Zeigen Sie, daB die Matrix \(A=\left(\begin{array}{rr}1 / \sqrt{2} & 1 / \sqrt{2} \\ -1 / \sqrt{2} & 1 / \sqrt{2}\end{array}\right)\) die fur cine orthogonale Matrix hinreichende Bedingung \(\mathbf{A}^{\mathbf{T}}=\mathbf{A}^{-1}\) erfullt.

To understand the notion of a matrix transpose, let’s start with the basics. A matrix transpose is simply a new matrix obtained by swapping the rows and columns of the original matrix. For any matrix \( A \), the transpose \( A^T \) is formed by interchanging rows with columns.

For example, consider the matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \). The transpose of this matrix, \( A^T \), would be \( \begin{pmatrix} a & c \ b & d \end{pmatrix} \).
Transposing a matrix is useful in various calculations, such as solving systems of linear equations and eigenvalue problems.

In our exercise, matrix \( A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \), its transpose is calculated by switching the rows and columns, resulting in \( A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \).

Next, let’s discuss the concept of a matrix inverse. The inverse of a matrix \( A \), denoted as \( A^{-1} \), is a matrix such that when multiplied by \( A \) results in the identity matrix. In mathematical terms, \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.

Not every matrix has an inverse. A matrix must be square (same number of rows and columns) and its determinant must not be zero. For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is calculated as:
\( A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \), where \( ad - bc \) is the determinant.

In the given problem, to find the inverse of matrix \( A \), we first calculate its determinant: \[ ad - bc = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - \left( \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} = 1 \].
Since the determinant is 1, we can safely proceed to find \( A^{-1} \): \( \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \), which matches the transpose in this particular case.

Understanding how to calculate the determinant is crucial in linear algebra. The determinant is a scalar value that can be computed from the elements of a square matrix. It provides important properties of the matrix, such as whether a matrix is invertible, and is denoted as \( det(A) \) or \( |A| \).

For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is given by \( ad - bc \). This formula may seem simple, but the concept is powerful and extends to larger matrices in more complex ways.

In our problem, the matrix \( A \) is \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \) with elements: \( a = \frac{1}{\sqrt{2}} \), \( b = \frac{1}{\sqrt{2}} \), \( c = -\frac{1}{\sqrt{2}} \), and \( d = \frac{1}{\sqrt{2}} \).
We calculate the determinant as follows:
\[ det(A) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} - ( \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt{2}} ) = \frac{1}{2} + \frac{1}{2} = 1 \].
Since the determinant is not zero, the matrix \( A \) is invertible.

Linear algebra is a branch of mathematics focused on vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.

Some of the basic operations in linear algebra involve matrices and determinants, eigenvalues, and eigenvectors. Understanding how to manipulate and work with matrices is fundamental in linear algebra because they can represent linear transformations and systems of linear equations.
Let’s connect this with the given exercise. In linear algebra, an orthogonal matrix is a matrix whose transpose is also its inverse: \( A^T = A^{-1} \). This property is quite valuable, as orthogonal matrices preserve length and orthogonality in transformations.
In our example, we verified that the transpose of a given matrix \( A \) is indeed equal to its inverse, confirming that it is orthogonal. This demonstrates the concepts of transposition and inversion in the context of linear algebra and highlights the importance of determinant calculation as well.