/*! 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 40 True or false (with a counterexa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 40

True or false (with a counterexample if false and a reason if true): (a) A 4 by 4 matrix with a row of zeros is not invertible. (b) A matrix with 1s down the main diagonal is invertible. (c) If \(A\) is invertible then \(A^{-1}\) is invertible. (d) If \(A^{\mathrm{T}}\) is invertible then \(A\) is invertible.

Short Answer

Expert verified
(a) True. (b) True. (c) True. (d) True.

Step by step solution

01

Analyze Part (a)

Determine whether a 4 by 4 matrix with a row of zeros is invertible. A matrix is invertible if its determinant is non-zero. A row of zeros in a matrix implies that its determinant is zero because the matrix cannot have full rank (rank less than the number of rows). Hence, such a matrix is not invertible. Therefore, the statement is **True**.
02

Analyze Part (b)

Examine whether a matrix with 1s down the main diagonal is invertible. A matrix with 1s on the main diagonal is an identity matrix. Identity matrices have determinants equal to 1, which is non-zero, indicating that they are always invertible. Thus, the statement is **True**.
03

Analyze Part (c)

Verify if the inverse of an invertible matrix, denoted as \(A^{-1}\), is also invertible.If a matrix \(A\) is invertible, then there exists a matrix \(A^{-1}\) such that \(AA^{-1} = I\), where \(I\) is the identity matrix. The property of invertibility implies that \(A^{-1}\) must also have an inverse, which is \(A\), because \((A^{-1})^{-1} = A\).Consequently, the statement is **True**.
04

Analyze Part (d)

Check if the transpose of a matrix \(A^{\mathrm{T}}\) being invertible implies that \(A\) itself is invertible.The invertibility of the transpose of a matrix implies that the original matrix is also invertible. This is because the determinant of \(A\) and \(A^{\mathrm{T}}\) are equal, and if one is non-zero, the other must also be non-zero, ensuring both are invertible.Therefore, the statement is **True**.

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.

Determinant of a Matrix
The determinant is a special number that you can calculate from a square matrix. It plays a key role in many areas of linear algebra, primarily in determining whether a matrix is invertible. When examining a matrix, the determinant tells us if the matrix has an inverse. If the determinant is zero, the matrix is not invertible, meaning there's no matrix you can multiply it by to get the identity. If the determinant is non-zero, then the matrix does have an inverse.

Calculating the determinant involves a specific formula that changes with the matrix size, but essentially it sums products of elements such that signs alternate. It's important to understand this concept because:
  • A zero determinant indicates linear dependence among rows or columns.
  • It affects solutions to systems of linear equations, as a zero determinant usually means no unique solutions.
Matrix Transpose
The transpose of a matrix, denoted as \(A^{\mathrm{T}}\), is a new matrix in which the rows and columns are swapped. This allows you to reflect the matrix over its main diagonal. For a matrix \(A\) with elements \(a_{ij}\), the transpose \(A^{\mathrm{T}}\) will have elements \(a_{ji}\).

The transpose has some interesting properties:
  • The determinant of \(A\) is equal to the determinant of \(A^{\mathrm{T}}\).
  • Transposing twice gets you back to the original matrix, \((A^{\mathrm{T}})^{\mathrm{T}} = A\).
  • The transpose of a product of two matrices equals the product of their transposes in reverse; \((AB)^{\mathrm{T}} = B^{\mathrm{T}}A^{\mathrm{T}}\).
These properties make the transpose particularly useful in proofs and transformations in linear algebra.
Identity Matrix
An identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere. It is denoted as \(I\) and plays a critical role in matrix multiplication. In the context of matrix inversion, the identity matrix acts like the number 1 in regular multiplication. Any matrix multiplied by its inverse results in the identity matrix: \(AA^{-1} = I\).

Some key features of the identity matrix include:
  • The identity matrix is always square (same number of rows and columns).
  • Its determinant is 1, indicating it is always invertible.
  • Multiplying any matrix by an identity matrix leaves the original matrix unchanged, \(AI = IA = A\).
Understanding the identity matrix helps you grasp the concept of invertibility and how it relates to solving matrix equations.
Matrix Rank
The rank of a matrix is an important number that tells us the maximum number of linearly independent row vectors or column vectors in the matrix. It relates directly to the concept of invertibility and the determinant.

High-ranked matrices are desirable as they guarantee full control over linear transformations. Here's why:
  • A square matrix is invertible if and only if its rank is equal to the number of rows (or columns), ensuring full rank.
  • The rank indicates the dimension of the column space or row space, giving insight into the span of vectors represented by the matrix.
  • In a practical sense, a matrix with full rank implies that its rows or columns form a basis of the space, which is crucial for solving systems of equations.
Understanding the rank of a matrix allows you to determine crucial properties such as its invertibility, solution spaces, and dependencies among vectors.

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

Suppose \(A\) is invertible and you exchange its first two rows to reach \(B\). Is the new matrix \(B\) invertible? How would you find \(B^{-1}\) from \(A^{-1}\) ?

Here is a new factorization of \(A\) into triangular times symmetric: Start from \(A=L D U\). Then \(A\) equals \(L\left(U^{T}\right)^{-1}\) times \(U^{\mathrm{T}} D U\). Why is \(L\left(U^{T}\right)^{-1}\) triangular? Its diagonal is all 1s. Why is \(U^{\mathrm{T}} D U\) symmetric?

(a) The matrix \(\left((A B)^{-1}\right)^{\mathrm{T}}\) comes from \(\left(A^{-1}\right)^{\mathrm{T}}\) and \(\left(B^{-1}\right)^{\mathrm{T}} .\) In what order? (b) If \(U\) is upper triangular then \(\left(U^{-1}\right)^{\mathrm{T}}\) is ________ triangular.

The less familiar form \(A=L P U\) exchanges rows only at the end: $$ A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 3 \\ 2 & 5 & 8 \end{array}\right] \rightarrow L^{-1} A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 3 & 6 \end{array}\right]=P U=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\left[\begin{array}{lll} 1 & 1 & 1 \\ 0 & 3 & 6 \\ 0 & 0 & 2 \end{array}\right] $$ What is \(L\) is this case? Comparing with \(P A=L U\) in Box 1 J, the multipliers now stay in place \(\left(\ell_{21}\right.\) is 1 and \(\ell_{31}\) is 2 when \(A=L P U\) ).

Write down the 3 by 3 finite-difference matrix equation ( \(h=\frac{1}{4}\) ) for $$ -\frac{d^{2} u}{d x^{2}}+u=x, \quad u(0)=u(1)=0 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

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.