anonymous
  • anonymous
Solve the following system by Gauss–Jordan elimination. 2x₁+5x₂+3x₃=35 12x₁+31x₂+20x₃=216
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I got x₁=5/2 x₂=6 x₃=t (free variable) The answer says I am wrong. Care to tell me what I did wrong?
anonymous
  • anonymous
anonymous
  • anonymous

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UnkleRhaukus
  • UnkleRhaukus
x_3=0
anonymous
  • anonymous
Right. But whats wrong with my other answers?
anonymous
  • anonymous
Here is what the online page looks like:
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anonymous
  • anonymous
Shouldn't the other variables be in terms of the free variable?
anonymous
  • anonymous
What did were your elimination steps?
anonymous
  • anonymous
It's difficult to post the elimination steps here. I checked with my calculator though. It's correct.
anonymous
  • anonymous
But yeah, that's what I think @wio . I don't know how to express it though.
UnkleRhaukus
  • UnkleRhaukus
when we check your solution , we see t must be 0 x₁=5/2 x₂=6 x₃=t _________ 2x₁+5x₂+3x₃=35 5+30+3t = 35 35 +3t = 35 _________ 12x₁+31x₂+20x₃ = 216 60 + 186 + 20t = 216 216 + 20t = 216
anonymous
  • anonymous
but t is a free variable is it not? SO it can be any value it wants.
anonymous
  • anonymous
\[ \begin{bmatrix} 2&5&3 \\ 12&31&20 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 35 \\ 216 \\ t \end{bmatrix} \]
anonymous
  • anonymous
Did you solve it like this?
anonymous
  • anonymous
No... How did you get 0,0 and 1?
anonymous
  • anonymous
I just added in the equation \(x_3=t\)
anonymous
  • anonymous
Sorry, that doesn't make sense.
anonymous
  • anonymous
I added in the equation: \[ 0x_1+ 0x_2 +1x_3 = t \]
anonymous
  • anonymous
Ohh I see.
anonymous
  • anonymous
All of the elimination you did before should work for the most part.
anonymous
  • anonymous
So I do a back substitution?
anonymous
  • anonymous
You first want to get the second row to have a zero in it's first colum
anonymous
  • anonymous
because x3=t so solve the others n terms of t.
anonymous
  • anonymous
then you can back substitute
anonymous
  • anonymous
1 second.
anonymous
  • anonymous
x₁=2.5+3.5t ?
anonymous
  • anonymous
x₂=6-2t?
anonymous
  • anonymous
yeah I'm getting that too
anonymous
  • anonymous
did you try it?
anonymous
  • anonymous
One attempt left :3 . Fingers crossed.
anonymous
  • anonymous
you can double check it you know
anonymous
  • anonymous
How?
anonymous
  • anonymous
plug in each equation into the equations
anonymous
  • anonymous
Assign a random number to t and see if it works?
anonymous
  • anonymous
like, plug in t for x_3 and 6-2t for x_2 into the equations
anonymous
  • anonymous
hmm actually I'm not quite sure how to double check it
anonymous
  • anonymous
let me look at wolfram
anonymous
  • anonymous
You can do this on wolfram? O_O .
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=2x%2B5y%2B3z%3D35%2C+12x%2B31y%2B20z%3D216%2C+z+%3D+t
anonymous
  • anonymous
wolfram agrees... it's just a matter of the stupid answer verify thing not being a wingspan
anonymous
  • anonymous
THANKS!!! It's right :) .
anonymous
  • anonymous
No problem. Matrix is your friend.
anonymous
  • anonymous
I love Linear algebra <3 .

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