Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 3 & -4 & -2 \\ -5 & 7 & -8 \\ -10 & 13 & -8 \end{array}\right] \mathbf{y} $$

Short answer

A: The general solution of the given matrix equation is: $$ \mathbf{y}(t) = (c_1 + c_2) \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} e^{-1t} $$ where \(c_1\), \(c_2\), and \(c_3\) are constants.

Step-by-step solution

Write down the given matrix equation.

We are given the matrix equation: $$ \mathbf{y}^{\prime} = \left[\begin{array}{rrr} 3 & -4 & -2 \\ -5 & 7 & -8 \\ -10 & 13 & -8 \end{array}\right] \mathbf{y} $$ We need to find the general solution, which can be written as \(\mathbf{y}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t}\), where \(\lambda_i\) are the eigenvalues, \(\mathbf{v}_i\) are the corresponding eigenvectors, and \(c_i\) are constants.

Find the eigenvalues of the matrix A.

To find the eigenvalues, we need to solve the characteristic equation given by \(\det(A - \lambda I) = 0\). For the provided matrix, this equation becomes: $$ \det \left[\begin{array}{ccc} 3 - \lambda & -4 & -2 \\ -5 & 7 - \lambda & -8 \\ -10 & 13 & -8 - \lambda \end{array}\right] = 0 $$ Computing the determinant, we get \((3 - \lambda)[(-8 - \lambda)(7 - \lambda) -(-8)(13)] -(-4)[-5((-8)- \lambda) - (-8)(-10)] -(-2)[-5(13) -(-10)(7 - \lambda)] = 0\). Simplifying this expression will lead to a cubic equation with eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = 1\), and \(\lambda_3 = -1\).

Find the eigenvectors corresponding to each eigenvalue.

To find the eigenvectors, we need to solve the system of equations \((A - \lambda I)\mathbf{v} = 0\) for each eigenvalue. For \(\lambda_1 = 1\): $$\left[\begin{array}{rrr} 2 & -4 & -2 \\ -5 & 6 & -8 \\ -10 & 13 & -9 \end{array}\right] \mathbf{v}_1 = 0$$ Putting this system in row echelon form gives us $\mathbf{v}_1 = k_1 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix}\(, where \)k_1$ is a constant. For \(\lambda_2 = 1\), the result will be the same as for the first eigenvalue since they are the same. For \(\lambda_3 = -1\): $$\left[\begin{array}{rrr} 4 & -4 & -2 \\ -5 & 8 & -8 \\ -10 & 13 & -7 \end{array}\right] \mathbf{v}_3 = 0$$ Putting this system in row echelon form gives us $\mathbf{v}_3 = k_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\(, where \)k_3$ is a constant.

Write down the general solution.

Using the eigenvalues and the corresponding eigenvectors, we can write the general solution of the given system as: $$ \mathbf{y}(t) = c_1 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_2 \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} e^{-1t} $$ We can simplify the solution by combining the eigenvectors corresponding to the same eigenvalue: $$ \mathbf{y}(t) = (c_1 + c_2) \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} e^{-1t} $$ This is the general solution of the given matrix equation.

Key concepts

Eigenvalues

When solving systems of differential equations, finding eigenvalues is a crucial step. The eigenvalues of a matrix give us important insight into the behavior of the system over time. To find the eigenvalues

• Set up the matrix equation \[ (A - \lambda I) = 0 \] where \( A \) is your original matrix and \( \lambda I \) is the identity matrix scaled by the eigenvalue.
• Calculate the determinant of \( (A - \lambda I) \) to form a polynomial equation, typically cubic for a 3x3 matrix.
• Solve this polynomial for \( \lambda \). Each solution is an eigenvalue.

In our example, solving the characteristic equation yields eigenvalues \( \lambda_1 = 1 \), \( \lambda_2 = 1 \), and \( \lambda_3 = -1 \). Eigenvalues describe whether solutions grow, decay, or oscillate.

Eigenvectors

Once eigenvalues are found, the next step is to determine the corresponding eigenvectors. Eigenvectors show the directions of stretching or compressing.

• For each eigenvalue \( \lambda \), solve \((A - \lambda I)\mathbf{v} = 0\).
• This involves setting up a system of equations and finding non-zero solutions \( \mathbf{v} \).
• Use row reduction to simplify and solve.

For example:- For \( \lambda_1 = 1 \) and \( \lambda_2 = 1 \), the eigenvector is \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} \).- For \( \lambda_3 = -1 \), the eigenvector is \( \mathbf{v}_3 = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \).The eigenvectors align with axes of transformation defined by \( A \).

Matrix Equations

Matrix equations are used to represent systems of differential equations compactly.In the context of differential equations, - \( \mathbf{y}' = A\mathbf{y} \) represents a system where differentiation is described by matrix multiplication.- The matrix \( A \) defines the coefficients affecting each variable.Here's how it works:

• A matrix equation translates a system of linear differential equations into matrix form.
• It serves to simplify and organize computations.

In our exercise, the matrix \(\left[\begin{array}{rrr}3 & -4 & -2 \-5 & 7 & -8 \-10 & 13 & -8\end{array}\right]\)acts upon the vector \( \mathbf{y} \) leading to the solution \( \mathbf{y}(t) \).

General Solution

The general solution of a matrix differential equation combines eigenvalues and eigenvectors.

• It is expressed as a linear combination of solutions: \[\mathbf{y}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + c_3 \mathbf{v}_3 e^{\lambda_3 t}\]
• \( c_1, c_2, c_3 \) are arbitrary constants determined by initial conditions.

In our example:- The repeated eigenvalue \( \lambda_1 = 1 \) appears in two terms, which means we can combine them.- Resulting in: \[\mathbf{y}(t) = (c_1 + c_2) \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} e^{1t} + c_3 \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} e^{-1t}\]This reflects the system's behavior as time progresses.