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Solve this linear system using matrices

  1. Jan 29, 2005 #1
    One of the questions on a handout was this:

    Solve the following linear system:

    A) Precisely B) With a calculator.

    Since this linear system contains irrational numbers, how would you solve it "precisely"? How would the answer be any different than if a calculator was used? Do I just leave alone the irrational square root terms and express the answer like that? Thanks for any help.
  2. jcsd
  3. Jan 29, 2005 #2
    Your solution may contain irrational numbers. Most calculators only display rational approximations (as some finite decimal).
    Unless you can find another way of expressing them that is not an approximation, yes. :)
  4. Jan 29, 2005 #3
    Maybe the professor is trying to show you what happens when you solve a system like that with a calculator. I don't know what happens when you try to solve that with a calculator, but since the above works for all x,y in R I would guess that the calculator can't represent this solution or something. The professor might therefore be trying to show you the dangers of using a calculator to solve systems of equations. Just a guess :/

    EDIT: Why it works for all x,y in R--multiply top by -sqrt(3)
  5. Jan 29, 2005 #4
    If you multiply the first equation by sqrt(3), you will see that the two become identical. This means that they are linearly dependant and there will be NO unique solution. There will actually be an infinite number of solutions. Maybe that is what your prof was trying to show.
  6. Jan 30, 2005 #5
    Parth Dave, so what about the calculator part? Would it hold true for the calculator part too? How would I know?

    Also, if I had problem like:

    [4*sqrt of 3 5 : 2 ]
    [5 sqrt of 13 : 1 ]

    What approach do I need to take here (same subject on solving precisely and solving by calculator)? I have to solve using determinants. He never covered this in class yet it's on the homework.
  7. Jan 30, 2005 #6
    Well if you have to solve it using determinants then you probably need to solve it using Cramer's rule.
  8. Jan 30, 2005 #7
    I tried that with the precise method by leaving alone the irrational numbers (keeping the square roots the way they are). And I end up with this really really long term that just doesn't seem right you know? Thanks for your help though.
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