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Matrix Application

  1. Jul 24, 2007 #1
    The question is "For single women in the poll, the percentage who said no exceeded the combined percentages for those who said yes and those who said not sure by 22%. If the percentage who said yes is doubled, it is 7% more than the percentage who said no. Find the percentage of single women who responded yes, no, and not sure."

    The chapter deals with matrices so I naturally tried to setup the problem as follows
    1 1 -1 | -22 (yes, no, not sure)
    2 0 -1 | 7

    The back of the book setups on an equations as follows
    x + z = y - 22
    2x = y + 7

    I tried to eliminate one of the variables, but then I always end up with two unknowns in the final equation. The main problem I'm running into is that the chapter provides examples that Gaussian Elimination works nicely on, but this method does not appear to work so nicely on this problem. Is the way I setup it up the first time correct & is there a particular method that would work better than another to solve the problem?

    Fyi, the back of the book as yes 34%, no 61%, & not sure 5% for the answer.
  2. jcsd
  3. Jul 24, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    What does this fact tell you about the original system of equations? This result is giving you a big hint about what you need to do... if you know how to read the hint.
  4. Jul 25, 2007 #3
    If you have a system of equations with three unknowns but only two equations, what do you know about the system(can there be a unique solution)? Do you know about free variables yet?

    Basically you reduce the system to echelon form, reduced echelon form if you like(easier as you will need to use back-substitution later if you don't, but some people like doing that for some odd reason), and the variables that do not appear as leading 1's in the system are called 'free variables', you can assign any number to them and get a solution, in other words the system has an infinite number of solutions. You can also solve the system by solving for the non-free variables in terms of the free variables.

    I am assuming that in your problem you find the general solution for the system by solving for the non-free variables in terms of the free variables, then plug in a value for the free variable z and find the solution you are looking for. Can you do it with this information?
  5. Jul 25, 2007 #4

    D H

    Staff: Mentor

    You have an underspecified system of equations. A unique solution cannot exist. To illustrate, suppose you had but two unknowns and the single equation 2x=y+7. What is the solution? It is a line, not a point. Some other piece of information is needed to reduce the problem to a single solution.

    In this case, such information exists. What else can you say about the percentages for yes, no, and not sure?
  6. Jul 25, 2007 #5
    Your trying to solve a simultaneous equation using a matrix, Yeah? Well, the co-efficient matrix has to be square, and the number of collumns in the co-efficient matrix has to equal the number of rows in the other one.

    It seems to me like there is information missing? Why does one equation have two variables, the other three?

    You need either two equations with two variables, or three equations with three variables.
  7. Jul 25, 2007 #6
    This here really did the trick.

    I set the matrix up as

    1 1 -1 | -22
    2 0 -1 | 7
    1 1 1 | 100

    Then broke it down to

    1 0 0 | 34
    0 1 0 | 5
    0 0 1 | 61

    So it worked, thanks for the tip, was working on this one for a long time & just couldn't solve the two equations by themselves, but that makes sense now why not.
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