Question 12.2: Solve the two simultaneous equations: f1(x1, x2) = x1² + 2x2......
Solve the two simultaneous equations:
f_{1}(x_{1}, x_{2}) = x_{1}^2 + 2x_{2} \ – \ 3 = 0
f_{2}(x_{1}, x_{2}) = x_{1}x_{2} \ – \ 3x_{2}^2 + 2 = 0
Let the initial values of x_{1} and x_{2} be 3 and 2, respectively.
Step 0:
x^{1} = x^{0}- \left|\begin{matrix} 6 & 2 \\ 2 & -9 \end{matrix} \right|^{-1} f(x^{0})
=\left|\begin{matrix} 3 \\ 2 \end{matrix} \right|- \left|\begin{matrix} \frac{9}{58} & \frac{2}{58} \\ \frac{2}{58} & -\frac{6}{58} \end{matrix} \right|\left|\begin{matrix} 10 \\ 4 \end{matrix} \right|=\left|\begin{matrix} 1.586 \\ 1.241 \end{matrix} \right|
Step 1:
x^{2} = x^{1}- \left|\begin{matrix} 3.172 & 2 \\ 1.241 & -5.86 \end{matrix} \right|^{-1} f(x^{1})
=\left|\begin{matrix} 1.586 \\ 1.241 \end{matrix} \right|- \left|\begin{matrix} 0.278 & 0.095 \\ 0.095 & -0.151 \end{matrix} \right|\left|\begin{matrix} 1.997 \\ -0.652 \end{matrix} \right|=\left|\begin{matrix} 1.092 \\ 1.024 \end{matrix} \right|
Step 2:
x^{3} = x^{2}- \left|\begin{matrix} 2.184 & 2 \\ 1.025 & -5.058 \end{matrix} \right|^{-1} f(x^{2})
=\left|\begin{matrix} 1.092 \\ 1.025 \end{matrix} \right|- \left|\begin{matrix} 0.386 & 0.153 \\ 0.078 & -0.167 \end{matrix} \right|\left|\begin{matrix} 0.242 \\ -0.033 \end{matrix} \right|=\left|\begin{matrix} 1.004 \\ 1.001 \end{matrix} \right|
Thus, the results are rapidly converging to the final values of x_{1} and x_{2} which are 1 and 1, respectively.