Question

In each exercise, the coefficient matrix \(A\) of the given linear system has a full set of eigenvectors and is therefore diagonalizable. (a) As in Example 4 , make the change of variables \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), where \(T^{-1} A T=D\). Reformulate the given problem as a set of uncoupled problems. (b) Solve the uncoupled system in part (a) for \(\mathbf{z}(t)\), and then form \(\mathbf{y}(t)=T \mathbf{z}(t)\) to obtain the solution of the original problem.\(\mathbf{y}^{\prime}=\left[\begin{array}{ll}6 & -6 \\ 2 & -1\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r}1 \\\ -3\end{array}\right]\)

Short answer

Answer: The solution to the initial value problem is \(\mathbf{y}(t) = \begin{bmatrix}-21e^{2t} + 22e^{3t} \\ -14e^{2t} + 11e^{3t}\end{bmatrix}\).

Step-by-step solution

Finding eigenvalues and eigenvectors of matrix A

First, we need to find the eigenvalues and eigenvectors of matrix \(A\), given by: \(A = \begin{bmatrix}6 & -6 \\ 2 & -1\end{bmatrix}\) Find the eigenvalues \(\lambda\) by solving the characteristic equation \(|A - \lambda I| = 0\): \(\begin{vmatrix}6-\lambda & -6 \\ 2 & -1-\lambda\end{vmatrix} = (6-\lambda)(-1-\lambda) - (-6)(2) = 0\) Solve the quadratic equation to find the eigenvalues: \(\lambda^{2} - 5\lambda + 8 = 0\) \(\lambda = 2, 3\) Now, find the eigenvectors for each eigenvalue: For \(\lambda = 2\): \((A - \lambda I)x = 0 \Rightarrow \begin{bmatrix}4 & -6 \\ 2 & -3\end{bmatrix}\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix}=0\) This yields the eigenvector \(v_{1}=\begin{bmatrix}3 \\ 2\end{bmatrix}\). For \(\lambda = 3\): \((A - \lambda I)x = 0 \Rightarrow \begin{bmatrix}3 & -6 \\ 2 & -4\end{bmatrix}\begin{bmatrix}x_{1} \\ x_{2}\end{bmatrix}=0\) This yields the eigenvector \(v_{2}=\begin{bmatrix}2 \\ 1\end{bmatrix}\).

Compute the inverse of matrix T

Next, we need to compute the inverse of matrix \(T\), which is formed by the eigenvectors found above: \(T = \begin{bmatrix}3 & 2 \\ 2 & 1\end{bmatrix}\) To find \(T^{-1}\), compute the adjoint of \(T\) and divide each element by the determinant of \(T\): Determinant of \(T\): \(\det(T) = (3)(1) - (2)(2) = -1\) Adjoint of \(T\): \(\operatorname{adj}(T) = \begin{bmatrix}1 & -2 \\ -2 & 3\end{bmatrix}\) \(T^{-1} = \frac{1}{\det(T)} \operatorname{adj}(T) = \begin{bmatrix}-1 & 2 \\ 2 & -3\end{bmatrix}\)

Solve the uncoupled system for z(t)

Now we can find \(\mathbf{z}(t)\) by applying the transformation \(\mathbf{z}(t) = T^{-1}\mathbf{y}(t)\). The given initial condition is \(\mathbf{y}(0) = \begin{bmatrix}1 \\ -3\end{bmatrix}\), so we have: \(\mathbf{z}(0) = T^{-1}\mathbf{y}(0) = \begin{bmatrix}-1 & 2 \\ 2 & -3\end{bmatrix}\begin{bmatrix}1 \\ -3\end{bmatrix} =\begin{bmatrix}-7 \\ 11\end{bmatrix}\) With diagonal matrix \(D = \begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}\), the uncoupled system is: \(\mathbf{z}^{\prime} = D\mathbf{z}\) The solutions for \(\mathbf{z}(t)\) are: \(z_1(t) = C_1e^{2t}\) with \(z_1(0) = C_1e^{0} = -7\), so \(C_1 =-7\) \(z_2(t) = C_2e^{3t}\) with \(z_2(0) = C_2e^{0} = 11\), so \(C_2 = 11\) Thus, we have \(\mathbf{z}(t) = \begin{bmatrix}-7e^{2t}\\11e^{3t}\end{bmatrix}\)

Find the solution y(t)

Finally, we can find the solution \(\mathbf{y}(t)\) using the transformation \(\mathbf{y}(t) = T\mathbf{z}(t)\): \(\mathbf{y}(t) = \begin{bmatrix}3 & 2 \\ 2 & 1\end{bmatrix}\begin{bmatrix}-7e^{2t} \\ 11e^{3t}\end{bmatrix} = \begin{bmatrix}(3)(-7e^{2t}) + (2)(11e^{3t}) \\ (2)(-7e^{2t}) + (1)(11e^{3t})\end{bmatrix}\) So the solution to the original problem is: \(\mathbf{y}(t) = \begin{bmatrix}-21e^{2t} + 22e^{3t} \\ -14e^{2t} + 11e^{3t}\end{bmatrix}\)

Key concepts

Eigenvalues and Eigenvectors

When dealing with matrices, particularly in the context of linear systems and transformations, understanding eigenvalues and eigenvectors is crucial. An eigenvalue of a matrix is a scalar that, when multiplying an eigenvector, results in the eigenvector being only scaled (not direction-changed) by the matrix. This is represented as: \ A \mathbf{v} = \lambda \mathbf{v} \, where \( A \) is a matrix, \( \lambda \) is an eigenvalue, and \( \mathbf{v} \) is an eigenvector.

To find eigenvalues, you solve the characteristic equation \(|A - \lambda I| = 0\), where \( I \) is the identity matrix. Finding the solutions to this equation provides the potential \( \lambda \) values. For each eigenvalue, you can then determine corresponding eigenvectors by evaluating \((A - \lambda I)\mathbf{x} = 0\).

Ultimately, eigenvalues and eigenvectors are fundamental in transforming systems and equations into simpler forms, such as diagonalizing matrices, which can substantially ease solving linear differential equations.

Linear Differential Equations

Linear differential equations involve derivatives of a function and can be expressed in matrix form. In our context, a linear differential equation system can be solved using diagonalizable matrices. This process involves transforming the system into simpler, computationally easier parts.

By diagonalizing the coefficient matrix \( A \) using its eigenvectors, we convert the system into a set of independent equations referred to as an uncoupled system. This is done using a transformation matrix \( T \), constructed from these eigenvectors. The original system \( \mathbf{y}' = A\mathbf{y} \) transforms into \( \mathbf{z}' = D\mathbf{z} \) by setting \( \mathbf{z} = T^{-1}\mathbf{y} \), where \( D \) is a diagonal matrix composed of the eigenvalues of \( A \).

Solving for \( \mathbf{z}(t) \) becomes straightforward because the equations are uncoupled and each involves only one variable.

Matrix Transformation

Matrix transformations are operations that allow us to change the basis or form of a system. They are significant in reducing a complex matrix into a simpler form, such as diagonal or identity matrices. Essentially, these transformations use matrices to convert one vector into another, maintaining key properties such as direction or magnitude in specific conditions.

In our context, with a system \( \mathbf{y}(t) \), we transform it by using a matrix \( T \), composed of eigenvectors, and find its inverse \( T^{-1} \). This helps simplify the differential equation’s complexity by leveraging the relationship \( T^{-1}AT = D \) where \( D \) is diagonal.

Once the original vector \( \mathbf{y}(t) \) is transformed to \( \mathbf{z}(t) \) using \( \mathbf{z}(t)=T^{-1}\mathbf{y}(t) \), solving \( \mathbf{z}(t) \) for the differential equation simplifies the problem considerably, as each differential equation in the transformed set no longer interacts with the others. After solving \( \mathbf{z}(t) \), the solution \( \mathbf{y}(t) \) can be retrieved by reverting the transformation with \( \mathbf{y}(t) = T\mathbf{z}(t) \), thus transforming back to the original system.