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Chapter 1: Problem 4

Begründen Sie (ohne Rechnung), warum die nachstehenden Determinanten verschwinden: a) \(\operatorname{det} A=\left|\begin{array}{rrr}1 & -2 & 3 \\ -4 & 8 & 0 \\\ 0,5 & -1 & 3\end{array}\right|\) b) \(\quad \operatorname{det} \mathbf{B}=\left|\begin{array}{rrr}1 & 0 & -2 \\\ 5 & 0 & 3 \\ 0 & 0 & 4\end{array}\right|\) c) \(\quad\) det \(C=\left|\begin{array}{llrl}1 & 4 & -3 & 6 \\ 0 & 2 & 3 & 8 \\\ 1 & 4 & -3 & 6 \\ 0 & 1 & 1 & 1\end{array}\right|\) d) \(\operatorname{det} \mathbf{D}=\left|\begin{array}{rrrr}6 & -3 & -15 & 24 \\\ 3 & 5 & -7 & 0 \\ -2 & 1 & 5 & -8 \\ 1 & 1 & 0 & 0\end{array}\right|\)

Short Answer

Expert verified
A has proportional rows. B has a column of zeros. C has identical rows. D's rows sum to zero.

Step by step solution

01

- Identifying Determinants That Disappear

For each matrix, examine its structure for properties that cause the determinant to be zero.
02

- Matrix A

Matrix A has proportional rows: The second row is -4 times the first row, which means the rows are linearly dependent. A matrix with linearly dependent rows has a determinant of zero.
03

- Matrix B

Matrix B has a row of zeros (second column is all zeros). This implies that there is no change in the volume represented by the determinant, causing it to be zero.
04

- Matrix C

Matrix C has two identical rows (first and third rows are the same). Since having identical rows means the rows are linearly dependent, the determinant is zero.
05

- Matrix D

Matrix D does not have any immediately apparent linearly dependent rows. However, further examination reveals that the sum of the first three rows equals zero, implying linear dependence, hence the determinant is zero.

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Key Concepts

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

linear dependence
Linear dependence in matrices is a fundamental concept in linear algebra. It occurs when one row or column of a matrix can be represented as a linear combination of other rows or columns. If any rows or columns in a matrix are linearly dependent, the determinant of that matrix is zero. For example, in Matrix A, the second row is exactly -4 times the first row. This makes the rows linearly dependent, ensuring that the determinant is zero. Recognizing linear dependence quickly allows for the assessment of matrix properties without calculation.
matrix properties
Several intrinsic properties of matrices can lead to useful insights, especially about their determinants. Knowing these properties helps in efficiently solving matrix-related problems. Key matrix properties include:
  • Proportional rows or columns, leading to linear dependence
  • Identical rows or columns, also leading to linear dependence
  • Presence of a row or column of zeros, which reduces the determinant to zero
  • Sum of rows or columns equaling zero, indicating linear dependence
Understanding these properties helps in quickly identifying matrices with a zero determinant, like in Examples A, B, C, and D provided in the solutions.
zero determinant
A zero determinant indicates that a matrix does not have an inverse and that its rows or columns are linearly dependent. This means the transformation represented by the matrix does not change the space's dimension, leading to volume collapse to zero. In our examples:
  • Matrix A: Second row is -4 times the first row
  • Matrix B: Second column consists entirely of zeros
  • Matrix C: First and third rows are identical
  • Matrix D: Sum of the first three rows is zero
In each case, the matrices have properties causing linear dependence, so their determinants are zero.
proportional rows
Proportional rows in a matrix signify linear dependence. When one row can be expressed as a scalar multiple of another, the matrix's determinant is zero. In Matrix A from our example, the second row is -4 times the first row. This proportionality indicates linear dependence. Because the rows depend on each other, the matrix cannot span its multidimensional space effectively, resulting in a zero determinant. Identifying such proportional rows is a quick way to diagnose determinant properties without performing detailed calculations.
identical rows
Identical rows in a matrix also lead to linear dependence and thus a zero determinant. If two rows are the same, one can be seen as a linear combination of the other with a scalar multiplier of one. Consider Matrix C in our example. The first and third rows are identical. This redundancy in the information conveyed by the matrix means it does not change the space it represents significantly, causing the determinant to be zero. Recognizing identical rows helps in quickly concluding that a matrix is singular and has a zero determinant.

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Most popular questions from this chapter

Zeigen Sie, daB die drei r?umlichen Vektoren $$ \mathbf{a}=\left(\begin{array}{l} 1 \\ 4 \\ 1 \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{r} 2 \\ -1 \\ 2 \end{array}\right) \quad \text { und } \quad c=\left(\begin{array}{r} 10 \\ 4 \\ 10 \end{array}\right) $$ komplanar sind, d.h. in einer gemeinsamen Ebene liegen.

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 \(\mathbf{A}\).

Gegeben sind die Matrizen \(\mathbf{A}=\left(\begin{array}{lll}2 & 4 & 1 \\ 1 & 3 & 5\end{array}\right), \quad \mathbf{B}=\left(\begin{array}{rrr}3 & 1 & 1 \\ 0 & 2 & 1 \\ -1 & 5 & 1\end{array}\right)\) und \(C=\left(\begin{array}{rrr}-2 & 0 & 3 \\ 2 & 5 & 1 \\\ -1 & 1 & 1\end{array}\right)\) Berechnen Sie (falls möglich): a) \((\mathbf{A} \cdot \mathbf{B}) \cdot \mathbf{C}\) b) \(\mathrm{A} \cdot(\mathrm{B} \cdot \mathrm{C})\) c) \(\quad A \cdot(B+C)^{T}\) d) \((\mathrm{A} \cdot \mathbf{B})^{\boldsymbol{T}}\)

F?r welchen Wert des Parameters \(\lambda\) sind die Vektoren $$ \mathbf{a}_{1}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad \mathbf{a}_{2}=\left(\begin{array}{r} -\lambda \\ 2 \\ 5 \end{array}\right) \text { und } \mathbf{a}_{3}=\left(\begin{array}{r} -2 \\ \lambda \\ 7 \end{array}\right) $$ linear abh?ngig?

Welche der folgenden 3-rcihigen Matrizen sind orthogonal? $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{rrr} 1 & 0 & 1 \\ 2 & -2 & 0 \\ 0 & -1 & 3 \end{array}\right), \quad \mathrm{B}=\frac{1}{3}\left(\begin{array}{rrr} 2 & 2 & 1 \\ 1 & -2 & 2 \\ 2 & -1 & -2 \end{array}\right) \\ &C=\left(\begin{array}{ccc} 0 & 1 / \sqrt{2} & -1 / \sqrt{2} \\ 0 & 1 / \sqrt{2} & 1 / \sqrt{2} \\ 1 & 0 & 0 \end{array}\right) \end{aligned} $$
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