Solution of the variables x 1 and x 2 for the following equations is to be obtained by employing the Newton Raphson iterative method. equation (i) 10x 2 sin x 1 - 0.8 = 0 equation (ii) 10x_{2}^{2}−10x_{2}\cos ⁡x_{1}−0.6 = 0 Assuming the initial values x 1 = 0.0 and x 2 = 1.0, the Jacobian matrix is

\left[\begin{array}{cc}10 & −0.8 \ \text{ 0} & −0.6 \ ]\end{array}

\left[\begin{array}{cc}10 & 0 \ \text{ 0} & 10 \ ]\end{array}

\left[\begin{array}{cc}0 & −0.8 \ \text{ 10} & −0.6 \ ]\end{array}

\left[\begin{array}{cc}10 & 0 \ \text{ 10} & −10 \ ]\end{array}

GATE Electrical Engineering (EE) 2011 Solved Paper

Question : 47 of 65

Marks: +1, -0

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}−10x_{2}\cos ⁡x_{1}−0.6 = 0

Assuming the initial values x₁ = 0.0 and x₂ = 1.0, the Jacobian matrix is

${[\table 10 , −0.8 ; \text" 0" , −0.6 ; ]}$
${[\table 10 , 0 ; \text" 0" , 10 ; ]}$
${[\table 0 , −0.8 ; \text" 10" , −0.6 ; ]}$
${[\table 10 , 0 ; \text" 10" , −10 ; ]}$

Validate

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