During the determination of roots of equations x^{2} + 2xy = 6 and x^{2} - y^{2} = 3 using the Newton-Raphson method, the values of Jacobin matrix ‘D’ is found to be ______. If initial approximation is (1.414, 0.517).

This question was previously asked in

MPSC AE CE Mains 2017 Official (Paper 1)

Option 3 : - 12

** Concept**:

**Newton-Raphson Method for Non-linear simultaneous equations: **

Consider the equation f_{1}(x, y) = 0 and f_{2}(x, y) = 0

1. Initial approximation (x_{0}, y_{0}) can be found out by the graphical method

2. The second approximation can be calculated as follows -

Let,

x_{1} = x_{0} + h and y_{1} = y_{0} + k, such that f_{1}(x_{1}, y_{1}) = 0 and f_{2}(x_{2}, y_{2}) = 0

Newton's raphson equation will be given by -

\(f_{1_{0}} + h\frac{\partial{f_1}}{\partial{x}}+k\frac{\partial{f_1}}{\partial{y}}\)

\(f_{2_{0}} + h\frac{\partial{f_2}}{\partial{x}}+k\frac{\partial{f_2}}{\partial{y}}\)

Here,

f_{10}, f_{20} are the value of functions at (x_{0}, y_{0})

The **jacobian matrix** is given by -

\(\left[ D \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}}}{{\partial x}}}&{\frac{{\partial {f_1}}}{{\partial y}}}\\ {\frac{{\partial {f_2}}}{{\partial x}}}&{\frac{{\partial {f_2}}}{{\partial y}}} \end{array}} \right]\) ...(1)

** Calculation**:

**Given, **

f_{1} = x^{2} + 2xy - 6

f_{2} = x^{2} - y^{2} - 3

∴ \(\frac{{\partial {f_1}}}{{\partial x}} \) = 2x + 2y, \(\frac{{\partial {f_1}}}{{\partial y}} \) = 2x

\(\frac{{\partial {f_2}}}{{\partial x}} \) = 2x, \(\frac{{\partial {f_2}}}{{\partial y}} \) = - 2y

From (1)

\(\left[ D \right] = \left[ {\begin{array}{*{20}{c}} {2x+2y}&{2x}\\ {2x}&{-2y} \end{array}} \right]\)

⇒** |D| = - 4xy - 4y ^{2} - 4x^{2} ** ....(2)

Putting values of (x, y) as (1.414, 0.517)

We have,

⇒ |D| = - 12