/*! 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 58 Suppose \(A\) is any \(3 \times ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 58

Suppose \(A\) is any \(3 \times 3\) non-singular matrix and \((A-3 I)(A-5 I)=O\), where \(l=I_{3}\) and \(O=O_{3}\). If \(\alpha A+\beta A^{-1}=4 I\), then \(\alpha+\beta\) is equal to [Online April 15, 2018] (a) 8 (b) 12 (c) 13 (d) 7

Short Answer

Expert verified
\(\alpha + \beta = 4\) satisfies all conditions.

Step by step solution

01

Understand the Problem

We are given a 3x3 non-singular matrix \(A\) and need to use the equation \((A-3I)(A-5I)=O\). We also have \(\alpha A + \beta A^{-1} = 4I\). We aim to find \(\alpha + \beta\).
02

Expand the Given Polynomial

The equation \((A-3I)(A-5I) = O\) implies that \(A\) is a solution to the quadratic polynomial \(x^2 - 8x + 15 = 0\), where \(x\) represents the eigenvalue. The roots of this polynomial are \(x = 3\) and \(x = 5\). This suggests that the eigenvalues of \(A\) are 3 and 5.
03

Property of Eigenvalues of Matrix Inverse

For a matrix \(A\), if it has eigenvalues \(\lambda_1, \lambda_2, \lambda_3\), the eigenvalues of \(A^{-1}\) are \(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}\). Given the eigenvalues of \(A\) are 3, 5, and possibly another eigenvalue (which must be 1 because \((A-3I)(A-5I)=O\) is zero with multiplicity 1), the eigenvalues of \(A^{-1}\) are \(\frac{1}{3}, \frac{1}{5}, 1\).
04

Solve the Matrix Equation

We have \(\alpha A + \beta A^{-1} = 4 I\). By considering eigenvalues, the relation becomes \(\alpha\lambda + \beta \frac{1}{\lambda} = 4\) for each eigenvalue \(\lambda\). Substitute \(\lambda = 3, 5, 1\) in sequence.
05

Use Eigenvalue Substitution

For \(\lambda = 3\), \(3\alpha + \frac{1}{3}\beta = 4\). For \(\lambda = 5\), \(5\alpha + \frac{1}{5}\beta = 4\). For \(\lambda = 1\), \(\alpha + \beta = 4\).
06

Solve the System of Equations

Solve these equations:1. \(3\alpha + \frac{1}{3}\beta = 4\) 2. \(5\alpha + \frac{1}{5}\beta = 4\) 3. \(\alpha + \beta = 4\).The third equation directly gives \(\alpha + \beta = 4\), satisfying all conditions since the system has consistent solutions based on eigenvalue properties.

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.

Eigenvalues
Eigenvalues are special numbers associated with a square matrix. In essence, if you have a matrix \(A\) and a non-zero vector \(v\), the matrix transforms the vector as \(Av = \lambda v\). Here, \(\lambda\) is the eigenvalue. Essentially, eigenvalues give you a sense of how the matrix scales when it acts on certain vectors.
  • Understanding eigenvalues is crucial because they only scale the vector without changing its direction.
  • The example equation \((A-3I)(A-5I) = O\) from the exercise indicates that the eigenvalues of matrix \(A\) are 3 and 5.
  • When you know the eigenvalues, you can make powerful assertions about the matrix. For instance, they help determine characteristics of the transformation matrix \(A\).
Remember, eigenvalues are not just random numbers but convey the essence of matrix behavior. In our case, they help solve equations related to the matrix and also relate to the matrix's invertibility and the scalar equations given in the original problem.
Matrix Inverse
A matrix inverse is akin to dividing in regular arithmetic. When you have a non-singular matrix \(A\), its inverse \(A^{-1}\) is a matrix that results in the identity matrix when multiplied by \(A\) itself, i.e., \(A \cdot A^{-1} = I\).
  • The exercise discusses \(A^{-1}\), critical because it features in the equation \(\alpha A + \beta A^{-1} = 4I\).
  • Understanding the inverse is crucial for solving equations involving matrices. It "undoes" the transformation implied by the matrix \(A\).
  • The relation with eigenvalues is also interesting. As detailed, if \(A\) has eigenvalues \(\lambda_1, \lambda_2, \lambda_3\), its inverse has eigenvalues \(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}\).
Thus, knowing about the inverse allows you to resolve equations and find relationships between variables, much like in regular algebra.
Polynomial Function
Polynomial functions in matrix algebra often help describe properties of matrices, like eigenvalues. In our problem, the term \((A-3I)(A-5I) = O\) indicates a polynomial in terms of the matrix \(A\).
  • This can be visualized as providing an equation: \(A^2 - 8A + 15I = O\), allowing determination of eigenvalues.
  • Through polynomials, one can find the characteristic equation of a matrix, which is essential in computing eigenvalues.
  • Eigenvalues root from such polynomial equations. Hence, solving them gives essential matrix features.
Working with polynomial equations gives powerful insights into behavior and characteristics of matrices, which otherwise would be hard to determine.
Non-Singular Matrix
A non-singular matrix is simply one that is invertible, meaning it has an inverse. For a \(3 \times 3\) matrix \(A\), if \(|A| eq 0\), it's non-singular.
  • Having a non-singular matrix assures us that solutions like \(Ax = b\) have unique solutions.
  • The term "non-singular" automatically signals that \(A\) can be expressed in terms of its inverse \(A^{-1}\).
  • In the exercise, knowing \(A\) is non-singular lets us safely work with its inverse in the equation \(\alpha A + \beta A^{-1} = 4I\).
Recognizing a matrix as non-singular opens a wide range of methods, particularly involving inverses, to unravel complex problems.

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

The set of all values of \(\lambda\) for which the system of linear equations \(x-2 y-2 z=\lambda x\) \(x+2 y+z=\lambda y\) \(-x-y=\lambda 2\) has a non-trivial solution: \(\quad\) [Jan. 12, 2019 (II)] (a) is a singleton (b) contains exactly two elements (c) is an empty set (d) contains more than two elements

if \(\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=a x-12\), then ' \(a\) ' is equal to: [Online April 11, 2015] (a) 24 (b) \(-12\) (c) \(-24\) (d) 12

The number of values of \(k\) for which the system of linear equations, \((k+2) x+10 y=k, k x+(k+3) y=k-1\) has no solution, is \(\quad\) [Online April 16, 2018] (a) Infinitely many (b) 3 (c) 1 (d) 2

If \(\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|=\alpha(a+b)(b+c)(c+a) \neq 0\) then \(\alpha\) is equal to \(\quad\) Online May 12, 2012] (a) \(a+b+c\) (b) \(a b c\) (c) 4 (d) 1

Let \(a, b, c\) be such that \(b(a+c) \neq 0\) if [2009] \(\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0\) then the value of \(n\) is : (a) any even integer (b) any odd integer (c) any integer (d) zero
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.