1. **State the problem:** Solve the system of equations using Gaussian elimination method: $$\begin{cases} 2x - 3y + 5z = 2 \\ x + 4y - 2z = 1 \\ 4x + 5y + z = 4 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 2 & -3 & 5 & 2 \\ 1 & 4 & -2 & 1 \\ 4 & 5 & 1 & 4 \end{array}\right]$$ 3. **Make the first pivot 1 by swapping row 1 and row 2:** $$\left[\begin{array}{ccc|c} 1 & 4 & -2 & 1 \\ 2 & -3 & 5 & 2 \\ 4 & 5 & 1 & 4 \end{array}\right]$$ 4. **Eliminate the first column below the pivot:** - Row 2 = Row 2 - 2*Row 1 - Row 3 = Row 3 - 4*Row 1 Calculations: Row 2: $(2 - 2*1, -3 - 2*4, 5 - 2*(-2), 2 - 2*1) = (0, -11, 9, 0)$ Row 3: $(4 - 4*1, 5 - 4*4, 1 - 4*(-2), 4 - 4*1) = (0, -11, 9, 0)$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 4 & -2 & 1 \\ 0 & -11 & 9 & 0 \\ 0 & -11 & 9 & 0 \end{array}\right]$$ 5. **Eliminate the second column below the pivot in row 2:** - Row 3 = Row 3 - Row 2 Calculations: Row 3: $(0 - 0, -11 - (-11), 9 - 9, 0 - 0) = (0, 0, 0, 0)$ Matrix now: $$\left[\begin{array}{ccc|c} 1 & 4 & -2 & 1 \\ 0 & -11 & 9 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$ 6. **Back substitution:** - From row 2: $-11y + 9z = 0 \Rightarrow -11y = -9z \Rightarrow y = \frac{9}{11}z$ - From row 1: $x + 4y - 2z = 1 \Rightarrow x = 1 - 4y + 2z$ Substitute $y$: $$x = 1 - 4\left(\frac{9}{11}z\right) + 2z = 1 - \frac{36}{11}z + 2z = 1 - \frac{36}{11}z + \frac{22}{11}z = 1 - \frac{14}{11}z$$ 7. **Solution:** The system has infinitely many solutions parameterized by $z$: $$\boxed{\left(x, y, z\right) = \left(1 - \frac{14}{11}z, \frac{9}{11}z, z\right), \quad z \in \mathbb{R}}$$