No, the matrix formed by the rows $\{(1,0),(0,-2),(0,1)\}$ is not in row echelon form. Here's why:

For a matrix to be in row echelon form, it must satisfy these conditions:

1. The leading (non-zero) entry in each row must be 1 (this is called the leading 1).
2. Each leading 1 must be to the right of the leading 1 in the row above it.
3. Any rows consisting entirely of zeros must be at the bottom of the matrix.

Let’s consider the matrix:

1 & 0 \\ 0 & -2 \\ 0 & 1 \end{pmatrix}$$ - The first row $$(1, 0)$$ has a leading 1, which is correct. - The second row $$(0, -2)$$ does not have a leading 1; it has a leading $$-2$$, which violates condition (1). - The third row $$(0, 1)$$ has a leading 1, but since the second row isn’t in the proper form, the overall matrix isn't in row echelon form. To convert this matrix to row echelon form, you would need to: - Divide the second row by $$-2$$ to make the leading coefficient in that row equal to 1. Once this is done, the matrix would look like: $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{pmatrix}$$ This form satisfies the row echelon conditions. Would you like further clarification, or do you have any questions? ### Related Questions: 1. How do you convert a matrix to row echelon form? 2. What is the difference between row echelon form and reduced row echelon form? 3. Can any matrix be transformed into row echelon form? 4. How does Gaussian elimination help achieve row echelon form? 5. What are the key properties of row echelon form? **Tip:** Row echelon form focuses on the structure of leading 1's, while reduced row echelon form additionally ensures each leading 1 is the only non-zero entry in its column.