 3x2 = 5
    1x1 + 2x2 = 3
    and I have to solve for x.

    Do the columns of the coefficient matrix of this system of linear equation, have the same meaning as the matrix above, vectors?
     
  2. jcsd
  3. May 5, 2015 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not certain what your question is but I think the answer is:

    Yes, the matrix equation corresponding to the system of equations
    ax+ by+ cz= p
    dx+ ey+ fz= q
    gx+ hy+ iz= r

    is
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}p \\ q \\ r\end{bmatrix}[/tex]
    so that, yes, the columns are the coefficients of the three unknown numbers.

    Here, to determine if < 3, 3, -4> is in the "column space" of the given matrix you have to determine if it is in the space spanned by <1, -4, -3>, <-3, 6, 7>, and < -4, 2, 6>, the columns written as individual vectors.
    That is the same as asking if there exist numbers, x, y, and z, such that a< 1, -4, -3>+ b<-3, 6, 7>+ c<-4, 2, 6>= <3, 3, -4> which is, in turn, the same as asking if there exist x, y, and z such that a- 3b- 4c= 3, -4a+ 6b+ 3c= 3, and -3a+ 7b- 4c= -4, a system of equations which is the same as the matrix equation
    [tex]\begin{bmatrix}1 & -3 & -4 \\ -4 & 6 & 3 \\ -3 & 7 & -4\end{bmatrix}\begin{bmatrix}x \\ y \\ x \\ y \end{bmatrix}= \begin{bmatrix}3 \\ 3 \\ -4 \end{bmatrix}[/tex]

    And, yes, you can attempt to solve that system of equation/matrix equation (the whole point is whether or not it can solved) by row reducing the "augmented matrix"
    [tex]\begin{bmatrix}1 & -3 & -4 & 3 \\ -4 & 6 & 3 & 3 \\ -3 & 7 & -4 & - 4\end{bmatrix}[/tex]
     
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