How do you solve for x using the expansion method?

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SUMMARY

The discussion focuses on solving the system of equations using the expansion method, specifically for the equations x + y + z = 1, x + y + 2z = 2, and x + y + 3z = 1. Participants conclude that the determinant of the corresponding matrix A is zero, indicating that there is no unique solution for x. The term "expansion method" may refer to Cramer's rule, which is used for calculating determinants in linear algebra.

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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.
 

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