if u=yz/x, v=zx/y, w=xy/z, show that delta(u,v,w)/delta(x,y,z)=4
Answers
Step-by-step explanation:
\underline{\textsf{Given:}}
Given:
\mathsf{u=\dfrac{yz}{x},\;v=\dfrac{zx}{y},\;w=\dfrac{xy}{z}}u=
x
yz
,v=
y
zx
,w=
z
xy
\underline{\textsf{To prove:}}
To prove:
\mathsf{\dfrac{\delta(u,v,w)}{\delta(x,y,z)}=4}
δ(x,y,z)
δ(u,v,w)
=4
\underline{\textsf{Solution:}}
Solution:
\textsf{We know that,}We know that,
\begin{gathered}\mathsf{\dfrac{\delta(u,v,w)}{\delta(x,y,z)}=|\begin{array}{ccc}\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{array}|}\end{gathered}
δ(x,y,z)
δ(u,v,w)
=∣
∂x
∂u
∂x
∂v
∂x
∂w
∂y
∂u
∂y
∂v
∂y
∂w
∂z
∂u
∂z
∂v
∂z
∂w
∣
\begin{gathered}\mathsf{\dfrac{\delta(u,v,w)}{\delta(x,y,z)}=|\begin{array}{ccc}\frac{-yz}{x^2}&\frac{z}{x}&\frac{y}{x}\\&&\\\frac{z}{y}&\frac{-xz}{y^2}&\frac{x}{y}\\&&\\\frac{y}{z}&\frac{x}{z}&\frac{-xy}{z^2}\end{array}|}\end{gathered}
δ(x,y,z)
δ(u,v,w)
=∣
x
2
−yz
y
z
z
y
x
z
y
2
−xz
z
x
x
y
y
x
z
2
−xy
∣
\textsf{Expanding along first row, we get}Expanding along first row, we get
\mathsf{=\dfrac{-yz}{x^2}(\dfrac{x^2yz}{y^2z^2}-\dfrac{x^2}{yz})-\dfrac{z}{x}(\dfrac{-xyz}{yz^2}-\dfrac{xy}{yz})+\dfrac{y}{x}(\dfrac{xz}{yz}+\dfrac{xyz}{y^2z})}=
x
2
−yz
(
y
2
z
2
x
2
yz
−
yz
x
2
)−
x
z
(
yz
2
−xyz
−
yz
xy
)+
x
y
(
yz
xz
+
y
2
z
xyz
)
\mathsf{=\dfrac{-yz}{x^2}(\dfrac{x^2}{yz}-\dfrac{x^2}{yz})-\dfrac{z}{x}(\dfrac{-x}{z}-\dfrac{x}{z})+\dfrac{y}{x}(\dfrac{x}{y}+\dfrac{x}{y})}=
x
2
−yz
(
yz
x
2
−
yz
x
2
)−
x
z
(
z
−x
−
z
x
)+
x
y
(
y
x
+
y
x
)
\mathsf{=\dfrac{-x^2yz}{x^2yz}+\dfrac{x^2yz}{x^2yz}-\dfrac{z}{x}(\dfrac{-2x}{z})+\dfrac{y}{x}(\dfrac{2x}{y})}=
x
2
yz
−x
2
yz
+
x
2
yz
x
2
yz
−
x
z
(
z
−2x
)+
x
y
(
y
2x
)
\mathsf{=0+2+2}=0+2+2
\mathsf{=4}=4
\implies\boxed{\mathsf{\bf\dfrac{\delta(u,v,w)}{\delta(x,y,z)}=4}}⟹
δ(x,y,z)
δ(u,v,w)
=4