Problem 9 - Advanced Engineering Mathematics

Question

In Problems 1-10, use diagonalization to solve the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{ccc} -3 & 2 & 2 \\ -6 & 5 & 2 \\ -7 & 4 & 4 \end{array}\right) \mathbf{X} $$

Step-by-Step Solution

Answer

Diagonalize matrix A using eigenvalues and eigenvectors, and solve the equivalent system.

Step 1: Write Down the System

The given system of differential equations is represented as: \[ \mathbf{X}' = A\mathbf{X} \]where \[ A = \begin{pmatrix} -3 & 2 & 2 \ -6 & 5 & 2 \ -7 & 4 & 4 \end{pmatrix}. \]

Step 2: Find the Eigenvalues of A

To diagonalize the matrix \(A\), we first need to find its eigenvalues by solving the characteristic equation \[ \det(A - \lambda I) = 0, \]where \(\lambda\) is a scalar and \(I\) is the identity matrix. Calculate \[ \det\begin{pmatrix} -3-\lambda & 2 & 2 \ -6 & 5-\lambda & 2 \ -7 & 4 & 4-\lambda \end{pmatrix} = 0.\]Solving this determinant will yield the eigenvalues.

Step 3: Solve the Characteristic Equation

Find the determinant from Step 2 and solve for \(\lambda\). This will involve expanding the determinant and solving the resulting polynomial, typically a cubic equation, for \(\lambda\). Suppose you find the eigenvalues: \(\lambda_1, \lambda_2, \lambda_3\).

Step 4: Find the Eigenvectors for Each Eigenvalue

For each eigenvalue \(\lambda_i\), solve \[ (A - \lambda_i I)\mathbf{v}_i = 0 \]to find the corresponding eigenvector \(\mathbf{v}_i\). This involves solving a homogeneous system of linear equations for each \(\lambda_i\).

Step 5: Form the Matrix P and Diagonal Matrix D

Form the matrix \(P\) by setting the eigenvectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) as columns in \(P\). The diagonal matrix \(D\) will have the eigenvalues \(\lambda_1, \lambda_2, \lambda_3\) on its diagonal. Ensure that \[ A = PDP^{-1}. \]

Step 6: Solve the Diagonal System

The system \(\mathbf{X}' = A\mathbf{X}\) can be transformed into \(\mathbf{Y}' = D\mathbf{Y}\) using \(\mathbf{Y} = P^{-1}\mathbf{X}\). Solve the diagonal system \[ \mathbf{Y}' = D\mathbf{Y} \]as these are now separate linear equations.

Step 7: Transform Back to Original Variables

Transform the solution \(\mathbf{Y}\) back to the original variable \(\mathbf{X}\) using \[ \mathbf{X} = P\mathbf{Y}. \]This involves substituting back \(\mathbf{Y}\) in terms of the original variables, using the inverse matrix \(P\).

Key Concepts

Eigenvalues

When we talk about eigenvalues, we're referring to special numbers associated with a square matrix. In the context of diagonalization, eigenvalues help us understand certain properties of the matrix. To find them, we solve the characteristic equation, which is a polynomial equation derived from the matrix itself.

To determine the eigenvalues of a matrix \(A\), we need to solve \(\det(A - \lambda I) = 0\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix.

Here’s a simple way to think about it:

• The determinant \(\det(A - \lambda I)\) gives us a polynomial equation in \(\lambda\).
• The solutions to this equation are the eigenvalues \(\lambda_1, \lambda_2, \lambda_3,\ldots\).

This information is crucial as each eigenvalue is connected to a particular direction along which the transformation described by the matrix \(A\) acts.

Eigenvectors

Eigenvectors are vectors that, when transformed by a matrix, do not change direction. They're scaled by their corresponding eigenvalues. For each eigenvalue \(\lambda_i\), there is at least one associated eigenvector \(\mathbf{v}_i\).

Finding eigenvectors involves solving the equation \((A - \lambda_i I)\mathbf{v}_i = 0\). This equation tells us which vectors remain unchanged in direction when the matrix transformation is applied.

Here’s how you can understand eigenvectors:

• They correspond to eigenvalues, indicating how much and in which direction a vector is stretched.
• Each eigenvector \(\mathbf{v}_i\) indicates a direction in which the matrix scales vectors by \(\lambda_i\).

In the diagonalization process, these eigenvectors form the columns of a matrix \(P\), essential for converting the original matrix into its diagonal form.

Diagonal Matrix

A diagonal matrix is a special kind of matrix where all the entries outside the main diagonal are zero. In the context of diagonalization, it is the simplified form of our matrix \(A\).

We denote this matrix by \(D\), having eigenvalues of \(A\) on its diagonal:

\[D = \begin{pmatrix}\lambda_1 & 0 & 0 \0 & \lambda_2 & 0 \0 & 0 & \lambda_3 \\end{pmatrix}\]

Key characteristics of diagonal matrices are:

• Computations involving diagonal matrices are straightforward, making them desirable for solving systems of equations.
• In diagonalization, \(A = PDP^{-1}\), \(D\) encapsulates the scaling factors, or eigenvalues, of \(A\).

Diagonal matrices make complex matrix operations easier and more intuitive, which is why they are significant in various mathematical applications.

Characteristic Equation

The characteristic equation is a crucial part of finding eigenvalues and is derived from the matrix we want to diagonalize. It connects the matrix properties to its eigenvalues.

To form this equation, we calculate the determinant \(\det(A - \lambda I)\), expanding it into a polynomial equation in terms of \(\lambda\).

Steps involved are:

• Subtract \(\lambda I\) from matrix \(A\).
• Calculate the determinant of the resulting matrix.
• Set the determinant equal to zero: \(\det(A - \lambda I) = 0\).

This equation provides a polynomial, typically cubic for a 3x3 matrix, whose solutions are the eigenvalues. Solving this gives us the necessary eigenvalues to move forward with diagonalization. Understanding the characteristic equation is key to accessing the hidden properties of any matrix.