Question

If x + y +z = u, y + z = v, z = uvw. Then the Jacobian
a (x,y,z)
is
a(u,v,w)



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Answer 1

Answer:

UV

Step-by-step explanation:

It's a lengthy explanation.

Answer 2

The value of Jacobian is $u^{2}v$

Step-by-step explanation:

Given,

=> x + y + z = u

y + z = uv

z = uvw

=> Converting the given equation in the terms of x ,y and z

* x = u - (y + z)

x = u - uv

x = u(1 - v)

* y + z = uv

y = uv - z

y =  uv - uvw

z = uvw

=> J = $\frac{d(x,y,z)}{d(u,v,w)}$

We know that , $J = \left[\begin{array}{ccc}\frac{dx}{du} &\frac{dx}{dv} &\frac{dx}{dw} \\\frac{dy}{du} &\frac{dy}{dv} &\frac{dy}{dw} \\\frac{dz}{du} &\frac{dz}{dv} &\frac{dz}{dw} \end{array}\right]$

Solving and substituting the required values in the above matrix ,

$J = \left[\begin{array}{ccc}\frac{d}{du}(u-uv) &\frac{d}{dv}(u-uv) &\frac{d}{dw}(u-uv) \\\frac{d}{du}(uv-uvw) &\frac{d}{dv}(uv-uvw) &\frac{d}{dw}(uv-uvw) \\\frac{d}{du}(uvw) &\frac{d}{dv}(uvw) &\frac{d}{dw}(uvw) \end{array}\right]$

$J = \left[\begin{array}{ccc}1-v&-u&0\\v-vw&u-uw&-uv\\vw&uw&uv\end{array}\right]$

$= > R_{2} - > R_{2} +R_{3}$

$J = \left[\begin{array}{ccc}1-v&-u&0\\v&u&0\\vw&uw&uv\end{array}\right]$

$= > R_{1} - > R_{1} +R_{2}$

$J = \left[\begin{array}{ccc}1&0&0\\v&u&0\\vw&uw&uv\end{array}\right]$

=> Solving the determinant,

1(u(uv) - 0)

$u^{2}v$

Hence the value of Jacobian is $u^{2}v$.