Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method equation (i) 10x2 sin x1 - 0.8 = 0 equation (ii) 10x_2^2 - 10x2 cos x1 - 0.6 = 0 Assuming the initial values x1 = 0.0 and x2 = 1.0, the Jacobian matrix is

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Numerical Methods

Solution of the variables x₁ and x₂ for the following equations is to be obtained by employing the Newton-Raphson iterative method
equation (i) 10x₂ sin x₁ - 0.8 = 0
equation (ii) 10$${\text{x}}_2^2$$ - 10x₂ cos x₁ - 0.6 = 0
Assuming the initial values x₁ = 0.0 and x₂ = 1.0, the Jacobian matrix is

A) \[\left[ {\begin{array}{*{20}{c}} {10}&{ - 0.8} \\ 0&{ - 0.6} \end{array}} \right]\] B) \[\left[ {\begin{array}{*{20}{c}} {10}&0 \\ 0&{10} \end{array}} \right]\] C) \[\left[ {\begin{array}{*{20}{c}} 0&{ - 0.8} \\ {10}&{ - 0.6} \end{array}} \right]\] D) \[\left[ {\begin{array}{*{20}{c}} {10}&0 \\ {10}&{ - 10} \end{array}} \right]\]

Answer & Solution

Correct Answer: Option B

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Solution of the variables x1 and x2 for the following equations...

Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method equation (i) 10x2 sin x1 - 0.8 = 0 equation (ii) 10$${\text{x}}_2^2$$ - 10x2 cos x1 - 0.6 = 0 Assuming the initial values x1 = 0.0 and x2 = 1.0, the Jacobian matrix is

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