Question

Let α, β, γ , and δ be constants. A transformation$T:R^2\Rightarrow R^2$where $x=\alpha u+\beta v$and $y=\gamma u+\delta v$, is called a linear transformation of $R^2$. Use this transformation to answer Exercises 53–55.

Assuming that the Jacobian is nonzero, find expressions for u and v as functions of x and y.

Short answer

$u=\left(\frac{\delta x-\beta y}{\delta \alpha -\beta \gamma }\right),v=\left(\frac{\gamma x-\alpha y}{\gamma \beta +\alpha \delta }\right)$

Step-by-step solution

Given information

The equations of transformations are

$x=\alpha u+\beta v;y=\gamma u+\delta v$

The objective is to find the expression of u and v in terms of x and y.

Find the expression for v

Determine the expression for v by eliminating u.

$$\begin{gathered} \gamma x-\alpha y=\gamma (\alpha u+\beta v)-\alpha (\gamma u+\delta v) \\ \gamma x-\alpha y=\gamma \alpha u+\gamma \beta v-\alpha \gamma u-\alpha \delta v \\ \gamma x-\alpha y=\gamma \beta v+\alpha \delta v \\ \gamma x-\alpha y=\left(\gamma \beta +\alpha \delta \right)v \\ v=\frac{\gamma x-\alpha y}{\gamma \beta +\alpha \delta } \end{gathered}$$

Find the expression for u

Determine the expression for u by eliminating v.

$$\begin{gathered} \delta x-\beta y=\delta (\alpha u+\beta v)-\beta (\gamma u+\delta v) \\ \delta x-\beta y=\delta \alpha u+\delta \beta v-\beta \gamma u-\beta \delta v \\ \delta x-\beta y=\delta \alpha u-\beta \gamma u \\ \delta x-\beta y=\left(\delta \alpha -\beta \gamma \right)u \\ u=\frac{\delta x-\beta y}{\delta \alpha -\beta \gamma } \end{gathered}$$