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Homework Help: Quick question about matrix operations.

  1. Jan 21, 2010 #1
    Say there are these systems of equations :

    x - 2y + z = 0
    2y - 8z = 8
    -4x + 5y + 9z = -9

    In matrix form, it can be represented like this :

    --
    [1 -2 1 0] < -- row 1
    [0 2 -8 8] < -- row 2
    [-4 5 9 -9] < -- row 3

    When we do elementary row operations, say on row3 = row3 + 4*row1. That changes
    row 3 to [0 -3 13 -9], if we substitute this for old row 3 we get a simpler matrix, buts
    slightly modified. My question is that, when we do a row operation, how is the
    resultant matrix essentially the same, i.e has the same solution set ?

    I mean when we do row3 += 4*row1, does that not alter row3 to become a different
    row with different solution ? Maybe I missed something in algebra class. Basically, I do
    not see why row3 = row3 + 4*row1, has the same solution as just row3 ? I mean
    don't we have to do this, row3 + 4*row1 = row4 + 4row1 ? How come, we are only adding
    it to one side ?
     
  2. jcsd
  3. Jan 22, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    A row operation essentially amounts to combining two equations to another. In your example, you're multiplying the first equation by 4 and adding it to the third.

    [tex]x-2y+z=0 \rightarrow 4x-8y+4z=0[/tex]

    and

    [tex]-4x+5y+9z=-9[/tex]
    [tex](-4x+4x)+(5y-8y)+(9z+4z)=-9+0[/tex]
    [tex]-3y+13z=-9[/tex]

    If you're willing to accept that doing the algebra doesn't change the solutions, then you should be willing to accept the row operations don't change the solutions.
     
  4. Jan 22, 2010 #3

    Mark44

    Staff: Mentor

    When you replace a row by itself + a multiple of another row, you get different values for that row. Although the rows are different from before, the system of equations that is represented by the rows has exactly the same solution set.

    Think about it in terms of the equations that the augmented matrix represents. For a simple example, consider

    x + y = 3
    x - y = 1

    If I add the first equation to the second, my system now becomes

    x + y = 3
    2x = 4

    (What I have done is replace the 2nd equation by itself + 1 times the first equation.)

    It can easily be seen that the 2nd equation yields x = 2. Substition into the first equation gives y = 1, so the solution of the system is (2, 1).

    Notice that x = 2, y = 1 is also the solution of the original system of equations. Again, even though the two systems of equation look different, they have the same solution set.

    There are three operations that you can apply to a system of equation or an augmented matrix that represents such a system:
    1. Replace a row/equation by a nonzero multiple of itself.
    2. Interchange two rows/equations.
    3. Add a multiple of one row/equation to another row/equation. (I.e., replace a row/equation by itself plus a nonzero multiple of another row/equation.)

    These operations give you a new system of equations/augmented matrix with exactly the same solution set.
     
  5. Jan 22, 2010 #4
    >>"What I have done is replace the 2nd equation by itself + 1 times the first equation."

    See thats the part thats bugging me. From middle school I have took that method as
    granted. Do you think that there if a formal proof of why that method would work? Or
    is it just drawn from intuitive reason and induction?

    >> "When you replace a row by itself + a multiple of another row, you get different values for that row"

    I can see why scaling a row would still have the same solution set, but we are adding a value to a row. I mean we can't
    just add a value to a row an consider it to be the same equation in different form, unless the value is cancelled by subtracting it
    again.
     
  6. Jan 22, 2010 #5

    Mark44

    Staff: Mentor

    Yes, all three of the operations can be proved.
     
  7. Jan 22, 2010 #6
    Ok, I will search for it. Thanks.
     
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