is an nxn matrix invertible if it can't be taken to row echelon form

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.

Explore the conditions under which an n×n matrix is invertible, with a focus on row echelon form and matrix rank. Understand how determinants and linear dependence impact matrix invertibility, and learn the key steps involved in matrix reduction. Suitable for undergraduate linear algebra students.

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