Solving for X inside of a 3X3Matrix

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The discussion focuses on solving a system of equations using Cramer's rule within a 3x3 matrix framework. The initial setup leads to the equation x(2x-3)-1(2)+2(1)=-4x, which simplifies to 2x^2+x=0. A mistake is identified in the division step, where the incorrect result of 2x^2+1x=0 is derived. The correct solutions found are x=0 and x=-1, but the user expresses frustration over the error in their calculations. Ultimately, the thread highlights the importance of careful manipulation of equations in matrix algebra.
smashbrohamme
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Using cramers rule.

the equation is.

x,1,2
1,x,3 all set to equal -4x
0,1,2


I made my 2x2 matrix like this

x|x,3| -1|1,3| +2|1,x|
|1,2| |0,2| |0,1|

eventually I ended up with x(2x-3)-1(2)+2(1)=-4x

condensed it down to 2x^2-3x=-4x

2x^2+1x=0
x(x+1)=0

x=0, or x=-1.

I go back to plug it in...and its wrong! :(
 
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When you had your initial equation, 2x^2+x=0 and you divided by 2, you got an incorrect result.
 
smashbrohamme said:
2x^2+1x=0
x(x+1)=0

Little mistake in this step...
 

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