Echelon form and set of solutions.

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  • #1
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Hello,

I have the following system of linear equations -

kx + 3y -z = 1
x + 2y - z = 2
-kx + y + 2z = -1

I have reduced it to

1 2 -1 : 2
0 1 1/4 : 0
0 0 (7-6k)/7k : (8-4k)/7

assuming k ≠ 0.

I would now like to be able to determine for what values of k will the system have no solutions, a single solution, and an infinite number of solutions.

Would it be correct to say that there would be no solution for k=7/6, and a single solution for every k ≠ 7/6 (therefore, the system will never have an infinite number of solutions)?
 

Answers and Replies

  • #2
Dick
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Hello,

I have the following system of linear equations -

kx + 3y -z = 1
x + 2y - z = 2
-kx + y + 2z = -1

I have reduced it to

1 2 -1 : 2
0 1 1/4 : 0
0 0 (7-6k)/7k : (8-4k)/7

assuming k ≠ 0.

I would now like to be able to determine for what values of k will the system have no solutions, a single solution, and an infinite number of solutions.

Would it be correct to say that there would be no solution for k=7/6, and a single solution for every k ≠ 7/6 (therefore, the system will never have an infinite number of solutions)?

Sounds correct to me.
 
  • #3
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What about k different than 0? I had to divide by k whilst reducing the matrix. How come it when k=0 the system still yields a solution? Is it possible that the system will never have an infinite number of solutions? (seems like something is amiss with my reduced matrix/method)
 
  • #4
Dick
Science Advisor
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What about k different than 0? I had to divide by k whilst reducing the matrix. How come it when k=0 the system still yields a solution? Is it possible that the system will never have an infinite number of solutions? (seems like something is amiss with my reduced matrix/method)

Sure you divided by k. You still have a k in the denominator. But you can get rid of it by multiplying the last row by k. If you really want to you can also put k=0 in the original matrix and check there is no problem. To get an infinite number of solutions you need a row of zeros. Can't happen here.
 
  • #5
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But what if I didn't perform the row operations properly? I went over them and yet mistakes are always possible. May you please confirm that I got it aright? I mean, that my reduced form is indeed the reduced form of that matrix?
 
  • #6
Dick
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But what if I didn't perform the row operations properly? I went over them and yet mistakes are always possible. May you please confirm that I got it aright? I mean, that my reduced form is indeed the reduced form of that matrix?

There's more than one reduced form, you can do it a number of different ways. But my way also showed no solutions for k=7/6.
 
  • #7
880
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Thank you very much, Dick
 

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