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Solve the following system by Gauss–Jordan elimination. 2x₁+5x₂+3x₃=35 12x₁+31x₂+20x₃=216

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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?

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Other answers:

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

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