/*! 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 30 Suppose \(P\) is invertible. Pro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 30

Suppose \(P\) is invertible. Prove that \(\left|P^{-1}\right|=|P|^{-1}\).

Short Answer

Expert verified
For an invertible matrix \(P\), we know that \(PP^{-1} = P^{-1}P = I\). Using the properties of determinants, we have: \(|(PP^{-1})| = |P||P^{-1}|\) Since \(PP^{-1} = I\), we have \(|I| = 1\), so: \(1 = |P||P^{-1}|\) Now solving for \(|P^{-1}|\), we get: \(\frac{1}{|P|} = |P^{-1}|\) Thus, we have proved that \(\left|P^{-1}\right| = |P|^{-1}\) for an invertible matrix \(P\).

Step by step solution

01

Recall the properties of determinants and inverse matrices

Recall the following properties of determinants: 1. \(|AB| = |A||B|\) for any two square matrices \(A\) and \(B\). 2. \(|I| = 1\), where \(I\) is the identity matrix. Also, recall the property of the inverse matrix: 3. For an invertible matrix \(P\), \(PP^{-1} = P^{-1}P = I\).
02

Calculate the determinant of the product of the matrix and its inverse

Using property 1 of determinants, calculate the determinant of the product of matrix \(P\) and its inverse \(P^{-1}\): \(|(PP^{-1})| = |P||P^{-1}|\) However, because of property 3 of inverse matrices, we know that \(PP^{-1} = I\). Thus, we have: \(|I| = |P||P^{-1}|\)
03

Solve for the determinant of the inverse matrix

We know from property 2 that \(|I| = 1\). Therefore: \(1 = |P||P^{-1}|\) Now, we need to solve for \(|P^{-1}|\). To do this, divide both sides of the equation by \(|P|\): \[\frac{1}{|P|} = \frac{|P||P^{-1}|}{|P|}\] Since \(P\) is invertible, its determinant is nonzero, so we can safely divide both sides. This simplifies to: \[\frac{1}{|P|} = |P^{-1}|\] Which is the proof we wanted to show. Thus, for an invertible matrix \(P\), its inverse determinant is equal to the inverse of its determinant: \(\left|P^{-1}\right| = |P|^{-1}\)

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.

Inverse Matrices
When discussing matrices, the term "inverse matrix" refers to a specific type of matrix. In matrix algebra, an inverse matrix is a matrix that, when multiplied by the original matrix, results in the identity matrix.
Think of it as the matrix equivalent of the number 1 in arithmetic. The importance of inverse matrices lies in their ability to "undo" the effect of the original matrix during multiplication.
  • For a matrix \(P\) to have an inverse \(P^{-1}\), it must be a square matrix with non-zero determinant \(|P|\).
  • If the matrix is invertible, the property \(PP^{-1} = P^{-1}P = I\) holds true, where \(I\) is the identity matrix with ones on its diagonal and zeros elsewhere.
The existence of an inverse is crucial in solving systems of linear equations. If a matrix can be inverted, you can multiply both sides of equilibrium equations to isolate and solve for unknowns easily.
Properties of Determinants
Determinants play a significant role in linear algebra, especially in determining whether a matrix is invertible. They are a scalar value calculated from the elements of a square matrix. Understanding the properties of determinants helps in solving various matrix operations.
  • Determinants are used to test if a matrix is invertible. Only matrices with a non-zero determinant have inverses.
  • One key property is that for any square matrices \(A\) and \(B\), \(|AB| = |A||B|\). This property is vital because it simplifies calculations involving matrix products.
  • The identity matrix, denoted as \(I\), has a determinant of 1, \(|I| = 1\). This is crucial because it means when a matrix multiplies its inverse, resulting in this identity matrix, the determinant is preserved as 1.
These properties are used heavily in proofs and practical applications, allowing mathematicians and engineers to simplify complex equations and verify solutions quickly.
Matrix Multiplication
Matrix multiplication is the process of multiplying two matrices to produce a third matrix. Unlike regular multiplication of numbers, multiplying matrices involves a set manner of combining rows and columns.
  • The product of two matrices \(A\) and \(B\) is a new matrix \(C\), where each element \(C_{ij}\) is calculated by taking the dot product of the \(i^{th}\) row of \(A\) and the \(j^{th}\) column of \(B\).
  • For matrix multiplication to be defined, the number of columns in the first matrix \(A\) must equal the number of rows in the second matrix \(B\).
  • Matrix multiplication is associative \((A(BC) = (AB)C)\) and distributive \((A(B + C) = AB + AC)\), but not commutative \((AB eq BA)\) in general. This means the order of multiplication matters.
Understanding matrix multiplication is essential for working with transformations in space and is crucial in fields like computer graphics, engineering, and physics for modeling and computations.

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

Let \(\sigma, \tau \in S_{n} .\) Show that \(\operatorname{sgn}(\tau \circ \sigma)=(\operatorname{sgn} \tau)(\operatorname{sgn} \sigma) .\) Thus, the product of two even or two odd permutations is even, and the product of an odd and an even permutation is odd.

Prove Theorem 8.13: Let \(F\) and \(G\) be linear operators on a vector space \(V\). Then (i) \(\operatorname{det}(F \circ G)=\operatorname{det}(F) \operatorname{det}(G)\), (ii) \(F\) is invertible if and only if \(\operatorname{det}(F) \neq 0\)

\(\operatorname{Let} A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) (a) Find adj \(A\) (b) Show that \(\operatorname{adj}(\operatorname{adj} A)=A\) (c) When does \(A=\operatorname{adj} A ?\)

Find \(\operatorname{det}(M)\) where \(M=\left[\begin{array}{lllll}3 & 4 & 0 & 0 & 0 \\ 2 & 5 & 0 & 0 & 0 \\ 0 & 9 & 2 & 0 & 0 \\ 0 & 5 & 0 & 6 & 7 \\ 0 & 0 & 4 & 3 & 4\end{array}\right]=\left[\begin{array}{lrrrr}3 & 4 & 0 & 0 & 0 \\\ \frac{2}{0} & 5 & 0 & 0 & 0 \\ 0 & \frac{5}{9} & \frac{0}{2} & \frac{0}{0} & -\frac{0}{0} \\ 0 & -\frac{2}{5} & \frac{2}{0} & \frac{0}{6} & -\frac{0}{7} \\ 0 & 0 & 4 & 3 & 4\end{array}\right]\) \(M\) is a (lower) triangular block matrix; hence, evaluate the determinant of each diagonal block: $$ \left|\begin{array}{ll} 3 & 4 \\ 2 & 5 \end{array}\right|=15-8=7, \quad|2|=2, \quad\left|\begin{array}{ll} 6 & 7 \\ 3 & 4 \end{array}\right|=24-21=3 $$ Thus, \(|M|=7(2)(3)=42\).

Find the parity of the permutations \(\sigma=32154, \tau=13524, \pi=42531\) in \(S_{5}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.