# During the determination of roots of equations x2 + 2xy = 6 and x2 - y2 = 3 using the Newton-Raphson method, the values of Jacobin matrix ‘D’ is found to be ______. If initial approximation is (1.414, 0.517).

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1. - 4
2. - 8
3. - 12
4. + 4

Option 3 : - 12

## Detailed Solution

Concept:

Newton-Raphson Method for Non-linear simultaneous equations:

Consider the equation f1(x, y) = 0 and f2(x, y) = 0

1. Initial approximation (x0, y0) can be found out by the graphical method

2. The second approximation can be calculated as follows -

Let,

x1 = x0 + h and y1 = y0 + k, such that f1(x1, y1) = 0 and f2(x2, y2) = 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,

f10, f20 are the value of functions at (x0, y0)

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,

f1 = x2 + 2xy - 6

f2 = x2 - y2 - 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 - 4y2 - 4x2      ....(2)

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

We have,

⇒ |D| = - 12
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