Question

Find the inverse of the transformation $$\begin{array}{rl}& x^{\prime }=2x-3y\\ & y^{\prime }=x+y\end{array}$$ that is, find $x,y$ in terms of $x^{\prime },y^{\prime }$. (Hint : Use matrices.) Is the transformation orthogonal?

Short answer

The inverse transformation is $x=\frac{1}{5}{x}^{\prime }+\frac{3}{5}{y}^{\prime }$ $y=-\frac{1}{5}{x}^{\prime }+\frac{2}{5}{y}^{\prime }$. The transformation is not orthogonal.

Step-by-step solution

Write the system of equations in matrix form

Express the given transformation using a matrix representation. The system of equations can be written as: $$\left(\begin{array}{c}{x}^{\prime }{y}^{\prime }\end{array}\right)=\left(\begin{array}{ccc}2& -31& 1\end{array}\right)\left(\begin{array}{c}xy\end{array}\right)$$

Find the inverse of the transformation matrix

To find the inverse transformation, first determine the inverse of the matrix: For a matrix $$A=\left(\begin{array}{ccc}a& bc& d\end{array}\right)\text{, the inverse is given by}{A}^{-1}=\frac{1}{ad-bc}\left(\begin{array}{ccc}d& -b-c& a\end{array}\right)$$ Using this, find the inverse of $$\left(\begin{array}{ccc}2& -31& 1\end{array}\right)$$

Calculate the inverse matrix

First calculate the determinant: $$ad-bc=(2)(1)-(-3)(1)=2+3=5\therefore A^{-1}=\frac{1}{5}\left(\begin{array}{ccc}1& 3-1& 2\end{array}\right)$$ The inverse matrix is: $$A^{-1}=\left(\begin{array}{ccc}\frac{1}{5}& \frac{3}{5}-\frac{1}{5}& \frac{2}{5}\end{array}\right)$$

Express $(x,y)$ in terms of $(x^{\prime },y^{\prime })$

Use the inverse matrix to transform $(x^{\prime },y^{\prime })$ back to $(x,y)$: $$\left(\begin{array}{c}xy\end{array}\right)=A^{-1}\left(\begin{array}{c}{x}^{\prime }{y}^{\prime }\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{5}& \frac{3}{5}-\frac{1}{5}& \frac{2}{5}\end{array}\right)\left(\begin{array}{c}{x}^{\prime }{y}^{\prime }\end{array}\right)$$ This gives: $$x=\frac{1}{5}{x}^{\prime }+\frac{3}{5}{y}^{\prime }$$ $$y=-\frac{1}{5}{x}^{\prime }+\frac{2}{5}{y}^{\prime }$$

Check if the transformation is orthogonal

To check orthogonality, determine if the original transformation matrix is orthogonal; that is if $$A^{-1}=A^T$$. Since $$\left(\begin{array}{ccc}2& -31& 1\end{array}\right)A^{-1}\ne A^T$$},

Conclusion

Thus, the transformation is not orthogonal.

Key concepts

inverse matrix

An inverse matrix is like the 'reverse' of a matrix. When you multiply a matrix by its inverse, you get the identity matrix (which is like the number 1 for matrices). To find the inverse of a 2x2 matrix, you first need to calculate its determinant. For a matrix \ \ A = $$\left(\begin{array}{ccc}a& bc& d\end{array}\right)$$, the determinant is calculated as $ad-bc$. The formula for the inverse is \ \ A^{-1} = \frac{1}{ad - bc} $$\left(\begin{array}{ccc}d& -b-c& a\end{array}\right)$$. So, by finding the determinant and using this formula, we can get the inverse matrix.

linear transformation

A linear transformation is a function that takes a vector and transforms it into another vector. It has to satisfy two main properties: additivity and homogeneity. This means if you have vectors u and v, and a scalar k, then $T(u+v)=T(u)+T(v)$ and $T(ku)=kT(u)$. Imagine moving and scaling shapes on a plane; that's what linear transformations do. They are often represented by matrices. In our example, the equations $x^{\prime }=2x-3y$ and $y^{\prime }=x+y$ form a linear transformation. Using matrices simplifies these operations, allowing us to find solutions and inverses.

orthogonality test

Orthogonality is like checking if transformations keep the properties of a shape (like distances and angles). For a matrix to be orthogonal, its transpose has to be equal to its inverse $A^{-1}=A^T$. This means the rows and columns of the matrix are orthogonal unit vectors. To test this, you can multiply the matrix by its transpose and see if you get the identity matrix. If not, the transformation is not orthogonal. In the given exercise, after finding the transpose and inverse, they do not match, confirming the transformation is not orthogonal.

determinant calculation

The determinant is a special number that can tell us a lot about a matrix. For a 2x2 matrix $A=\left(\begin{array}{ccc}a& bc& d\end{array}\right)$, the determinant is calculated as $ad-bc$. The determinant helps in finding the inverse of a matrix and also tells if a matrix is invertible. If the determinant is zero, the matrix does not have an inverse. In the exercise, the determinant is found to be 5 (since $2\cdot 1-(-3)\cdot 1=5$), which is non-zero, so the inverse exists. This determinant is crucial in finding the inverse matrix.