Problem 10 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rr} -3 & -10 \\ 5 & 11 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x}^{\prime} = A\mathbf{x}\) with the given matrix \(A = \begin{bmatrix} -3 & -10 \\ 5 & 11 \end{bmatrix}\) is \(\mathbf{x}(t) = C_1 e^{(4+3i)t}\mathbf{v_1} + C_2 e^{(4-3i)t}\mathbf{v_2}\), where \(\mathbf{v_1} = \begin{bmatrix} 1 \\ \frac{-7-3i}{10} \end{bmatrix}\), \(\mathbf{v_2} = \begin{bmatrix} 1 \\ \frac{-7+3i}{10} \end{bmatrix}\), and \(C_1\) and \(C_2\) are constants depending on initial conditions.

Step 1: Find Eigenvalues of Matrix A

To find the eigenvalues of the matrix A, we need to solve the characteristic equation given by: \[|A - \lambda I| = 0\] Where \(\lambda\) represents the eigenvalues and I is the identity matrix. So, for the given matrix \(A = \begin{bmatrix} -3 & -10 \\ 5 & 11 \end{bmatrix}\), the characteristic equation becomes, \[| \begin{bmatrix} -3-\lambda & -10 \\ 5 & 11-\lambda \end{bmatrix}| = \lambda^2 - 8\lambda + 67 = 0.\]

Step 2: Solve the Characteristic Equation

We have the equation \(\lambda^2 - 8\lambda + 67 = 0\). We need to find the real or complex roots of this equation. Using the quadratic formula, we get \[\lambda = \frac{8 \pm \sqrt{8^2 - 4(67)}}{2} = 4 \pm 3i\] The eigenvalues are \(\lambda_1 = 4+3i\) and \(\lambda_2 =4-3i\).

Step 3: Find the Eigenvectors

Now that we have the eigenvalues, we need to find the corresponding eigenvectors for each. ### Eigenvector for \(\lambda_1=4+3i\) ### Substitute \(\lambda_1\) into the matrix equation \((A-\lambda I)\mathbf{v}=0\), we get \[\begin{bmatrix} -7-3i & -10 \\ 5 & 7-3i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}\] Solve for x and y: \[(-7 - 3i)x -10y = 0\] \[5x + (7 - 3i)y = 0\] Solve for y in terms of x: \[y=\frac{-7 - 3i}{10}x\] The eigenvector for \(\lambda_1 = 4 + 3i\) is \[\mathbf{v_1} =\begin{bmatrix} 1 \\ \frac{-7-3i}{10} \end{bmatrix}.\] ### Eigenvector for \(\lambda_2=4-3i\) ### Substitute \(\lambda_2\) into the matrix equation \((A-\lambda I)\mathbf{v}=0\), we get \[\begin{bmatrix} -7+3i & -10 \\ 5 & 7+3i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}\] Solve for x and y: \[(-7 + 3i)x -10y = 0\] \[5x + (7 + 3i)y = 0\] Solve for y in terms of x: \[y=\frac{-7 + 3i}{10}x\] The eigenvector for \(\lambda_2 = 4 - 3i\) is \[\mathbf{v_2} =\begin{bmatrix} 1 \\ \frac{-7+3i}{10} \end{bmatrix}.\]

Step 4: Find the General Solution

Now that we have the eigenvalues and eigenvectors, we can find the general solution of the given linear system: \[\mathbf{x}(t) = C_1 e^{(4+3i)t}\mathbf{v_1} + C_2 e^{(4-3i)t}\mathbf{v_2}\] Where \(C_1\) and \(C_2\) are constants that depend on the initial conditions of the system.

Key Concepts

Eigenvalues

To grasp the concept of eigenvalues, imagine a special number linked to a matrix that offers insights into its behavior. In simpler terms, eigenvalues help us study systems of linear equations. They provide information about transformations and stability in systems. For a square matrix \( A \), you calculate eigenvalues by solving the characteristic equation:

• The characteristic equation is \(|A - \lambda I| = 0\).
• In this matrix, \(\lambda\) represents the eigenvalues, and \( I \) is the identity matrix of the same size as \( A \).

By solving this equation for the given matrix, you determine the eigenvalues. These solutions could be real or complex numbers, and they describe the possible scaling factors by which the matrix can stretch or shrink vectors.

Eigenvectors

Once you identify the eigenvalues of a matrix, the next step is to discover the corresponding eigenvectors. Think of eigenvectors as special directions that remain consistent after a transformation by the matrix. For a given eigenvalue \( \lambda \), the eigenvector \( \mathbf{v} \) satisfies the equation \((A - \lambda I) \mathbf{v} = 0\). The goal is to solve this system of equations to find \( \mathbf{v} \). In this case:

• Each eigenvalue leads to a unique solution for \( \mathbf{v} \) which remains unchanged in direction when multiplied by the matrix.
• The eigenvectors might be complex, just like the eigenvalues, as they describe vectors in certain vector spaces with complex numbers.

Identifying eigenvectors is crucial as they define the principal axis of transformations, helping to simplify and solve complex systems.

Complex Numbers

Complex numbers are numbers that consist of a real part and an imaginary part. They are pivotal when dealing with certain mathematical fields like solving polynomials and linear systems involving certain matrices. A complex number is written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, with \( i \) being the imaginary unit. Why are they important?

• They allow for a full set of solutions in equations where real numbers fail, such as the square root of negative numbers.
• In the context of linear systems, eigenvalues can frequently be complex, indicating oscillatory behavior in dynamic systems.

Understanding complex numbers broadens the scope of mathematical analysis and allows us to solve problems that would otherwise be insolvable using only real numbers.

Characteristic Equation

The characteristic equation is a key concept in finding the eigenvalues of a matrix, a process that determines the stability and dynamics of systems described by linear equations. For a square matrix \( A \), the characteristic equation is determined by setting \[ |A - \lambda I| = 0 \] where \( |...| \) denotes the determinant and \( I \) is the identity matrix. Solving this equation involves finding the roots, which are the eigenvalues.Here's what you need to remember:

• The characteristic equation is often a polynomial equation whose degree is the same as the size of the matrix.
• The solutions (eigenvalues) of this equation can be real or complex, hinting at various physical properties of the system.
• In our scenario, it resulted in a quadratic equation, easily solved using the quadratic formula, offering insight into the system's behavior.

Understanding the characteristic equation is vital as it aids in recognizing how the system evolves over time or reacts to different conditions.