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Let αi,βi,andγibe constants for i = 1, 2, and 3. A transformation T:R3→R3 defined by,

role="math" x=α1u+β1v+γ1wy=α2u+β2v+γ2wz=α3u+β3v+γ3w

is called a linear transformation of R3. Prove that this transformation takes a plane ax + by + cz = d in the xyz coordinate system to a plane in the uvw-coordinate system if the Jacobian of the transformation is nonzero.

Short Answer

Expert verified

It is proven that the coefficients of the plane in uvw- system are non zero, and hence represent a plane, if Jacobian is non-zero.

Step by step solution

01

Given information

The equations of transformations are,

x=α1u+β1v+γ1wy=α2u+β2v+γ2wz=α3u+β3v+γ3w

The objective is to determine the transformation of a lineax+by+cz=din xyz- coordinate system to uvw- system.

02

Find the Jacobian

The Jacobian of a transformation using partial derivatives is computed as,

∂(x,y,z)∂(u,v,w)=∂x∂u∂y∂u∂z∂u∂x∂v∂y∂v∂z∂v∂x∂w∂y∂w∂z∂w∂(x,y,z)∂(u,v,w)=α1α2α3β1β2β3γ1γ2γ3

To transform the equation of plane in uvw- system , substitute the equations of transformations in the equation of line.

ax+by+cz=0(aα1+bα2+cα3)u+(aβ1+bβ2+cβ3)v+(aγ1+bγ2+cγ3)w=d

03

Proof

For the purpose of proving the statement, let us assume the converse.

Let us consider that both coefficients are zero.

Write the system of equations in matrix form.

α1α2α3β1β2β3γ1γ2γ3abc=000

The determinant of the matrix is the Jacobian of the transformation.

Hence, the system is consistent if Jacobian is non-zero.

Since the system has only trivial solution. This is not an accepted solution for a given plane.

Hence, the system has not any real solution, if the Jacobian is non-zero.

This means the system of equations does not exist.

Thus, our assumption is proved to be incorrect.

Therefore, the coefficients of the plane in uvw- system are non zero, and hence represent a plane, if Jacobian is non-zero.

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