anonymous
  • anonymous
Applied Linear Algebra Question, I'm posting question give me one second
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
okay so I got to the point where I dont know what to do next
anonymous
  • anonymous
I have to find values of "a" where the system has no solution, Unique Solution, and infinte solutions

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anonymous
  • anonymous
I got the system to row echelon form
anonymous
  • anonymous
lt me draw what i got out
anonymous
  • anonymous
|dw:1442684208867:dw|
anonymous
  • anonymous
Do I need to put this into reduce echelon form?
anonymous
  • anonymous
Actually yea I think that would help
anonymous
  • anonymous
let me do that on paper, first brb
anonymous
  • anonymous
oh wait made a mistake in the drawing, the "3a+10" has a denominator of "2a-4"
anonymous
  • anonymous
|dw:1442685001288:dw|
anonymous
  • anonymous
If a is not equal to 2, then we have a unique solution Plugging in a = 2 in the second we get 1 -2 3 3 0 -4+2a 3-3a 10-3a 0 0 1 -1 the middle row second term is non zero, so you can solve for it, and that produces a triangular system with a unique solution. Try a number.
IrishBoy123
  • IrishBoy123
find the determinant and set it to zero and that might save a lot of typing
IrishBoy123
  • IrishBoy123
ie set to zero and solve for a
anonymous
  • anonymous
I assume you mean the determinant of the coefficient matrix.
anonymous
  • anonymous
the determinant of the coefficient matrix is 12 - 6a the matrix has a unique solution as long as the determinant is non zero. 12 - 6a = 0 12 = 6a 12/6 = a if a ≠ 2 then the system has a unique solution (because the determinant is non-zero)
anonymous
  • anonymous
if a = 2 you have a contradiction, can you see why?
anonymous
  • anonymous
Case 1: "a" does not equal to 2, give a uniqure solution case 2: "a" equals 2, gives no Solution Case 3: there is never a infinite solution is that correct?
IrishBoy123
  • IrishBoy123
yeah, work that way :p if \(a \ne 2\) you have a solution so set a to 2 and do your reduction, see where you go
anonymous
  • anonymous
It will be no solution if u get set a = 2
anonymous
  • anonymous
correct
anonymous
  • anonymous
So would it be right to say that this system does not have infinite solutions?
anonymous
  • anonymous
So it has no solution when a=2, a unique solution a does not =2, but not infinite solutions right? or when a does not equal to 2 it will have infinite solution as well?
anonymous
  • anonymous
anyone there?
anonymous
  • anonymous
it will never have infinite solutions
anonymous
  • anonymous
what I'm confused about is that if a does not equal to 2, will it have infinite solutions
anonymous
  • anonymous
oh ok, can you tell me why though?
anonymous
  • anonymous
if a does not equal to 2 it will have a unique solution if a is equal to 2 it will have no solution (because of contradiction) there is no case of 'infinite solutions'
anonymous
  • anonymous
oooh I see, lol thanks, I blanked out for a second there. Thank you so much!
anonymous
  • anonymous
Your welcome.
IrishBoy123
  • IrishBoy123
|dw:1442690538127:dw|
IrishBoy123
  • IrishBoy123
|dw:1442690850922:dw|

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