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In Problems \(37-40\), use diagonalization to solve the given system. \(\mathbf{X}^{\prime}=\left(\begin{array}{rr}5 & -2 \\ 21 & -8\end{array}\right) \mathbf{X}+\left(\begin{array}{l}6 \\\ 4\end{array}\right)\)

Short Answer

Expert verified
Diagonalize the matrix, solve for eigenvalues/eigenvectors, then transform and solve the system.

Step by step solution

01

Matrix A and Matrix B Identification

First, we identify the given system: \( \mathbf{X}^{\prime} = A \mathbf{X} + \mathbf{B} \), where \( A = \begin{pmatrix} 5 & -2 \ 21 & -8 \end{pmatrix} \) and \( \mathbf{B} = \begin{pmatrix} 6 \ 4 \end{pmatrix} \). We will perform operations on matrix \( A \) to solve this system.
02

Find Eigenvalues of Matrix A

To diagonalize the matrix \( A \), we first find its eigenvalues by solving the characteristic equation \( |A - \lambda I| = 0 \), where \( I \) is the identity matrix of the same size as \( A \). Thus, for \( A \) we solve: \[ \left| \begin{pmatrix} 5-\lambda & -2 \ 21 & -8-\lambda \end{pmatrix} \right| = (5 - \lambda)(-8 - \lambda) - (-2)(21) = 0 \]
03

Compute Characteristic Polynomial

The determinant we compute is: \((5 - \lambda)(-8 - \lambda) + 42 = \lambda^2 + 3\lambda - 82 = 0\). Solve this quadratic equation using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a=1, b=3, c=-82 \).
04

Solve the Quadratic Equation for Eigenvalues

Using the quadratic formula, we find \( \lambda \): \[ \lambda = \frac{-3 \pm \sqrt{3^2 - 4\cdot1\cdot(-82)}}{2} = \frac{-3 \pm \sqrt{9 + 328}}{2} = \frac{-3 \pm \sqrt{337}}{2} \]. The eigenvalues are \( \lambda_1 = \frac{-3 + \sqrt{337}}{2} \) and \( \lambda_2 = \frac{-3 - \sqrt{337}}{2} \).
05

Find Eigenvectors for Each Eigenvalue

For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvectors. Substitute \( \lambda_1 \) and \( \lambda_2 \) one at a time into \( A - \lambda I \) and solve for \( \mathbf{v} \).
06

Construct Matrix P and D

Using the normalized eigenvectors, form matrix \( P \) and the diagonal matrix \( D \) with eigenvalues on the diagonal. Matrix \( P \) is formed by placing each eigenvector as columns, and \( D = \text{diag}(\lambda_1, \lambda_2) \).
07

Solve the Original System

Since \( A = PDP^{-1} \), rewrite the system \( \mathbf{X}^{\prime} = A\mathbf{X} + \mathbf{B} \) in terms of \( P \). Let \( \mathbf{X} = P\mathbf{Y} \), then \( \mathbf{Y}^{\prime} = D\mathbf{Y} + P^{-1}\mathbf{B} \). Solve this to find \( \mathbf{Y}(t) \) using the diagonal matrix \( D \), and convert back to \( \mathbf{X}(t) = P\mathbf{Y}(t) \).
08

Consider the Non-Homogeneous Part

The non-homogeneous term \( \mathbf{B} \) results in an additional particular solution. Combine this with the homogeneous solution of \( \mathbf{Y}(t) \) to get the complete solution \( \mathbf{X}(t) = \mathbf{X}_{homogeneous}(t) + \mathbf{X}_{particular}(t) \).
09

Formulate the Final Solution

Based on all computed components, express the final solution for the system \( \mathbf{X}(t) \) that involves both the homogeneous and particular solutions. Verify by substituting back into the original differential system.

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Key Concepts

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

Eigenvalues
In the context of diagonalization, eigenvalues serve as a foundational concept. These values, usually denoted by \( \lambda \), are crucial for simplifying complex matrix equations. To compute eigenvalues, we work with the characteristic equation \( |A - \lambda I| = 0 \). This involves the determinant of the matrix obtained by subtracting \( \lambda \) times the identity matrix from our original matrix \( A \). This process essentially finds scalar values, \( \lambda \), where the matrix \( A - \lambda I \) becomes non-invertible (i.e., determinant is zero). For our given matrix \( A \): - The characteristic polynomial is determined as \( \lambda^2 + 3\lambda - 82 = 0 \). - We solve this quadratic equation using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides the solutions \( \lambda_1 = \frac{-3 + \sqrt{337}}{2} \) and \( \lambda_2 = \frac{-3 - \sqrt{337}}{2} \). These eigenvalues are essential for constructing the diagonal matrix \( D \) in the diagonalization process.
Eigenvectors
Eigenvectors are non-zero vectors that, when multiplied by a matrix \( A \), result in a scaled version of the original vector, where the scaling factor is the associated eigenvalue. For each eigenvalue found, corresponding eigenvectors are calculated by solving \( (A - \lambda I)\mathbf{v} = 0 \). This equation asserts that the matrix \( A - \lambda I \) applied to vector \( \mathbf{v} \) equals the zero vector.To solve for eigenvectors in our specific example:- Use the eigenvalues \( \lambda_1 \) and \( \lambda_2 \) to set up individual equations.- Solve \( (A - \lambda I)\mathbf{v} = 0 \) for each \( \lambda \) using substitution or row reduction methods to find each \( \mathbf{v} \).The set of all eigenvectors forms a basis used to construct a matrix \( P \) in diagonalization, with each column of \( P \) being an eigenvector.
Matrix Diagonalization
Diagonalization is a method of simplifying matrices to make computations more manageable. It transforms a given square matrix \( A \) into a diagonal matrix \( D \) that is similar to \( A \), with the help of an invertible matrix \( P \).The process involves:- Finding the eigenvalues and eigenvectors of \( A \).- Constructing matrix \( P \) using the eigenvectors as columns.- Forming the diagonal matrix \( D \) with the eigenvalues placed on the diagonal.In mathematical terms, diagonalization means expressing \( A = PDP^{-1} \).In this form,- Matrix \( D \) is diagonal and contains eigenvalues as its entries.- Matrix \( P \) holds the eigenvectors that allow us to revert diagonalized operations back to the context of \( A \).Using diagonalization simplifies solving differential systems, as computing powers of diagonal matrices is computationally much simpler.
Non-Homogeneous Systems
The differential equation provided involves non-homogeneous systems, characterized by an extra term \( \mathbf{B} \) in the system \( \mathbf{X}^{\prime} = A\mathbf{X} + \mathbf{B} \). This non-homogeneous aspect requires additional steps for finding solutions beyond the typical homogeneous solution.In such systems, solving involves:- Separating the problem into homogeneous and non-homogeneous parts.- Solving the homogeneous system using diagonalization.- Finding a particular solution that satisfies the non-homogeneous equation.This is achieved by using the transformation \( \mathbf{X} = P\mathbf{Y} \), leading to \( \mathbf{Y}^{\prime} = D\mathbf{Y} + P^{-1}\mathbf{B} \).The general solution for \( \mathbf{X}(t) \) is a combination of:- The homogeneous solution \( \mathbf{X}_{homogeneous}(t) \), derived from the behavior without \( \mathbf{B} \).- A particular solution \( \mathbf{X}_{particular}(t) \), addressing the influence of \( \mathbf{B} \).Incorporating both solutions provides a comprehensive view of the system's behavior.

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