# How do you solve for x using the expansion method?

## Homework Statement

solve for x using the expansion method

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

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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tiny-tim
Homework Helper
Hi haengbon!

I'm not familiar with the "expansion method", but simple subtraction on …
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!

Mark44
Mentor
tiny-tim's not alone. I haven't heard of this method, either, so the work you show is a complete mystery to me.

vela
Staff Emeritus
Homework Helper
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.