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# Linear algebra / pivot position (conceptual) watch

1. Solve the non-homogeneous system
x(1) + 2x(2) = 3
2x(1) + 4x(2) = 6

After you reduce the matrix represented by this system, you get
[1 2 | 3]
[0 0 | 0]

My question is, why is there a solution, namely
[3] + x(2)[-2]
[0]______[1]

if there is no pivot in the reduced augmented matrix given above? I thought if there isn't a pivot in every row, then b is not a solution to a matrix equation?

Edit: Does a solution exist to the above reduced matrix given above because the theorem-for which every row in a matrix must have a pivot in its row otherwise b isn't a solution-only applies for some b, not just all b's? And therefore, the reduced matrix above is a legitimate solution because the matrix doesn't violate the existence and uniqueness theorem?
2. (Original post by bluesoul227)
Solve the non-homogeneous system
x(1) + 2x(2) = 3
2x(1) + 4x(2) = 6

After you reduce the matrix represented by this system, you get
[1 2 | 3]
[0 0 | 0]

My question is, why is there a solution, namely
[3] + x(2)[-2]
[0]______[1]

if there is no pivot in the reduced augmented matrix given above? I thought if there isn't a pivot in every row, then b is not a solution to a matrix equation?

Edit: Does a solution exist to the above reduced matrix given above because the theorem-for which every row in a matrix must have a pivot in its row otherwise b isn't a solution-only applies for some b, not just all b's? And therefore, the reduced matrix above is a legitimate solution because the matrix doesn't violate the existence and uniqueness theorem?
Before starting of matrix reduction you should to recognise
that there is only one equation here. The second one is the first
equation multiplied by 2.
So this system includes dependent equations, which you can see
as a zero row in the reduced matrix.
So it is true that unique solution does not exissts, but
infinitely many solutions exist.

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