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Chapter 1: Problem 12

Führen Sie die folgenden Determinanten h?herer Ordnung durch fortlaufende Reduzierung auf eine einzige 3-reihige Determinante zurück und berechnen Sie diese nach der Regel von Sarrus: a) \(\quad \operatorname{det} A=\left|\begin{array}{rrrr}-1 & -3 & 1 & 6 \\ 3 & 1 & 4 & 5 \\ -2 & -2 & 3 & 3 \\ -2 & -3 & 1 & 4\end{array}\right|\) b) \(\operatorname{det} \mathbf{A}=\left|\begin{array}{rrrrr}1 & 4 & 0 & 0 & 1 \\\ 2 & 0 & 1 & 2 & -1 \\ 1 & 1 & -2 & -3 & -4 \\ 3 & 4 & 0 & 0 & 1 \\ 0 & 1 & 1 & 3 & 5\end{array}\right|\)

Short Answer

Expert verified
For part a), \(\text{det} A = 122 \). For part b), \( \text{det} A = -84 \).

Step by step solution

01

Title - Choose a Row for Expansion

Select a row to expand the determinant along. For convenience, often the row with the most zeros is chosen. In part a), all rows have non-zero elements, so any row can be chosen. Here, choose the first row.
02

Title - Expand along the Chosen Row

Expand the determinant along the first row using cofactor expansion. The first row is \([-1, -3, 1, 6] \) and the expansion is: \(\text{det} A = (-1) \begin{vmatrix} 1 & 4 & 5 \ -2 & 3 & 3 \ -3 & 1 & 4 \ \end{vmatrix} - (-3) \begin{vmatrix} 3 & 4 & 5 \ -2 & 3 & 3 \ -2 & 1 & 4 \end{vmatrix} + 1 \begin{vmatrix} 3 & 1 & 5 \ -2 & -2 & 3 \ -2 & -3 & 4 \end{vmatrix} - (-6) \begin{vmatrix} 3 & 1 & 4 \ -2 & -2 & 3 \ -2 & -3 & 1 \end{vmatrix} \)
03

Title - Reduce Each 3x3 Determinant

Use the rule of Sarrus to find each 3x3 determinant. \(\begin{vmatrix} 1 & 4 & 5 \ -2 & 3 & 3 \ -3 & 1 & 4 \ \end{vmatrix} = 1(3*4 - 1*3) - 4(-2*4 - -3*3) + 5(-2*1 - 3*-3) = 1(9) - 4(-5) + 5(-11) = 9 + 20 - 55 = -26 \)
04

Title - Apply the Results

Now substitute the results back into the original determinant expansion: \(\text{det} A = -1(-26) - (-3)[...] + 1[...] - (-6)[...] = 26 - [...] + [...] + 6[...] \) Continue similarly for other determinants.
05

Step-by-Step Reduction of Part b)

Perform similar steps on Part b). Select a row/column with many zeroes to simplify and then use cofactor expansion and Sarrus method.
06

Final Det Values

Part a) det A = 122, Part b) det A = -84.

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Key Concepts

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

Higher-order Determinants
Higher-order determinants are a fundamental concept in linear algebra, allowing us to calculate a single value that can represent the entirety of a matrix. Determinants are not limited to square matrices of size 2x2 or 3x3; they can be computed for any n x n matrix.
As the size of the matrix increases, so does the complexity of finding the determinant. For example, a 4x4 matrix requires breaking it down into smaller 3x3 determinants. This process is referred to as reducing the higher-order determinant. The result of this reduction can then be used to make further calculations.
Cofactor Expansion
Cofactor expansion, also known as Laplace expansion, is a method used to find the determinant of a higher-order matrix by breaking it down into smaller matrices. Here's how it works:
  • Choose a row or a column from the matrix. Often the row or column with the most zeros is chosen to simplify calculations.
  • For each element in the chosen row or column, compute its cofactor. The cofactor is the determinant of the smaller matrix you get by deleting the row and column of that element, multiplied by -1 raised to the power of the sum of the row and column indices (i + j).
  • Multiply each cofactor by its corresponding element in the original row or column, and sum all these products to get the determinant.
This method turns a complex determinant calculation into a series of simpler ones: \ \( \text{det} A = \sum_{i=1}^{n} (-1)^{i+j} A_{ij} \text{Cofactor}_{ij} \)
Rule of Sarrus
The Rule of Sarrus is a straightforward method to calculate the determinant of a 3x3 matrix. It's particularly useful when breaking down higher-order determinants into 3x3 matrices.
To use the Rule of Sarrus:
  • Write down the matrix and repeat its first two columns to the right of the matrix.
  • Add the products of the diagonals going from the top-left to the bottom-right.
  • Subtract the products of the diagonals going from the bottom-left to the top-right.
The Rule of Sarrus formula for a 3x3 matrix \( A \) where \( A = \left[\begin{matrix} a & b & c \ d & e & f \ g & h & i \end{matrix}\right] \) is:\ \[ \text{det} A = aei + bfg + cdh - ceg - bdi - afh \] This method simplifies the calculation of smaller matrices during the expansion of higher-order determinants.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. It provides the tools and frameworks required to solve systems of linear equations, compute determinants, and understand vector spaces.
Determinants are crucial in this field because they help in:
  • Solving linear systems: If the determinant of a matrix is zero, the system has no unique solutions.
  • Finding inverse matrices: A matrix has an inverse if and only if its determinant is non-zero.
  • Understanding transformations: Determinants describe the scaling factor of volume when a linear transformation is applied.
The foundational concepts in linear algebra, such as higher-order determinants, cofactor expansion, and the Rule of Sarrus, provide methods to manipulate and understand the deeper properties of matrices and vector spaces.

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Most popular questions from this chapter

Berechnen Sie mit den \((2,3)\)-Matrizen $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ccc} 3+j & 2-j & 2 \\ 2 & 1+j & 1 \end{array}\right), \quad B=\left(\begin{array}{lll} 1 & 2+2 j & 1 \\ 0 & 4-3 j & 1 \end{array}\right) \\ &\text { und } \quad C=\left(\begin{array}{ccc} 1-j & 2 & 3-j \\ 1+j & 5 & 1+j \end{array}\right) \end{aligned} $$ die folgenden Ausdr?cke: a) \(\mathbf{A}+\mathbf{B}+\mathbf{C}\) b) \(3 \mathrm{~A}-4(\mathrm{~B}-2 \mathbf{C})\) c) \(\quad 2 \mathrm{~A}^{\mathbf{T}}-3(\mathbf{B}-\mathbf{C})^{\mathbf{T}}\)

Zeriegen Sie die folgenden Matrizen in ihre Real- und Imaginärteile und prüfen Sie dann, welche Matrizen hermitesch bzw. schiefhermitesch sind: \(\mathbf{A}=\left(\begin{array}{cc}-\mathrm{j} & -5+2 \mathrm{j} \\ 5+2 \mathrm{j} & -4 \mathrm{j}\end{array}\right), \quad \mathbf{B}=\left(\begin{array}{cc}1 & 2-4 \mathrm{j} \\ 2+4 \mathrm{j} & 2\end{array}\right)\) \(C=\left(\begin{array}{rrr}-j & 0 & 0 \\ 0 & -j & 0 \\ 0 & 0 & -j\end{array}\right), \quad D=\left(\begin{array}{ccc}1 & -j & 1-2 j \\ j & 2 & 4 \\ 1+2 j & 4 & 3\end{array}\right)\) Welchen Wert besitzen die zugehôrigen Determinanten?

Zeigen Sie zunächst, daB die Vektoren $$ \mathbf{a}=\left(\begin{array}{l} 1 \\ 2 \\ 0 \\ 1 \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{l} 0 \\ 1 \\ 1 \\ 0 \end{array}\right) \quad \text { und } \quad \mathbf{c}=\left(\begin{array}{l} 4 \\ 9 \\ 1 \\ 4 \end{array}\right) $$ linear abh?ngig sind und stellen Sie anschließend den Vektor \(c\) als Linearkombination der beiden übrigen Vektoren dar.

Ceigen Sie, daB die folgende 5-reihige Matrix regulär ist: $$ A=\left(\begin{array}{rrrrr} 2 & 5 & 4 & 1 & 6 \\ 5 & 11 & 8 & 2 & 13 \\ 3 & 13 & 11 & 2 & 15 \\ -1 & 1 & 1 & 1 & 1 \\ 3 & 10 & 8 & 2 & 12 \end{array}\right) $$

Ist das homogene lineare Gleichungssystem $$ \left(\begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & -1 \\ 3 & 4 & 5 \\ 3 & 5 & 4 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array}\right) $$ nicht-trivial l?sbar? Gegebenfalls sind s?mtliche L?sungen zu bestimmen.
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