Question

Find the general solution. $${\mathbf{y}}^{\prime }=\left[\begin{array}{rrr}-2& -12& 10\\ 2& -24& 11\\ 2& -24& 8\end{array}\right]\mathbf{y}$$

Short answer

Question: Calculate the general solution of the given system of linear differential equations using the matrix representation: $$\left(\begin{array}{rrr}-2& -12& 10\\ 2& -24& 11\\ 2& -24& 8\end{array}\right)$$.

Step-by-step solution

Calculate the eigenvalues and eigenvectors

To find the eigenvalues and eigenvectors, we set up the following equation: $$(\mathbf{A}-\lambda \mathbf{I})\mathbf{x}=\mathbf{0}$$ where $$\mathbf{A}$$ is the given matrix, $$\lambda$$ is the eigenvalue, $$\mathbf{I}$$ is the identity matrix and $$\mathbf{x}$$ is the eigenvector. Calculate the determinant of the matrix $$(\mathbf{A}-\lambda \mathbf{I})$$ and find the eigenvalues by solving the characteristic equation: $$\text{det}(\mathbf{A}-\lambda \mathbf{I})=0$$ Calculate the eigenvalues: $$\text{det}\left(\begin{array}{rrr}-2-\lambda & -12& 10\\ 2& -24-\lambda & 11\\ 2& -24& 8-\lambda \end{array}\right)=0$$ Solve the above equation to find the eigenvalues, and for each eigenvalue, substitute it back into the matrix equation $$(\mathbf{A}-\lambda \mathbf{I})\mathbf{x}=\mathbf{0}$$ to find the corresponding eigenvectors.

Construct the general solution

Now, use the matrix exponential $$\mathbf{Y}(t)=\text{exp}(\mathbf{A}t)$$ to find the general solution for the system of linear differential equations: $${\mathbf{y}}^{\prime }=\mathbf{A}\mathbf{y}$$ Assuming $$\mathbf{A}$$ is diagonalizable, the matrix exponential can be calculated using the eigendecomposition of $$\mathbf{A}$$ : $$\mathbf{Y}(t)=\mathbf{P}\text{exp}(\mathbf{D}t){\mathbf{P}}^{-1}$$ where $$\mathbf{D}$$ is the diagonal matrix with the eigenvalues on its diagonal, and $$\mathbf{P}$$ is the matrix with the eigenvectors as columns. If $$\mathbf{A}$$ is not diagonalizable, use the Jordan form of the matrix and its decomposition $$\mathbf{A}=\mathbf{P}\mathbf{J}{\mathbf{P}}^{-1}$$ to compute the matrix exponential, then $$\mathbf{Y}(t)=\mathbf{P}\text{exp}(\mathbf{J}t){\mathbf{P}}^{-1}$$ The general solution of the given system of differential equations is: $$\mathbf{y}=\mathbf{Y}(t)\mathbf{c}$$ where $$\mathbf{c}$$ is an arbitrary constant vector.

Key concepts

Eigenvalues

Understanding eigenvalues is fundamental when solving systems of linear differential equations. They are values of $\lambda$ for which the equation $(\mathbf{A}-\lambda \mathbf{I})\mathbf{x}=\mathbf{0}$ has non-trivial solutions. Here, $\mathbf{A}$ is a given square matrix, $\mathbf{I}$ is the identity matrix of the same size, and $\mathbf{x}$ is the eigenvector corresponding to the eigenvalue.
To find eigenvalues, we calculate the determinant of $\mathbf{A}-\lambda \mathbf{I}$, set it to zero, and solve the resulting characteristic equation. This is because the determinant of a matrix gives information on whether a matrix is invertible, and setting it to zero guarantees non-trivial solutions for $\mathbf{x}$. Such solutions are crucial as they form the basis for the space where the linear transformations act.
In the original exercise, performing this operation gives us the eigenvalues needed to further analyze the differential equations' behavior.

Matrix Exponential

The matrix exponential, noted as $\text{exp}(\mathbf{A}t)$, is an extension of the exponential function to square matrices and plays an essential role in solving linear differential equations systems.
For a matrix $\mathbf{A}$, the exponential $\text{exp}(\mathbf{A}t)$ translates a system of first-order linear differential equations into a form where solutions can be naturally expressed. This transformation leverages the power of exponentials in handling continuous change.
In practical terms, especially when dealing with differential equations, the matrix exponential allows us to express the solution $\mathbf{y}(t)=\text{exp}(\mathbf{A}t)\mathbf{c}$ where $\mathbf{c}$ is a constant vector. In the given problem, calculating the matrix exponential involves understanding eigendecomposition, as it simplifies the computation in circumstances where $\mathbf{A}$ is diagonalizable.

Eigendecomposition

Eigendecomposition involves decomposing a matrix $\mathbf{A}$ into its canonical form. It is expressed as $\mathbf{A}=\mathbf{P}\mathbf{D}{\mathbf{P}}^{-1}$, where $\mathbf{D}$ is a diagonal matrix containing eigenvalues, and $\mathbf{P}$ is a matrix whose columns are formed by corresponding eigenvectors.
This decomposition is particularly useful in simplifying complex matrix operations, like computing the matrix exponential. When $\mathbf{A}$ is diagonalizable, finding $\mathbf{Y}(t)=\text{exp}(\mathbf{A}t)=\mathbf{P}\text{exp}(\mathbf{D}t){\mathbf{P}}^{-1}$ becomes straightforward because the exponential of a diagonal matrix is easily obtained by exponentiating the diagonal elements.
In scenarios where $\mathbf{A}$ is not diagonalizable, alternative methods like Jordan decomposition might be used. However, eigendecomposition remains a preferable strategy in many problems due to its computational efficiency and simplicity.

Linear Algebra

Linear Algebra provides the foundational tools and language to understand structures like matrices and vectors, crucial for analyzing systems of linear differential equations.
This branch deals with concepts such as vector spaces, matrix operations, and transformations. It is the underpinning of finding eigenvalues and eigenvectors, performing matrix decompositions, and understanding complex systems like the one given in the exercise.
Every step in the original problem, from finding the eigenvalues to calculating the matrix exponential, relies on principles and techniques from linear algebra. Understanding the properties of matrices and their role in transformations allows one to translate abstract solutions into concrete answers for differential equations.