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March 26, 2017

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Consider the system
x-2y=k-1
4x+(1-k^2)y=8

For what values of k does the system have
a. unique solution?
b. infinitely many solutions?
c. no solution?

  • Linear Algebra - ,

    rearranging things a bit, you have

    y = x/2 + (1-k)/2
    y = 4x/(k^2-1) + 8/(1-k^2)

    for infinitely many solution, we need
    y = mx+b
    y = mx+c
    where b=c

    same slope:
    1/2 = 4/(k^2-1)
    k^2-1 = 8
    k^2 = 9
    k = ±3

    b=c:
    (1-k)/2 = 8/(1-k^2)
    (1-k)(1-k^2) = 16
    k=3
    So, if k=3 there are infinitely many solutions.

    If k = -3, there are no solutions.

    For any other value of k, there is one solution.

  • Linear Algebra - ,

    Can you perform using matrix? Thank you very much

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