Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -4 & 0 & -1 \\ -1 & -3 & -1 \\ 1 & 0 & -2 \end{array}\right] \mathbf{y} $$

Short answer

After calculating the determinant, we get: $$ (-4-\lambda)((-3-\lambda)(-2-\lambda)) - (-1)(-1)(-2-\lambda) = 0 $$ Now, we need to solve this equation for λ. Simplifying the equation, we obtain: $$ (\lambda^3 + 9\lambda^2 + 26\lambda + 24) - 2\lambda = 0 $$ $$ \lambda^3 + 9\lambda^2 + 24\lambda + 24 = 0 $$ By solving the cubic equation, we find the eigenvalues: λ₁ = -6, λ₂ = -3, and λ₃ = 0. #tag_title#Step 2: Find the eigenvectors for each eigenvalue#tag_content# Now, we have to find the eigenvectors corresponding to each eigenvalue. Let's start with λ₁ = -6. We need to solve the equation (A - λ₁I)v = 0: $$ \left[\begin{array}{rrr} 2 & 0 & -1 \\ -1 & 3 & -1 \\ 1 & 0 & 4 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to λ₁ = -6: v₁ = [1, 1, 2]. Next, let's find the eigenvector for λ₂ = -3: $$ \left[\begin{array}{rrr} -1 & 0 & -1 \\ -1 & 0 & -1 \\ 1 & 0 & 1 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to λ₂ = -3: v₂ = [1, 0, 1]. Finally, let's find the eigenvector for λ₃ = 0: $$ \left[\begin{array}{rrr} -4 & 0 & -1 \\ -1 & -3 & -1 \\ 1 & 0 & -2 \end{array}\right] \left[\begin{array}{r} x \\ y \\ z \end{array}\right] =0 $$ By solving this system of linear equations, we find the eigenvector corresponding to λ₃ = 0: v₃ = [1, 1, -3]. #tag_title#Step 3: Write the general solution using eigenvalues and eigenvectors#tag_content# Now that we have eigenvalues and their corresponding eigenvectors, we can write the general solution to the given system of linear differential equations as follows: $$ \begin{aligned} X(t) = c_1e^{-6t}\begin{bmatrix}1\\1\\2\end{bmatrix} + c_2e^{-3t}\begin{bmatrix}1\\0\\1\end{bmatrix} + c_3e^{0t}\begin{bmatrix}1\\1\\-3\end{bmatrix}. \end{aligned} $$ Here, c₁, c₂, and c₃ are arbitrary constants determined by the initial conditions.

Step-by-step solution

Find the eigenvalues of the matrix

To find the eigenvalues, we need to solve the characteristic equation |A - λI| = 0, where A is the given matrix and λ represents the eigenvalues. So, firstly let's form the matrix (A - λI): $$ \left[\begin{array}{rrr} -4-\lambda & 0 & -1 \\ -1 & -3-\lambda & -1 \\ 1 & 0 & -2-\lambda \end{array}\right] $$ Now, we need to calculate the determinant of this matrix and set it to zero to find the eigenvalues: $$ \begin{vmatrix} -4-\lambda & 0 & -1 \\ -1 & -3-\lambda & -1 \\ 1 & 0 & -2-\lambda \end{vmatrix} =0 $$

Key concepts

Matrix Eigenvalues

When dealing with systems of linear differential equations, one important concept is that of eigenvalues of matrices. Eigenvalues provide significant insights into the properties of a matrix and help determine the behavior of the system defined by the differential equations.

• An eigenvalue is a scalar that indicates by how much the transformation defined by the matrix stretches or shrinks vectors.
• To find eigenvalues, we often subtract a multiple of the identity matrix from the original matrix, forming the matrix \( A - \lambda I \), where \( \lambda \) represents the eigenvalues.
• The eigenvalues are found by solving the equation \( |A - \lambda I| = 0 \), which ensures transformations only stretch or shrink, aligned to an eigenvector, rather than rotating or skewing the input vector.

In simpler terms, eigenvalues give us special numbers that tell us about the intrinsic scaling effect of a linear transformation, which distinguishes the nature and stability of solutions to differential equations.

Characteristic Equation

The characteristic equation is a crucial part of understanding differential equations involving matrices. The matrix transformation equation is derived from considering eigenvalues.

• To derive the characteristic equation, you need to take the determinant of \( A - \lambda I \) and set it to zero: \( |A - \lambda I| = 0 \).
• This results in a polynomial equation in \( \lambda \), which represents the possible eigenvalues.
• The roots of this polynomial give you the eigenvalues. The degree of the polynomial is equal to the size of the matrix, which, in this case, leads to a cubic polynomial because we have a 3x3 matrix.

Think of the characteristic equation as a way to "unlock" the matrix, revealing all its vital characteristics as eigenvalues. Understanding the roots of this equation is fundamental in analyzing the system's dynamic behavior.

General Solution

Once we obtain the eigenvalues and associated eigenvectors from the differential equations, our goal is to find the most complete and robust general solution for the system.

• The general solution comprises all possible specific solutions for the system of differential equations.
• Using the eigenvalues and often corresponding eigenvectors, we can form the solution, typically involving exponential functions.
• The general solution formulates an expression that demonstrates how the solution evolves over time for different initial conditions.

In essence, the general solution blending eigenvalues, eigenvectors, and exponential growth or decay, provides a comprehensive picture of potential behaviors. It covers starting points that differ but adhere to the same dynamic rules of the system, presenting in full view how systems grow, contract, or cycle under the given conditions.