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Just curious

  1. Nov 19, 2011 #1
    An augmented matrix scaled by a number also means the solutions set is scaled by that same number. I believe this is true due to it basically being the same as elementary row operations preformed on each row. Unless it is a zero scalar in which case you lose all conditions. Is my method of thinking about this correct? Its not homework but I think it would help me understand a concept.
     
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  3. Nov 19, 2011 #2

    HallsofIvy

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    No, that's not true. If you have an augmented matrix, say
    [tex]\begin{bmatrix}a & b & c & j \\ d & e & f & k \\ g & h & i & l\end{bmatrix}[/tex]
    That is equivalent to the three equations ax+ by+ cz= j, dx+ ey+ fz= k, gx+ hy+ iz= l.
    It is also equivalent to the matrix equation
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}j \\ k \\ l\end{bmatrix}[/tex]

    "Scaling" the augmented matrix by a number, multiplying every row by that number, gives multiples of the same equations which have the same solutions. If you multiply the coefficient matrix in the second form by a constant, but not the right hand side, that will divide the soluutions by the number.
     
    Last edited: Nov 20, 2011
  4. Nov 20, 2011 #3

    Deveno

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    one can look at the 1x1 case:

    [a][x] =

    if we multiply by 2, let's say, we get:

    [2a][x] = [2b].

    suppose a ≠ 0. then in either case, we find:

    x = b/a, the solution to the 2nd equation is NOT x = 2(b/a).

    it is exactly the same for

    Ax = b, where A is a matrix, and b is a vector.

    if A is invertible, x = A-1b.

    and 2Ax = 2b has the same vector solution.

    in other words, multiplying an equation on both sides by a number,

    doesn't change the solution set for that equation.

    if ax + by + cz = d, then for any number m:

    m(ax + by + cz) = md, for the same x,y and z.
     
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