If an $n \times n$ matrix cannot be transformed into row echelon form, it is not invertible.

To explain why:

1. A matrix is invertible if and only if its determinant is non-zero.
2. When performing row operations to reduce a matrix to row echelon form, if you encounter a row of all zeros, this means that the matrix has linearly dependent rows.
3. A matrix with linearly dependent rows has a determinant of zero, meaning it is singular and thus non-invertible.

Thus, if you cannot fully reduce a matrix to row echelon form without hitting a row of all zeros, this implies the matrix does not have full rank (its rank is less than $n$) and is therefore not invertible.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How does the determinant of a matrix relate to its invertibility?
2. What are the steps for reducing a matrix to row echelon form?
3. How does the rank of a matrix determine if it's invertible?
4. What does it mean for a matrix to have linearly dependent rows?
5. Can non-square matrices be invertible?

Tip: A quick way to check if a matrix is invertible is to calculate its determinant. If it's zero, the matrix is not invertible.