Question

given u=yzx , v= zxy, w =xyz then the value of d(u,v,w)/d(x,y,z) is (a) 0 (b) 1 (c) 2 (d) 4​

Answer 1

SOLUTION

GIVEN

u = yzx , v = zxy, w = xyz

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{ = \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

(a) 0

(b) 1

(c) 2

(d) 4

EVALUATION

Hence the required Jacobian

$\displaystyle \sf{ = \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

$= \displaystyle \begin{vmatrix} \frac{ \partial u}{ \partial x} & \frac{ \partial u}{ \partial y} & \frac{ \partial u}{ \partial z}\\ \\ \frac{ \partial v}{ \partial x} & \frac{ \partial v}{ \partial y} & \frac{ \partial v}{ \partial z} \\ \\ \frac{ \partial w}{ \partial x} & \frac{ \partial w}{ \partial y} & \frac{ \partial w}{ \partial z} \end{vmatrix}$

$= \displaystyle \begin{vmatrix} yz & zx & xy\\ \\ yz & xz & xy \\ \\ yz & xz & xy \end{vmatrix}$

= 0 ( Each row are identical )

FINAL ANSWER

Hence the correct option is (a) 0

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Answer 2

Solution ⤵️

Given ⤵️

u = yzx , v = zxy, w = xyz

To Find ⤵️

Choose the correct option:-

$\displaystyle \sf{ = \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

(a) 0

(b) 1

(c) 2

(d) 4

Calculation ⤵️

$\therefore\sf {the\: required\: Jacobian}$

$\displaystyle \sf{ = \frac{ \partial (u,v,w)}{ \partial (x,y,z)} }$

$= \displaystyle \begin{vmatrix} \frac{ \partial u}{ \partial x} & \frac{ \partial u}{ \partial y} & \frac{ \partial u}{ \partial z}\\ \\ \frac{ \partial v}{ \partial x} & \frac{ \partial v}{ \partial y} & \frac{ \partial v}{ \partial z} \\ \\ \frac{ \partial w}{ \partial x} & \frac{ \partial w}{ \partial y} & \frac{ \partial w}{ \partial z} \end{vmatrix}$

$= \displaystyle \begin{vmatrix} yz & zx & xy\\ \\ yz & xz & xy \\ \\ yz & xz & xy \end{vmatrix}$

= 0 ( Each row are identical )

$\therefore \small \sf{Your\:Answer:-}$

Therefore the correct option is⤵️

option (a) 0.

$\small{\textbf{\textsf{{\color{navy}\:{Hope}}\:{\purple{it}}\:{\pink{helps}}\:{\color{pink}{you!!♡}}}}}$