1. **State the problem:** Solve the system of equations using the Gauss elimination method: $$\begin{cases} 2x + y + 6z = 14 \\ 3x + 5y + 2z = 21 \\ 4x + y - z = 1 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 2 & 1 & 6 & 14 \\ 3 & 5 & 2 & 21 \\ 4 & 1 & -1 & 1 \end{array}\right]$$ 3. **Make the first pivot 1 by dividing row 1 by 2:** $$R_1 \to \frac{1}{2}R_1 = \left[1 \quad \frac{1}{2} \quad 3 \quad 7\right]$$ 4. **Eliminate the first column entries below the pivot:** - For row 2: $R_2 \to R_2 - 3R_1 = \left[3 - 3(1), 5 - 3(\frac{1}{2}), 2 - 3(3), 21 - 3(7)\right] = \left[0, \frac{7}{2}, -7, 0\right]$ - For row 3: $R_3 \to R_3 - 4R_1 = \left[4 - 4(1), 1 - 4(\frac{1}{2}), -1 - 4(3), 1 - 4(7)\right] = \left[0, -1, -13, -27\right]$ 5. **Make the second pivot 1 by dividing row 2 by $\frac{7}{2}$:** $$R_2 \to \frac{2}{7}R_2 = \left[0, 1, -\frac{14}{7}, 0\right] = \left[0, 1, -2, 0\right]$$ 6. **Eliminate the second column entry in row 3:** $R_3 \to R_3 + R_2 = \left[0, -1 + 1, -13 + (-2), -27 + 0\right] = \left[0, 0, -15, -27\right]$ 7. **Make the third pivot 1 by dividing row 3 by -15:** $$R_3 \to -\frac{1}{15}R_3 = \left[0, 0, 1, \frac{27}{15}\right] = \left[0, 0, 1, \frac{9}{5}\right]$$ 8. **Back substitution:** - From row 3: $z = \frac{9}{5}$ - From row 2: $y - 2z = 0 \Rightarrow y = 2z = 2 \times \frac{9}{5} = \frac{18}{5}$ - From row 1: $x + \frac{1}{2}y + 3z = 7 \Rightarrow x = 7 - \frac{1}{2}y - 3z = 7 - \frac{1}{2} \times \frac{18}{5} - 3 \times \frac{9}{5} = 7 - \frac{9}{5} - \frac{27}{5} = 7 - \frac{36}{5} = \frac{35}{5} - \frac{36}{5} = -\frac{1}{5}$ 9. **Final solution:** $$\boxed{\left(x, y, z\right) = \left(-\frac{1}{5}, \frac{18}{5}, \frac{9}{5}\right)}$$