Problem 28 - Differential Equations and Linear Algebra

Question

An \(n \times n\) matrix \(A\) is called orthogonal if \(A^{T}=A^{-1}\) Show that the given matrices are orthogonal. $$A=\left[\begin{array}{cc} \sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2 \end{array}\right]$$

Step-by-Step Solution

Answer

To show that the given matrix \(A = \left[\begin{array}{cc}\sqrt{3} / 2 & 1 / 2 \\ -1 / 2 & \sqrt{3} / 2\end{array}\right]\) is orthogonal, we need to prove that its transpose \(A^T\) is equal to its inverse \(A^{-1}\). After calculating the transpose and inverse of A, we have: $$A^T = \left[\begin{array}{cc}\sqrt{3} / 2 & -1 / 2 \\ 1 / 2 & \sqrt{3} / 2\end{array}\right]$$ and $$A^{-1} = \left[\begin{array}{cc}\sqrt{3} / 2_ -1 / 2 \\ 1 / 2 & \sqrt{3} / 2\end{array}\right]$$ Since both \(A^T\) and \(A^{-1}\) are equal, we can conclude that the matrix A is orthogonal.

Step 1: Calculate the transpose of the matrix A

First, let's find the transpose of the given matrix A. The transpose of a matrix is obtained by switching its rows with its columns. So, for our given matrix: $$A=\left[\begin{array}{cc} \sqrt{3} / 2 & 1 / 2 \\\ -1 / 2 & \sqrt{3} / 2 \end{array}\right]$$ The transpose of the matrix A, denoted by \(A^T\), is: $$A^T=\left[\begin{array}{cc} \sqrt{3} / 2 & -1 / 2 \\\ 1 / 2 & \sqrt{3} / 2 \end{array}\right]$$

Step 2: Calculate the inverse of the matrix A

Now, let's find the inverse of the given matrix A. For a 2x2 matrix $$\left[\begin{array}{cc} a & b \\\ c & d \end{array}\right]$$, the inverse is given by: $$A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{cc} d & -b \\\ -c & a \end{array}\right]$$ Applying this formula to our matrix A: $$A^{-1} = \frac{1}{(\sqrt{3} / 2)(\sqrt{3} / 2) - (1 / 2)(-1 / 2)} \left[\begin{array}{cc} \sqrt{3} / 2 & -(1 / 2) \\\ 1 / 2 & \sqrt{3} / 2 \end{array}\right]$$ Simplifying the determinant (the scalar part): $$\frac{3/4 - 1/4}{2} = \frac{1}{2}$$ So the inverse matrix becomes: $$A^{-1}=\left[\begin{array}{cc} \sqrt{3} / 2 & -1 / 2 \\\ 1 / 2 & \sqrt{3} / 2 \end{array}\right]$$

Step 3: Compare the transpose and the inverse matrices

Now we have to compare the calculated transpose \(A^T\) and inverse \(A^{-1}\) matrices to see if they are equal. We already have: $$A^T=\left[\begin{array}{cc} \sqrt{3} / 2 & -1 / 2 \\\ 1 / 2 & \sqrt{3} / 2 \end{array}\right]$$ and $$A^{-1}=\left[\begin{array}{cc} \sqrt{3} / 2 & -1 / 2 \\\ 1 / 2 & \sqrt{3} / 2 \end{array}\right]$$ Since the matrices are equal, we can conclude that the given matrix A is orthogonal, as \(A^T=A^{-1}\).

Key Concepts

Matrix Transpose

A matrix transpose is a simple yet important operation in linear algebra. To transpose a matrix, you switch its rows and columns. What was a row becomes a column and vice versa. This operation is denoted by a superscript 'T'.

For example, consider a matrix:

• Matrix: \[A = \begin{bmatrix} \sqrt{3}/2 & 1/2 \-1/2 & \sqrt{3}/2 \end{bmatrix}\]
• Transpose: \[A^T = \begin{bmatrix} \sqrt{3}/2 & -1/2 \1/2 & \sqrt{3}/2 \end{bmatrix}\]

By transposing, you maintain the size of the matrix but change its orientation. This has especially useful consequences when determining if a matrix is orthogonal. An orthogonal matrix must have its transpose equal its inverse.

Matrix Inverse

The inverse of a matrix is akin to the reciprocal of a number. For a square matrix, the inverse matrix, when multiplied by the original matrix, yields an identity matrix. However, not all matrices have inverses.

To find the inverse of a 2x2 matrix:

• Consider matrix \[A = \begin{bmatrix} a & b \c & d \end{bmatrix}\]
• Its inverse is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \-c & a \end{bmatrix}\]

The term \(ad-bc\) is the matrix's determinant. For our matrix used in the exercise, the calculated inverse is:\[A^{-1} = \begin{bmatrix} \sqrt{3}/2 & -1/2 \1/2 & \sqrt{3}/2 \end{bmatrix}\] This inverse matches the transpose, confirming orthogonality.

Determinants

A determinant is a special number that can be calculated from a square matrix. It provides valuable properties about the matrix such as invertibility.

For a 2x2 matrix \[A = \begin{bmatrix} a & b \c & d \end{bmatrix}\], the determinant is calculated as \(ad - bc\).

A matrix will not have an inverse if its determinant equals zero; in other words, it is singular. In contrast, a non-zero determinant yields an invertible matrix, as seen in the given exercise where

• The determinant is:\[\frac{3}{4} - \left(-\frac{1}{4}\right) = \frac{1}{2}\]

This non-zero determinant confirmed that our matrix is invertible. It further corroborates the matrix's orthogonality as it matches its own transpose when inverted.