How do you solve for x using the expansion method?

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The discussion focuses on solving the system of equations using the expansion method, which involves calculating the determinant of a matrix derived from the equations. Participants express confusion over the term "expansion method," with some suggesting it may relate to Cramer's rule. The calculations reveal that the determinant equals zero, indicating that there is no unique solution for x. The conversation highlights the complexity of the method and the potential for multiple values of z, leading to the conclusion that the system is undefined. Overall, the thread emphasizes the challenges in applying the expansion method to this particular set of equations.
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Homework Statement




solve for x using the expansion method

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

Homework Equations


none


The Attempt at a Solution



1 1 1 1 1
1 1 2 1 1
1 1 3 1 1


1 1 1 1 1
1 1 2 1 1
1 1 3 1 1


1 1 1 1 1
1 1 2 1 1
1 1 3 1 1

(1)(1)(3) + (1)(2)(1) + (1)(1)(1) - (3)(1)(1) + (1)(2)(1) + (1)(1)(1)
(3)+(2)+(1) - (3)(2)(1)
=6 - 6
= 0

would this be undefined then?
 
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Hi haengbon! :smile:

I'm not familiar with the "expansion method", but simple subtraction on …
haengbon said:
x + y + z =1
x + y + 2z =2
x + y + 3z = 1

… gives both z = 1 and z = 0, so clearly there are no solutions! :wink:
 
tiny-tim's not alone. I haven't heard of this method, either, so the work you show is a complete mystery to me.
 
I think the OP is writing the system in terms of a matrix A where Ax=b and is trying to calculate the determinant of A. In this case, one gets

det A = (1)(1)(3)+(1)(2)(1)+(1)(1)(1)-(1)(1)(1)-(1)(2)(1)-(1)(1)(3) = 3+2+1-1-2-3 = 0

so there's no unique solution for x.

Perhaps "expansion method" refers to Cramer's rule.
 
That's the method that I learned for calculating the determinate in HS. You copy the first column over on the Right side so the diagonals are all nice and in line...then you calculate the number as shown above...forgot the + signs on the last line there.
(3)+(2)+(1) - (3)(2)(1) should have + in between the last 3 2 1.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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