1. **Problem Statement:** Solve the system of linear equations using the Relaxation Method starting with the initial vector $(0,0,0)$. 2. **Method Overview:** The Relaxation Method iteratively updates variables by correcting the largest residual divided by the diagonal coefficient. 3. **Given:** Diagonal coefficients: $a_{11}=6$, $a_{22}=-7$, $a_{33}=-8$ 4. **Iteration 0:** Start with $\mathbf{x}=(0,0,0)$. Residuals: $R_1=11$, $R_2=10$, $R_3=-15$ Largest residual magnitude: $|R_3|=15$ Update $x_3$: $$dx_3=\frac{R_3}{a_{33}}=\frac{-15}{-8}=1.875$$ New vector: $(0,0,1.875)$ 5. **Iteration 1:** Residuals: $R_1=9.125$, $R_2=8.125$, $R_3=0$ Largest residual magnitude: $|R_1|=9.125$ Update $x_1$: $$dx_1=\frac{9.125}{6}=1.5208$$ New vector: $(1.5208,0,1.875)$ 6. **Iteration 2:** Residuals: $R_1=0$, $R_2=6.6042$, $R_3=-3.0416$ Largest residual magnitude: $|R_2|=6.6042$ Update $x_2$: $$dx_2=\frac{6.6042}{-7}=-0.9435$$ New vector: $(1.5208,-0.9435,1.875)$ 7. **Iteration 3:** Residuals: $R_1=-2.8303$, $R_2=0$, $R_3=-2.0981$ Largest residual magnitude: $|R_1|=2.8303$ Update $x_1$: $$dx_1=\frac{-2.8303}{6}=-0.4717$$ New vector: $(1.0491,-0.9435,1.875)$ 8. **Iteration 4:** Residuals: $R_1=0$, $R_2=0.4714$, $R_3=-1.1547$ Largest residual magnitude: $|R_3|=1.1547$ Update $x_3$: $$dx_3=\frac{-1.1547}{-8}=0.1443$$ New vector: $(1.0491,-0.9435,2.0193)$ 9. **Iteration 5:** Residuals: $R_1=-0.1444$, $R_2=0.3271$, $R_3=0$ Largest residual magnitude: $|R_2|=0.3271$ Update $x_2$: $$dx_2=\frac{0.3271}{-7}=-0.0467$$ New vector: $(1.0491,-0.9902,2.0193)$ 10. **Iteration 6:** Residuals: $R_1=-0.2845$, $R_2=0$, $R_3=0.0464$ Largest residual magnitude: $|R_1|=0.2845$ Update $x_1$: $$dx_1=\frac{-0.2845}{6}=-0.0474$$ New vector: $(1.0017,-0.9902,2.0193)$ 11. **Iteration 7:** Residuals: $R_1=0$, $R_2=0.0476$, $R_3=0.1412$ Largest residual magnitude: $|R_3|=0.1412$ Update $x_3$: $$dx_3=\frac{0.1412}{-8}=-0.0177$$ New vector: $(1.0017,-0.9902,2.0016)$ 12. **Conclusion:** The solution converges to approximately: $$X_1 \approx 1, \quad X_2 \approx -1, \quad X_3 \approx 2$$ This iterative process continues until the residuals are within the desired tolerance, confirming the solution.