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Chapter 1: Problem 1

Aufgabe \(1.4\) Man berechne die beiden folgenden Determinanten (vgl. dazu auch Abschnitt 1.6.1): A. \(\Delta=\left|\begin{array}{ccc}1 & 0 & -1 \\ 3 & 1 & -3 \\ 1 & 2 & -2\end{array}\right|\) B. \(\Delta=\left|\begin{array}{rrrr}1 & 2 & -4 & 4 \\ 3 & 6 & 9 & 0 \\ -2 & 8 & 1 & 9 \\ 4 & 2 & -2 & 0\end{array}\right|\).

Short Answer

Expert verified
A: -1, B: -972

Step by step solution

01

Calculate Determinant A

To find the determinant of a 3x3 matrix, we use the formula for determinant calculation: \[\Delta = a(ei-fh) - b(di-fg) + c(dh-eg),\] where \[\begin{array}{ccc}a & b & c \d & e & f \g & h & i \end{array}\] is the matrix.For matrix A: \[\Delta = \left|\begin{array}{ccc}1 & 0 & -1 \ 3 & 1 & -3 \ 1 & 2 & -2\end{array}\right|\]Applying the formula:\[\Delta = 1(1 \cdot -2 - 2 \cdot -3) - 0 + (-1)(3 \cdot 2 - 1 \cdot 1)\]Calculating each part:\(1(-2 + 6) = 1(4) = 4\)\((-1)(6 - 1) = (-1)(5) = -5\)Thus, \[\Delta = 4 - 5 = -1\].
02

Calculate Determinant B using Laplace Expansion

To find the determinant of a 4x4 matrix, we can use cofactor expansion along the first row or any row/column. Here, we use the first row:Matrix B:\[\Delta = \left|\begin{array}{rrrr}1 & 2 & -4 & 4 \ 3 & 6 & 9 & 0 \ -2 & 8 & 1 & 9 \ 4 & 2 & -2 & 0\end{array}\right|\]Using the first row, \(\Delta = 1 \cdot M_{11} - 2 \cdot M_{12} + (-4) \cdot M_{13} + 4 \cdot M_{14}\). Where \(M_{ij}\) is the determinant of the \(3 \times 3\) minor matrix formed by removing the \(i\)-th row and \(j\)-th column.Calculating each minor determinant:1. \(M_{11} = \left|\begin{array}{ccc}6 & 9 & 0 \ 8 & 1 & 9 \ 2 & -2 & 0\end{array}\right| = 6(-2 \cdot 9 - 1 \cdot 0) - 9(8 \cdot 0 - 1 \cdot 2) + 0\) Simplifies to: \(6(-18) - 9(-2) = -108 + 18 = -90\).2. \(M_{12} = \left|\begin{array}{ccc}3 & 9 & 0 \ -2 & 1 & 9 \ 4 & -2 & 0\end{array}\right| = 3(-2 \cdot 0 - 1 \cdot 9) - 9(-2 \cdot 0 - 1 \cdot 4) + 0\) Simplifies to: \(3(-9) + 9(4) = -27 + 36 = 9\).3. \(M_{13} = \left|\begin{array}{ccc}3 & 6 & 0 \ -2 & 8 & 9 \ 4 & 2 & 0\end{array}\right| = 3(2 \cdot 9 - 8 \cdot 0) - 6(4 \cdot 9 - 0) + 0\) Simplifies to: \(3 \cdot 18 - 6 \cdot 36 = 54 - 216 = -162\).4. \(M_{14} = \left|\begin{array}{ccc}3 & 6 & 9 \ -2 & 8 & 1 \ 4 & 2 & -2\end{array}\right| = 3(8 \cdot -2 - 1 \cdot 2) - 6(-2 \cdot -2 - 4 \cdot 1) + 9(-2 \cdot 2 - 4 \cdot 8)\) Simplifies to: \(3(-16 - 2) - 6(4 - 4) + 9(-4 - 32)\) = \(3(-18) - 6(0) + 9(-36)\) = \(-54 - 324\) = \(-378\).Final determinant:\[\Delta = 1(-90) - 2(9) + (-4)(-162) + 4(-378)\]= \(-90 - 18 + 648 - 1512\)= \(-972\).

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Key Concepts

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

Matrix Calculations
Matrix calculations form the backbone of many aspects of linear algebra. These calculations are used to find solutions to systems of equations, among many other applications. In a 3x3 matrix determinant calculation like in Determinant A, we apply a specific formula to simplify our process.

To find the determinant of a 3x3 matrix, we use the formula:
  • Identify the elements of the matrix as: \( \begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array} \).
  • Apply the formula: \[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \].
  • Calculate each term separately and then combine.
These steps reduce what seems like a complex calculation into manageable parts. For larger matrices like a 4x4, using cofactor expansion鈥攚hich we'll discuss shortly鈥攕implifies the process even further.
Cofactor Expansion
Cofactor expansion is a powerful tool for calculating the determinant of larger matrices, like 4x4 matrices. It simplifies the process by breaking down a large determinant into smaller parts that are easier to handle.

The method involves selecting one row or column and expanding it into minors and cofactors. Each minor is a smaller matrix, obtained by removing the row and column of a chosen element. Here鈥檚 a step-by-step breakdown:
  • Select a row or column to expand鈥攗sually the one with the most zeros if possible.
  • For each element in that row or column, calculate the minor determinant, which is the determinant of the 3x3 matrix that remains after removing the elements' row and column.
  • Alternate the signs starting with a positive for the first element. For a first row expansion, use the pattern: positive, negative, positive, negative.
  • Compute the final determinant by summing all the products of the elements and their cofactor expansions.
By using cofactor expansion, you transform a complex determinant problem into a series of simpler calculations.
Linear Algebra
Linear algebra is a field of mathematics centered around vector spaces and linear mappings between these spaces. It involves various numerical tools used to solve linear systems, perform transformations, and handle data representations. Determinants are one of the many tools in linear algebra.

Determinants provide vital information about a matrix:
  • They determine whether a matrix has an inverse (a non-zero determinant indicates an invertible matrix).
  • They give insight into the scaling factor of linear transformations represented by the matrix.
  • The sign of the determinant can indicate orientation (positive for preserving orientation, negative for reversing it).
In solving problems in linear algebra, determinants help verify if a system of equations has a unique solution, infinitely many solutions, or no solution at all. This makes them crucial in understanding and working with linear equations and transformations.

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Most popular questions from this chapter

Man bestimme den Rang der Matrizen (vgl. auch den Absch nitt 1.6.3) A. \(\left(\begin{array}{rrrrr}0 & 3 & 6 & 9 & -3 \\ 0 & 0 & 2 & -4 & 8 \\ 1 & 2 & -3 & 4 & 5 \\ 1 & 5 & 5 & 9 & 10\end{array}\right)\). B. \(\left(\begin{array}{rrrr}1 & -3 & 2 & 0 \\ 4 & -11 & 10 & -1 \\ -2 & 8 & -5 & 3\end{array}\right)\)

A. Sind die folgenden Gleichu ngen nach der Summationskon vention zul盲ssig? 1\. \(A_{i} B_{j j}=C_{i} D_{k k}\), 2\. \(A_{i} B_{i}=C_{i} D_{j}\) 3\. \(A_{i} B_{i}=C_{i} D_{\underline{i}}\), 4\. \(A_{\underline{\underline{i}}} B_{i}=C_{i}\) 5\. \(A_{m} B_{n}=C_{m n}\) 6\. \(A_{i} B_{k}=C_{i} D_{j}\), 7\. \(\quad A_{i} B_{k}=A_{k} B_{i}\) 8\. \(\mu_{m n} A_{m}=c_{n} F_{i i}\) 9\. \(A=\alpha_{\underline{m}} C_{m m}\), 10\. \(A_{i i}=B_{i j} C_{j j i}\). B. Man schreibe die zul盲ssigen Gleichungen f眉r \(N=2\) aus.

Man vereinfache die folgenden Ausd r?cke: A. \(\delta_{i j} \delta_{j k_{1}}\) B. \(\delta_{i 2} \delta_{i k} \delta_{3 k}\) C. \(\delta_{1 k} A_{k_{1}}\) D. \(\delta_{i 2} \delta_{j k} A_{i j}\). 4\. Analog beweist man $$ \lambda_{\underline{m}} A_{i \ldots j m p \ldots q} \delta_{m n}=\lambda_{\underline{n}} A_{i \ldots j n p \ldots q} $$

Man setze ein A. \(u_{i}=A_{i k} n_{k}\) in \(\varphi=u_{k} v_{k_{1}}\) B. \(u_{i}=B_{i j} v_{j}\) und \(C_{i j}=p_{i} q_{j}\) in \(w_{i}=C_{m i} u_{m_{1}}\) C. \(\phi=\tau_{i j} d_{i j}, \quad \tau_{i j}=\eta\left(\frac{\partial v_{i}}{\partial x_{j}}+\frac{\partial v_{j}}{\partial x_{i}}\right), \quad d_{i j}=\frac{1}{2}\left(\frac{\partial v_{i}}{\partial x_{j}}+\frac{\partial v_{j}}{\partial x_{i}}\right)\) und \(q_{i}=-\kappa \frac{\partial T}{\partial x_{i}}(\kappa\) konstant \()\) in \(\rho T \frac{\mathrm{D} s}{\mathrm{D} t}=\phi-\frac{\partial q_{i}}{\partial x_{i}}\) Man multipliziere die Klammern aus.

Man bestimme den Rang der Matrizen (vgl. auch den Absch nitt 1.6.3) A. \(\left(\begin{array}{rrrrr}0 & 3 & 6 & 9 & -3 \\ 0 & 0 & 2 & -4 & 8 \\ 1 & 2 & -3 & 4 & 5 \\ 1 & 5 & 5 & 9 & 10\end{array}\right)\). B. \(\left(\begin{array}{rrrr}1 & -3 & 2 & 0 \\ 4 & -11 & 10 & -1 \\ -2 & 8 & -5 & 3\end{array}\right)\)
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