/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 (a) \((\mathbf{A B})^{T}=\left(\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 23

(a) \((\mathbf{A B})^{T}=\left(\begin{array}{rr}7 & 10 \\ 38 & 75\end{array}\right)^{T}=\left(\begin{array}{rr}7 & 38 \\ 10 & 75\end{array}\right)\) (b) \(\mathbf{B}^{T} \mathbf{A}^{T}=\left(\begin{array}{rr}5 & -2 \\ 10 & -5\end{array}\right)\left(\begin{array}{rr}3 & 8 \\ 4 & 1\end{array}\right)=\left(\begin{array}{rr}7 & 38 \\ 10 & 75\end{array}\right)\)

Short Answer

Expert verified
Yes, both parts confirm the property: \((\mathbf{A B})^T = \mathbf{B}^T \mathbf{A}^T\).

Step by step solution

01

Understand the given matrix operation

We are given two parts to the problem related to matrix transposes and multiplications. The task is to verify and understand the equivalence mentioned in both parts of the problem.
02

Solve part (a)

The problem states that \[(\mathbf{A B})^{T}=\left(\begin{array}{rr}7 & 10 \ 38 & 75\end{array}\right)^{T}=\left(\begin{array}{rr}7 & 38 \ 10 & 75\end{array}\right)\]This means that the transpose of the product of matrices \(\mathbf{A}\) and \(\mathbf{B}\) equals a specified matrix. Here, the transpose operation applied on \(\mathbf{A B}\) swaps the rows and columns of the original resultant matrix \(\mathbf{A B}\).
03

Verify the transpose property

Recall that for matrices \(\mathbf{A}\) and \(\mathbf{B}\), \[(\mathbf{A B})^T = \mathbf{B}^T \mathbf{A}^T\]The matrices shown in part (a) and (b) ensure that both expressions, \((\mathbf{A B})^T\) and \(\mathbf{B}^T \mathbf{A}^T\), are equal, consistently showing the transpose property at work.
04

Solve part (b)

In part (b), we need to confirm that \[\mathbf{B}^{T} \mathbf{A}^{T}=\left(\begin{array}{rr}5 & -2 \ 10 & -5\end{array}\right)\left(\begin{array}{rr}3 & 8 \ 4 & 1\end{array}\right)=\left(\begin{array}{rr}7 & 38 \ 10 & 75\end{array}\right)\]By performing matrix multiplication between \(\mathbf{B}^T\) and \(\mathbf{A}^T\), calculate each element:- First row, first column: \((5 \cdot 3) + (-2 \cdot 4) = 15 - 8 = 7\)- First row, second column: \((5 \cdot 8) + (-2 \cdot 1) = 40 - 2 = 38\)- Second row, first column: \((10 \cdot 3) + (-5 \cdot 4) = 30 - 20 = 10\)- Second row, second column: \((10 \cdot 8) + (-5 \cdot 1) = 80 - 5 = 75\)This confirms that the result is \(\left(\begin{array}{rr}7 & 38 \ 10 & 75\end{array}\right)\), thereby verifying the transpose product and confirming it matches part (a).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra. It involves taking two matrices and producing a third matrix, called the product. To multiply two matrices, \( \mathbf{A} \) and \( \mathbf{B} \), their dimensions must align. Specifically, the number of columns in the first matrix must equal the number of rows in the second. This results in a new matrix where each element is computed as the dot product of a row in \( \mathbf{A} \) and a column in \( \mathbf{B} \).
  • To form the element at row \( i \) and column \( j \) of the product matrix, multiply the corresponding elements of row \( i \) from \( \mathbf{A} \) with column \( j \) from \( \mathbf{B} \), and then sum up the results.
  • This operation can be expressed as: \( C_{ij} = \sum_{k} A_{ik} B_{kj} \).
  • Matrix multiplication is not commutative, meaning \( \mathbf{A} \mathbf{B} \) does not necessarily equal \( \mathbf{B} \mathbf{A} \).
The step-by-step method of matrix multiplication helps visualize how each element contributes to the final product. This technique is crucial for understanding more complex matrix operations.
Transpose Property
The transpose of a matrix involves swapping its rows and columns, effectively reflecting the matrix over its main diagonal. The transpose of a matrix \( \mathbf{A} \) is denoted as \( \mathbf{A}^{T} \). This operation is particularly useful when verifying properties of matrix products.
  • A critical property of transposes is: \( ( \mathbf{A} \mathbf{B})^{T} = \mathbf{B}^{T} \mathbf{A}^{T} \).
  • This means the transpose of a product of two matrices equals the product of their transposes, but in the reverse order.
  • Transposing does not change the eigenvalues of a matrix, though the nature and implication of eigenvectors might change.
This property is often used to simplify expressions and proofs in linear algebra, verifying equivalence of different matrix operations as shown in the original exercise.
Matrix Operations
Matrix operations encompass a variety of functions including addition, subtraction, multiplication, and transposition.
  • Addition/Subtraction: For matrices to be added or subtracted, they must have the same dimensions. Each element is then added or subtracted component-wise.
  • Multiplication: As previously explained, requires careful alignment of rows and columns, resulting in a new matrix with a different potential dimension than either operand.
  • Transpose: Flipping a matrix over its diagonal, switching its row and column indices.
Matrix operations often serve as the foundation for applying more complex computational methods in mathematics and engineering. The rules governing these operations ensure consistency and reliability across various calculations, reflecting the interconnected principles of linear algebra.
Linear Algebra
Linear algebra is an area of mathematics focused on vector spaces and linear mappings between these spaces. Matrices play a key role in this field as they represent linear transformations.
  • Vector spaces consist of vectors and allow operations such as addition and scalar multiplication, following specific axioms like distributiveness and associativity.
  • Matrices can represent these linear transformations, making it easy to shift, rotate, or scale vectors in a space.
  • Linear algebra underpins many computational algorithms like those used in computer graphics and data sciences.
Understanding linear algebra concepts enables the solving of systems of equations, performing transformations, and analyzing multidimensional datasets. Its applications extend to diverse areas such as engineering, physics, computer science, and beyond.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\left(\begin{array}{r}-7 \\ 17 \\\ -6\end{array}\right)+\left(\begin{array}{r}-1 \\ 1 \\\ 4\end{array}\right)-\left(\begin{array}{r}2 \\ 8 \\\ -6\end{array}\right)=\left(\begin{array}{r}-10 \\ 10 \\ 4\end{array}\right)\)

\(\mathbf{A}^{T}=\left(\begin{array}{rr}2 & -3 \\ 4 & 2\end{array}\right) ; \quad\left(\mathbf{A}^{T}\right)^{T}=\left(\begin{array}{rr}2 & 4 \\ -3 & 2\end{array}\right)=\mathbf{A}\)

The singular points of \(\left(x^{2}-25\right) y^{\prime \prime}+2 x y^{\prime}+y=0\) are -5 and \(5 .\) The distance from 0 to either of these points is \(5 .\) The distance from 1 to the closest of these points is 4.

Substituting \(y=\sum_{n=0}^{\infty} c_{n} x^{n}\) into the differential equation we have $$\begin{aligned} y^{\prime \prime}+(\sin x) y &=\sum_{n=2}^{\infty} n(n-1) c_{n} x^{n-2}+\left(x-\frac{1}{6} x^{3}+\frac{1}{120} x^{5}-\cdots\right)\left(c_{0}+c_{1} x+c_{2} x^{2}+\cdots\right) \\ &=\left[2 c_{2}+6 c_{3} x+12 c_{4} x^{2}+20 c_{5} x^{3}+\cdots\right]+\left[c_{0} x+c_{1} x^{2}+\left(c_{2}-\frac{1}{6} c_{0}\right) x^{3}+\cdots\right] \\ &=2 c_{2}+\left(6 c_{3}+c_{0}\right) x+\left(12 c_{4}+c_{1}\right) x^{2}+\left(20 c_{5}+c_{2}-\frac{1}{6} c_{0}\right) x^{3}+\cdots=0 \end{aligned}$$ Thus $$\begin{array}{c} 2 c_{2}=0 \\ 6 c_{3}+c_{0}=0 \\ 12 c_{4}+c_{1}=0 \\ 20 c_{5}+c_{2}-\frac{1}{6} c_{0}=0 \end{array}$$ and $$\begin{aligned} c_{2} &=0 \\ c_{3} &=-\frac{1}{6} c_{0} \\ c_{4} &=-\frac{1}{12} c_{1} \\ c_{5} &=-\frac{1}{20} c_{2}+\frac{1}{120} c_{0} \end{aligned}$$ Choosing \(c_{0}=1\) and \(c_{1}=0\) we find $$c_{2}=0, \quad c_{3}=-\frac{1}{6}, \quad c_{4}=0, \quad c_{5}=\frac{1}{120}$$ and so on. For \(c_{0}=0\) and \(c_{1}=1\) we obtain $$c_{2}=0, \quad c_{3}=0, \quad c_{4}=-\frac{1}{12}, \quad c_{5}=0$$ and so on. Thus, two solutions are $$y_{1}=1-\frac{1}{6} x^{3}+\frac{1}{120} x^{5}+\cdots \quad \text { and } \quad y_{2}=x-\frac{1}{12} x^{4}+\cdots$$

Let \(k=n+2\) so that \(n=k-2\) and $$\sum_{n=1}^{\infty} n c_{n} x^{n+2}=\sum_{k=3}^{\infty}(k-2) c_{k-2} x^{k}$$.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Measurements

Read Explanation

Radiation

Read Explanation

Physical Quantities And Units

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.