Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) \mathbf{x} $$

Short answer

The particular fundamental matrix satisfying the initial condition is: $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{2t}+2e^{4t} & -e^{2t}+e^{4t} \\ 3e^{2t}+2e^{4t} & -3e^{2t}+e^{4t} \end{pmatrix} $$

Step-by-step solution

Find eigenvalues and eigenvectors of the matrix A

We have the matrix A: $$ A = \begin{pmatrix} 5 & -1 \\ 3 & 1 \end{pmatrix} $$ To find the eigenvalues, we will solve the characteristic equation: $$ det(A - \lambda I) = \begin{vmatrix} 5-\lambda & -1 \\ 3 & 1-\lambda \end{vmatrix} = (5-\lambda)(1-\lambda) - (-1)(3) = \lambda^2 - 6\lambda + 8. $$ The eigenvalues are the roots of this quadratic equation, which are \(\lambda_1 = 2\) and \(\lambda_2 = 4\). Next, we find the eigenvectors corresponding to these eigenvalues: For \(\lambda_1 = 2\): $$ (A - 2I)\mathbf{v_1} = \begin{pmatrix} 3 & -1 \\ 3 & -1 \end{pmatrix}\mathbf{v_1} = \mathbf{0} \Rightarrow \mathbf{v_1} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ For \(\lambda_2 = 4\): $$ (A - 4I)\mathbf{v_2} = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix}\mathbf{v_2} = \mathbf{0} \Rightarrow \mathbf{v_2} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

Obtain the general solution of the system

Using the eigenvalues and eigenvectors, we can obtain the general solution of the system: $$ \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{4t} $$

Construct the fundamental matrix

Now, we will create the fundamental matrix by putting the solutions as columns: $$ \Psi(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} $$

Find the particular fundamental matrix satisfying \(\Phi(0) = 1\)

To find the particular fundamental matrix \(\mathbf{\Phi}(t)\) satisfying the initial condition \(\Phi(0) = \mathbf{1}\), we will multiply the fundamental matrix \(\Psi(t)\) by a constant matrix \(\mathbf{C}\): $$ \mathbf{\Phi}(t) = \Psi(t) \cdot \mathbf{C} $$ Evaluate \(\Psi(0)\) first: $$ \Psi(0) = \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} $$ Now, we want \(\Phi(0) = \Psi(0) \cdot \mathbf{C} = \mathbf{1}\), so we solve for \(\mathbf{C}\): $$ \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \cdot \mathbf{C} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Rightarrow \mathbf{C} = \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} $$ Now, we have the particular fundamental matrix \(\mathbf{\Phi}(t)\): $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} e^{2t}+2e^{4t} & -e^{2t}+e^{4t} \\ 3e^{2t}+2e^{4t} & -3e^{2t}+e^{4t} \end{pmatrix} $$ So the fundamental matrix is: $$ \Psi(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} $$ And the particular fundamental matrix satisfying the initial condition is: $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{2t}+2e^{4t} & -e^{2t}+e^{4t} \\ 3e^{2t}+2e^{4t} & -3e^{2t}+e^{4t} \end{pmatrix} $$

Key concepts

Eigenvalues

Eigenvalues are fundamental in understanding how matrices affect transformation. When dealing with a matrix equation like \[ A\mathbf{x} = \lambda \mathbf{x} \]the value \( \lambda \) represents the eigenvalue. In simple terms, it measures the factor by which a corresponding vector (eigenvector) is stretched during the transformation facilitated by matrix \( A \).
To find the eigenvalues of a matrix, one needs to solve its characteristic equation. For a 2x2 matrix like this one:\[A = \begin{pmatrix} 5 & -1 \3 & 1 \end{pmatrix}\]the determinant of \( A - \lambda I \) is calculated by:\[det(\begin{pmatrix} 5-\lambda & -1 \3 & 1-\lambda \end{pmatrix}) = \lambda^2 - 6\lambda + 8\]
This quadratic equation results in eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = 4 \). Understanding eigenvalues helps to determine system stability and dynamics side effects in differential equations.

Eigenvectors

Eigenvectors accompany eigenvalues in defining the characteristics of a matrix's transformation. For a given eigenvalue \( \lambda \), the eigenvectors \( \mathbf{v} \) satisfy the equation \[ (A - \lambda I)\mathbf{v} = \mathbf{0} \]This means after applying the matrix to the eigenvector, the result is merely a scaled version of the original eigenvector, with the scale factor being the eigenvalue itself.

• For \( \lambda_1 = 2 \), using matrix \( A \) yields eigenvector:\[ \begin{pmatrix} 1 \3 \end{pmatrix} \]
• For \( \lambda_2 = 4 \), the resulting eigenvector is:\[ \begin{pmatrix} 1 \1 \end{pmatrix} \]

These eigenvectors help form the general solution to the system of differential equations. They indicate directions in space along which the linear transformation by matrix \( A \) occurs.

System of Differential Equations

A system of differential equations contains multiple equations with more than one unknown function. These are quite common in modeling several real-world phenomena. In our problem, the system's primary equation is represented as:\[\mathbf{x}^{\prime} = \begin{pmatrix} 5 & -1 \3 & 1 \end{pmatrix} \mathbf{x}\]where \( \mathbf{x} \) reflects the state of the system over time, the matrix tells us how \( \mathbf{x} \) changes.
Solving this system involves uncovering a general solution, \( \mathbf{x}(t) \), curated by its eigenvalues and eigenvectors:\[\mathbf{x}(t) = c_1 \begin{pmatrix} 1 \3 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \1 \end{pmatrix} e^{4t}\]This illustrates how each component of the state vector evolves as time \( t \) advances.

Characteristic Equation

The characteristic equation is a valuable tool when dealing with square matrices. It is essential for identifying the eigenvalues of the matrix. The equation typically arises from setting the determinant of \( A - \lambda I \), where \( I \) is the identity matrix, equal to zero:\[\det(A - \lambda I) = 0\]In our specific context using:\[A = \begin{pmatrix} 5 & -1 \3 & 1 \end{pmatrix}\]we derive the characteristic equation as:\[\lambda^2 - 6\lambda + 8 = 0\]This quadratic formula can be solved to find eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = 4 \). These roots are crucial for analyzing the dynamics of a system of differential equations, assisting in constructing solutions like the fundamental matrix.