Zeigen Sie, daB die Zeilen- bzw. Spaltenvektoren der 3-reihigen Matrix $$ A=\left(\begin{array}{ccc} 2 / \sqrt{5} & -1 / \sqrt{30} & -1 / \sqrt{6} \\ 1 / \sqrt{5} & 2 / \sqrt{30} & 2 / \sqrt{6} \\ 0 & 5 / \sqrt{30} & -1 / \sqrt{6} \end{array}\right) $$ ein orthonormiertes Vektorsystem bilden, die Matrix \(\mathbf{A}\) daher orthogonal ist. Bestimmen Sie die inverse Matrix \(\mathbf{A}^{-1}\) sowie die Determinante von \(\mathbf{A}\).

Matrix orthonormality is a fundamental concept in linear algebra. A matrix is orthonormal if its row vectors (or column vectors) are both orthogonal to each other and normalized. First, to check orthogonality, we compute the dot product of every pair of different row vectors or column vectors. If the dot product is zero, they are orthogonal. Secondly, we must ensure each vector has a length (magnitude) of 1, confirming they are normalized. For matrix orthonormality:

• Orthogonality: The vectors are perpendicular, i.e., their dot product is zero.
• Normalization: Each vector's length is one.

When both of these conditions hold true, the matrix is orthonormal, which implies some helpful properties like the matrix being orthogonal.

The dot product, also known as the scalar product, of two vectors measures how much the vectors point in the same direction. It is calculated by multiplying corresponding components of the vectors and then summing those products. For vectors \(\textbf{u} = [u_1, u_2, u_3]\) and \(\textbf{v} = [v_1, v_2, v_3]\), the dot product is: \[ \textbf{u} \bullet \textbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \] The dot product is zero if the vectors are orthogonal (perpendicular). For example, when verifying the orthogonality of matrix \(A\), we computed the dot product of row 1 and row 2 as follows: \[2 / \sqrt{5} \times 1 / \sqrt{5} + (-1 / \sqrt{30}) \times 2 / \sqrt{30} + (-1 / \sqrt{6}) \times 2 / \sqrt{6} = 0 \] This zero result indicates that these rows are orthogonal.

The inverse of a matrix \((A^{-1})\) is a matrix that, when multiplied by the original matrix \((A)\), results in the identity matrix \(I\). In mathematical terms, \[A \times A^{-1} = A^{-1} \times A = I \] For an orthogonal matrix, finding the inverse is straightforward because the inverse of an orthogonal matrix is simply its transpose. This property simplifies calculations and is mathematically expressed as: \[ A^{-1} = A^T \] In the given exercise, since matrix \(A\) is orthogonal, we directly use the transpose of \(A\) to find \(A^{-1}\). So, \[ A^{-1} = \left(\begin{array}{ccc} 2 / \sqrt{5} & 1 / \sqrt{5} & 0 \ -1 / \sqrt{30} & 2 / \sqrt{30} & 5 / \sqrt{30} \ -1 / \sqrt{6} & 2 / \sqrt{6} & -1 / \sqrt{6} \end{array}\right) \]

The determinant of a matrix is a scalar value that gives important insights into the matrix's properties. For example, it indicates whether the matrix is invertible (if the determinant is zero, the matrix is not invertible). To calculate it for a 3x3 matrix \(\textbf{A}\): \[ det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] However, for orthogonal matrices, a special property simplifies this: their determinant is always \(\textbf{\textpm1}\). In the given exercise, since matrix \(\textbf{A}\) is orthogonal, its determinant is calculated directly: \[ det(A) = \left| \begin{array}{ccc} 2 / \sqrt{5} & -1 / \sqrt{30} & -1 / \sqrt{6} \ 1 / \sqrt{5} & 2 / \sqrt{30} & 2 / \sqrt{6} \ 0 & 5 / \sqrt{30} & -1 / \sqrt{6} \end{array} \right| = 1 \] This result reaffirms that \(\textbf{A}\) is an orthogonal matrix.