(a) Verify that a 1 b 1 c 1 d 1 a 2 b 2 c 2 d 2 = a 1 a 2 + b 1 c 2 a 1 b 2 + b 1 d 2 c 1 a 2 + d 1 c 2 c 1 b 2 + d 1 d 2 (b) If x = x u , υ , y = y u , υ is a one-to-one transformation, then u = u x , y , υ = υ x , y . Assuming the necessary differentiability, use the result in part (a)and the chain rule to show that ∂ x , y ∂ u , υ ⋅ ∂ u , υ ∂ x , y = 1
(a) Verify that a 1 b 1 c 1 d 1 a 2 b 2 c 2 d 2 = a 1 a 2 + b 1 c 2 a 1 b 2 + b 1 d 2 c 1 a 2 + d 1 c 2 c 1 b 2 + d 1 d 2 (b) If x = x u , υ , y = y u , υ is a one-to-one transformation, then u = u x , y , υ = υ x , y . Assuming the necessary differentiability, use the result in part (a)and the chain rule to show that ∂ x , y ∂ u , υ ⋅ ∂ u , υ ∂ x , y = 1
a
1
b
1
c
1
d
1
a
2
b
2
c
2
d
2
=
a
1
a
2
+
b
1
c
2
a
1
b
2
+
b
1
d
2
c
1
a
2
+
d
1
c
2
c
1
b
2
+
d
1
d
2
(b) If
x
=
x
u
,
υ
,
y
=
y
u
,
υ
is a one-to-one transformation, then
u
=
u
x
,
y
,
υ
=
υ
x
,
y
.
Assuming the necessary differentiability, use the result in part (a)and the chain rule to show that
∂
x
,
y
∂
u
,
υ
⋅
∂
u
,
υ
∂
x
,
y
=
1
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Linear Equation | Solving Linear Equations | What is Linear Equation in one variable ?; Author: Najam Academy;https://www.youtube.com/watch?v=tHm3X_Ta_iE;License: Standard YouTube License, CC-BY