Final answer:
The given linear system is not in echelon form because it does not meet the necessary properties, such as having leading coefficients in a staircase pattern and the rows ordered correctly.
Step-by-step explanation:
To determine if a linear system is in echelon form, several properties need to be checked:
- All nonzero rows are above any rows of all zeroes.
- Each leading coefficient (also known as a pivot) of a nonzero row is to the right of the leading coefficient of the row above it.
- All entries in a column below a leading coefficient are zeroes.
The given linear system:
- 3x₁ + 2x₂ + 5x₃ = 0
- -6x₃ = -6
- -x₂ - 5x₃ = 11
Upon inspection, we can see:
- Row one has its first leading coefficient at x₁, which is fine.
- Row two should ideally follow a pattern that has the leading coefficient at or to the right of the previous row's leading coefficient, but it actually jumps back since there's no x₁ or x₂ term, violating the second property.
- Row three also violates the pattern because it starts with x₂, and it should be positioned above row two.
Therefore, the given system is not in echelon form due to the incorrect ordering that violates the properties of echelon systems. To be in echelon form, the rows would need to be reordered such that the leading coefficients align appropriately. As it stands, the equations require rearrangement to meet the definition of echelon form.