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What is M? system of linear equations

  1. Oct 29, 2003 #1
    what is M?
    x1+x2=2x3+4x4=5
    2x1+2x2=3x3+x4=3
    3x1+3x2-4x3-2x4=1

    why does M={ 1 1 -2 4 5 }
    2 2 -3 1 3
    3 3 -4 -2 1

    i know how they got this but how do you get
    ~{ 1 1 -2 4 5 } { 1 1 0 -10 -9
    0 0 1 -7- 7 ~~~ 0 0 1 -7 -7 ??????? HOW?
    0 0 2 -14 -14} 0 0 0 0 0
     
  2. jcsd
  3. Oct 29, 2003 #2

    HallsofIvy

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    Staff Emeritus
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    What is Q? What is P?

    Come on, If you don't know enough to say WHAT the problem is, you are in real trouble!

    I absolutely guarentee that the problem in your textbook doesn't say just "What is M?" without any explanation. You are given as system of linear equations and asked to find a MATRIX (which you happened to have called M- You could just as easily have called it "Fred" and asked "What is Fred?") consisting of the coefficients of the unknowns in the system of equations.

    It looks to me like you haven't even copied the problem correctly. Did your equations really have those "=" in the middle? Do you really have 5 equations or 3? I'm betting 3.

    Since you have 3 equations if 4 unknowns, you system will look like this when it is converted to "matrix" form:
    [ a b c d][x1] [5]
    [ e f g h][x2]= [3]
    [ i j k l][x3] [1]
    [x4]

    Now multiply the matrices on the lefthand side. What do you get?
    (You should have learned how to multiply matrices long ago if you are expected to do problems like this.) Compare the rows of the product matrix with the system of equation you are given. What do a, b, c, etc. have to be in order that the rows and equations are the same?
    It should be obvious that the numbers in each row have to be exactly the coefficients in the equations. Once you have that, just add the "right hand side" of the equations as a fifth column.

    Do you know what row operations are? Certainly if your textbook is doing problems like this "row operations" should be in the previous chapter or section. Look them up! Go to your instructor and throw yourself on his/her mercy!

    Starting from [1 1 -2 4 5]
    [2 2 3 1 3]
    [3 3 -4 -2 1]
    your objective is to arrive at [1 0 0 * *]
    [0 1 0 * *]
    [0 0 1 * *]
    which would correspond to x1= * etc.

    There are many different ways to do this. Just as you work with an entire equation at a time when solving equations, so you work with an entire row at a time (row operations). First: subtract twice the first row from the second: [ 1 1 -2 4 5 ]
    [2-2(1) 2-2(1)3-2(-2) 1-2(4) 3-2(5)]
    [3 3 -4 -2 1 ]
    or [1 1 -2 4 5]
    [0 0 7 -7 -7]
    [3 3 -4 -2 -1]. The purpose was to get that "0" in the second row, first column.

    Now subtract 3 time the first row from the third row to get a 0 in the first column of the third row: the result is
    [1 1 -2 4 5]
    [0 0 7 -7 -7]
    [0 0 2 -14 -16]
    Hmmm those "0"s in the second row look like trouble!
     
  4. Oct 29, 2003 #3
    o no sorry i got it...i was watching MIT's linear algebra lectures and i know...thanks.
     
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