Answer to Question #300532 in Calculus for Aquib

Question

If ux² + y² +2²,v=x+y+z₁w=xy+yz+zx, then the value of the jacobian is d(u,v,w)/d(x,y,z)

Answer 1

Solution:

$u=x² + y² +z²,v=x+y+z,w=xy+yz+zx$

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

$=\begin{vmatrix}2x&2y&2z \\1&1&1 \\y+z&x+z&x+y\end{vmatrix} \\=-2\begin{vmatrix}1&1&1 \\x&y&z \\y+z&x+z&x+y\end{vmatrix} \\=-2[y(x+y)-z(x+z)-x(x+y)+z(y+z)+x(x+z)-y(y+z)] \\=-2[xy+y^2-xz-z^2-x^2-xy+yz+z^2+x^2+xz-y^2-yz] \\=-2(0) \\=0$