# Chapter 2 - Systems of Linear Equations - 2.2 Direct Methods for Solving Linear Systems - Exercises 2.2 - Page 79: 9

Row Echelon Form: $$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ Reduced Row Echelon Form: $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

#### Work Step by Step

We're given $$\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix}$$ First, swap rows 1 and 3, to get the row echelon form: $$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ Now, subtract row 3 from rows 1 and 2: $$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ And subtract row 2 from row 1: $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ The matrix is now in reduced row echelon form.

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