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Matrices-back substitution

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    I have attached the question

    2. Relevant equations



    3. The attempt at a solution

    I think the answer is no solution because there is 5 variables but only 3 equations. Is that correct?
     

    Attached Files:

  2. jcsd
  3. Oct 31, 2011 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, that is not the reason. Look at the final row; it is shorthand for an equation involving x1, x2, x3, x4, x5. What is the equation?

    RGV
     
  4. Oct 31, 2011 #3
    do you mean 0x1 + 0x2 + 0x3 + 0x4 + 0x5 = 4?
     
  5. Oct 31, 2011 #4
    if the third is also an equation that means there must be an answer right? So because we have to assign variables and we have 5unknowns and 3 equations that must mean two values will be variables right?? so the answer has to be B yeah??
     
  6. Oct 31, 2011 #5

    Mark44

    Staff: Mentor

    For what values of the variables x1, x1, x2, x3, and x4 will this be a true statement?


    Not necessarily. There are three possibilities for a system of equations (which are here represented by an augmented matrix):
    1) a unique solution
    2) multiple solutions
    3) no solution.
     
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