Find the inverse of the transformation ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{y}$, that is, find x, y in terms of ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}$.

The given expressions are $x^{\text{'}}=2x=3y,y^{\text{'}}=x+y$.

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function f(s) is the piecewise-continuous and exponentially-restricted real function f(t)

The given equations can be represented in matrix form as:

$$\begin{gathered} \left(\frac{x^{\text{'}}}{y\text{'}}\right)=\left(2x-3y \\ x+y\right) \\ =\left(2-3 \\ 11\right)\left(x \\ y\right) \\ \left(x \\ y\right)={\left(2-3 \\ 11\right)}^{-1}\left(x^2 \\ {y}^2\right) \\ =\frac{1}{\left|2-3 \\ 11\right|}\left(13 \\ -12\right)\left(x^{\text{'}} \\ {y}^2\right) \end{gathered}$$

Solve further

$$\begin{gathered} =\frac{1}{5}\left(13 \\ -12\right)\left(x^y \\ y\right) \\ =\left(\frac{1}{5}{x}^{\text{'}}+\frac{3}{5}{y}^{\text{'}} \\ \frac{-1}{5}{x}^{\text{'}}+\frac{2}{5}{y}^{\text{'}}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} x=\frac{1}{5}{x}^{\text{'}}+\frac{3}{5}{y}^{\text{'}} \\ y=\frac{-1}{4}{x}^{\text{'}}+\frac{2}{5}{y}^{\text{'}} \end{gathered}$$

The transformation is not orthogonal since the determinant of the transformation matrix is not equal to $\pm 1$, and

It's obvious that the inverse of the matrix $$\begin{gathered} \left(2-3 \\ 11\right) \end{gathered}$$is not equal to its transpose.