1. **Problem Statement:** Test the consistency of the system and solve if consistent: $$\begin{cases} x + y + z = 3 \\ x + 2y + 3z = 4 \\ x + 4y + 9z = 6 \end{cases}$$ 2. **Step 1: Write the augmented matrix** $$\left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 1 & 2 & 3 & 4 \\ 1 & 4 & 9 & 6 \end{array}\right]$$ 3. **Step 2: Use row operations to find echelon form** - Subtract Row 1 from Row 2: $$R_2 \to R_2 - R_1: \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ \cancel{1} - \cancel{1} & 2 - 1 & 3 - 1 & 4 - 3 \\ 1 & 4 & 9 & 6 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 1 & 4 & 9 & 6 \end{array}\right]$$ - Subtract Row 1 from Row 3: $$R_3 \to R_3 - R_1: \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ \cancel{1} - \cancel{1} & 4 - 1 & 9 - 1 & 6 - 3 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 3 & 8 & 3 \end{array}\right]$$ - Subtract 3 times Row 2 from Row 3: $$R_3 \to R_3 - 3R_2: \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & \cancel{3} - 3 \times \cancel{1} & 8 - 3 \times 2 & 3 - 3 \times 1 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 2 & 0 \end{array}\right]$$ 4. **Step 3: Check for consistency** No row has the form $[0\ 0\ 0\ |\ c]$ with $c \neq 0$, so the system is consistent. 5. **Step 4: Back substitution to solve** - From Row 3: $2z = 0 \Rightarrow z = 0$ - From Row 2: $y + 2z = 1 \Rightarrow y + 0 = 1 \Rightarrow y = 1$ - From Row 1: $x + y + z = 3 \Rightarrow x + 1 + 0 = 3 \Rightarrow x = 2$ 6. **Final solution:** $$\boxed{(x, y, z) = (2, 1, 0)}$$