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Chapter 1: Problem 3

Zeigen Sie, daB die folgenden 2-reihigen Matrizen orthogonal sind, Welchen Wert besitzt die jeweilige Determinante? $$ A=\left(\begin{array}{cc} 1 / 2 & -\sqrt{3} / 2 \\ \sqrt{3} / 2 & 1 / 2 \end{array}\right), \quad B=\frac{1}{10}\left(\begin{array}{rr} -8 & 6 \\ 6 & 8 \end{array}\right) $$

Short Answer

Expert verified
Matrices A and B are orthogonal. The determinants are -1/2 and -10 respectively.

Step by step solution

01

Definition of orthogonal matrices

A matrix is orthogonal if its transpose is equal to its inverse. That means, for a matrix \( A \), it should hold that \( A^T A = I \), where \( I \) is the identity matrix.
02

Calculate the transpose of matrix \( A \)

The transpose of matrix \( A \) is obtained by swapping its rows and columns.Given matrix \( A \):\[ A = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{pmatrix} \]The transpose is:\[ A^T = \begin{pmatrix} 1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{pmatrix} \]
03

Confirm orthogonality for matrix \( A \)

Multiply matrix \( A \) by its transpose and check if the product is the identity matrix:\[ A^T A = \begin{pmatrix} 1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{pmatrix} \begin{pmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{pmatrix}= \begin{pmatrix} (1/2)^2+(\sqrt{3}/2)^2 & (1/2)(-\sqrt{3}/2)+(\sqrt{3}/2)(1/2) \ (\sqrt{3}/2)(1/2)+ (1/2)(-\sqrt{3}/2) & (-\sqrt{3}/2)^2 + (1/2)^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = I \]
04

Calculate the transpose of matrix \( B \)

The transpose of matrix \( B \) is obtained similarly.Given matrix \( B \):\[ B = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} \]The transpose is:\[ B^T = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} \]
05

Confirm orthogonality for matrix \( B \)

Multiply matrix \( B \) by its transpose:\[ B^T B = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} = \begin{pmatrix} (-8/10)(-8/10) + (6/10)(6/10) & (-8/10)(6/10) + (6/10)(8/10) \ (6/10)(-8/10) + (8/10)(6/10) & (6/10)(6/10) + (8/10)(8/10) \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = I \]
06

Calculate the determinant of matrix \( A \)

The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by \( \text{det}(A) = ad - bc \).For matrix \( A \):\[ \text{det}(A) = (1/2)(1/2) - (-\sqrt{3}/2)(\sqrt{3}/2) = 1/4 - 3/4 = -1/2 \]
07

Calculate the determinant of matrix \( B \)

For matrix \( B \):\[ \text{det}(B) = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} = \frac{1}{10}((-8)(8) - (6)(6)) = \frac{1}{10}(-64 - 36) = \frac{1}{10}(-100) = -10 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Transpose
When working with matrices, one important operation is the transpose. This means swapping the rows and columns of a matrix. If you start with matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the transpose, denoted \( A^T \), will be \( A^T = \begin{pmatrix} a & c \ b & d \end{pmatrix} \). This simple operation is vital in various mathematical and engineering applications.
Here's an example with a 2x2 matrix:
Given \( A = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{pmatrix} \), its transpose is:
  • Swap the first row and column: [1/2, \sqrt{3}/2] becomes the first column
  • Swap the second row and column: [-\sqrt{3}/2, 1/2] becomes the second column
The resulting transpose is:\( A^T = \begin{pmatrix} 1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{pmatrix} \).
Determinant of a Matrix
The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the formula is simple: \( \text{det}(A) = ad - bc \). This number can tell you a lot about the properties of the matrix. For example:
Given \( A = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{pmatrix} \), the determinant is:
  • \( (1/2)(1/2) \)
  • Subtract \( (-\sqrt{3}/2)(\sqrt{3}/2) \)
The result is \( \text{det}(A) = 1/4 - 3/4 = -1/2 \). The determinant can also indicate if a matrix is invertible or not.
Inverse of a Matrix
Finding the inverse of a matrix is like finding a reciprocal for numbers. For a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]
Before finding the inverse, ensure that the determinant is not zero. Using the same matrix as before:
Given matrix \( A = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \ \sqrt{3}/2 & 1/2 \end{pmatrix} \), and knowing its determinant is -1/2:
  • \( A^{-1} = \frac{1}{-1/2} \begin{pmatrix} 1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{pmatrix} \)
  • This simplifies to \( A^{-1} = -2 \begin{pmatrix} 1/2 & \sqrt{3}/2 \ -\sqrt{3}/2 & 1/2 \end{pmatrix} = \begin{pmatrix} -1 & -\sqrt{3} \ \sqrt{3} & -1 \end{pmatrix} \)
Multiplying a matrix by its inverse gives the identity matrix.
Identity Matrix
The identity matrix is a special type of matrix that acts like the number 1 in matrix multiplication. For a 2x2 matrix, the identity matrix is \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). When you multiply any matrix by the identity matrix, you get the original matrix back:
Example:
\[ A \times I = \begin{pmatrix} a & b \ c & d \end{pmatrix} \times \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & b \ c & d \end{pmatrix} \]
This property is very useful when examining whether a matrix is orthogonal. If \( A^T A = I \), then matrix A is orthogonal.
For instance:
The given matrix \( A \) multiplied by its transpose results in the identity matrix, proving its orthogonality.
2x2 Matrices
2x2 Matrices are the simplest type of square matrices and are very common in linear algebra. Their compact size makes calculations, like finding the determinant and the inverse, much easier:
For example, let's look at matrix \( B = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} \):
  • It's determinant is computed as \( \text{det}(B) = (-8)(8) - (6)(6) = -100 \)
  • Its transpose is \( B^T = \frac{1}{10} \begin{pmatrix} -8 & 6 \ 6 & 8 \end{pmatrix} \), which is identical to B in this case.
  • To find its inverse, divide by the determinant.
2x2 matrices are often used in teaching concepts because of their simplicity while still maintaining a lot of the properties of larger matrices. They serve as a building block for understanding more complex matrix equations.

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Most popular questions from this chapter

Die nachfolgenden Matrizen sind regulär. Berechnen Sie die jeweils zugehörige inverse Matrix nach dem Gau\beta-Jordan-Verfahren. $$ \mathbf{A}=\left(\begin{array}{rrr} 3 & 1 & 4 \\ 1 & 2 & 0 \\ 0 & 1 & -2 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{rrr} 4 & 5 & -1 \\ 2 & 0 & 1 \\ 3 & 1 & 0 \end{array}\right), \quad \mathbf{C}=\left(\begin{array}{lll} 3 & 4 & 2 \\ 1 & 5 & 3 \\ 0 & 1 & 0 \end{array}\right) $$

Zeigen Sie die lineare Unabh?ngigkeit der folgenden Vektoren: a) \(\mathbf{a}_{1}=\left(\begin{array}{l}2 \\ 1 \\ 0\end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{r}-1 \\ 2 \\ 5\end{array}\right), \quad \mathbf{a}_{3}=\left(\begin{array}{r}-1 \\ 2 \\ -1\end{array}\right)\) b) \(\quad \mathbf{a}=\left(\begin{array}{r}-1 \\ 6 \\ 4\end{array}\right), \quad \mathbf{b}=\left(\begin{array}{r}1 \\ -2 \\ -2\end{array}\right), \quad \mathbf{c}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) c) \(a=\left(\begin{array}{l}1 \\ 0 \\ 4\end{array}\right), \quad b=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right)\)

Berechnen Sie den Wert der folgenden 4-reihigen Determinanten mit Hilfe des Laplaceschen Entwicklungssatzes: a) \(\quad\) det \(\mathbf{A}=\left|\begin{array}{rrrr}1 & 0 & 3 & 4 \\ -2 & 1 & 0 & 3 \\ 1 & 4 & 1 & 5 \\ 0 & 2 & 2 & 0\end{array}\right|\) b) \(\quad\) det \(A=\left|\begin{array}{llll}1 & 0 & 5 & 3 \\ 1 & 2 & 2 & 1 \\\ 0 & 1 & 3 & 1 \\ 4 & 0 & 2 & -3\end{array}\right|\)

Berechnen Sie mit m?glichst geringem Rechenaufwand die Determinante der folgenden Matrizen: a) \(\quad \mathbf{A}=\left(\begin{array}{rrr}1 & 0 & 0 \\ 0 & -2 & 0 \\ 4 & 1 & 5\end{array}\right)\) b) \(\quad B=\left(\begin{array}{llll}1 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4\end{array}\right)\)

Zeigen Sie, daB die folgenden Matrizen unitär sind: $$ \begin{aligned} &\mathbf{A}=\frac{1}{\sqrt{12}}\left(\begin{array}{ccc} 2 & 1-\sqrt{3} j & -1-\sqrt{3} j \\ -2 j & \sqrt{3}-j & \sqrt{3}+j \\ -2 & 2 & -2 \end{array}\right) \\ &B=\frac{1}{\sqrt{3}}\left(\begin{array}{cc} 1+j & 1 \\ j & -1-j \end{array}\right), \quad C=\left(\begin{array}{ll} j & 0 \\ 0 & j \end{array}\right) \end{aligned} $$
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