Question

In Exercises \(1-16\) find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The general solution to the system of ODEs is given by: $$ \mathbf{y} = c_1 e^{t} (-3, 2, 1) + c_2 e^{2t} (1, 0, -1) + c_3 e^{3t} (1, 1, 1) $$

Step-by-step solution

Compute the eigenvalues of the matrix

Compute the eigenvalues of the given matrix A: $$ A=\left[\begin{array}{rrr} 3 & -3 & 1 \\\ 0 & 2 & 2 \\\ 5 & 1 & 1 \end{array}\right] $$ To find the eigenvalues \(\lambda\), we need to solve the characteristic equation: $$ \text{det}(A-\lambda I) = 0 $$ where I is the identity matrix. The determinant is as follows: $$ \text{det}\begin{vmatrix} 3-\lambda & -3 & 1 \\\ 0 & 2-\lambda & 2 \\\ 5 & 1 & 1-\lambda \end{vmatrix}= 0 $$ Upon computing the determinant, we find that \(\lambda_1 = 1\), \(\lambda_2 = 2\), \(\lambda_3 = 3\).

Compute the eigenvectors

Now, we need to find the eigenvectors associated with each eigenvalue. We do this by solving the equation \((A-\lambda I)\mathbf{v}=0\) for each eigenvalue. For \(\lambda_1 = 1\): $$ (A-\lambda_1 I)\mathbf{v}_1=\left[\begin{array}{rrr} 2 & -3 & 1 \\\ 0 & 1 & 2 \\\ 5 & 1 & 0 \end{array}\right]\mathbf{v}_1 = 0 $$ Solving this system gives us the eigenvector \(\mathbf{v}_1 = (-3, 2, 1)^T\). For \(\lambda_2 = 2\): $$ (A-\lambda_2 I)\mathbf{v}_2=\left[\begin{array}{rrr} 1 & -3 & 1 \\\ 0 & 0 & 2 \\\ 5 & 1 & -1 \end{array}\right]\mathbf{v}_2 = 0 $$ Solving this system gives us the eigenvector \(\mathbf{v}_2 = (1, 0, -1)^T\). For \(\lambda_3 = 3\): $$ (A-\lambda_3 I)\mathbf{v}_3=\left[\begin{array}{rrr} 0 & -3 & 1 \\\ 0 & -1 & 2 \\\ 5 & 1 & -2 \end{array}\right]\mathbf{v}_3 = 0 $$ Solving this system gives us the eigenvector \(\mathbf{v}_3 = (1, 1, 1)^T\).

Construct the general solution

Now that we have the eigenvalues and associated eigenvectors, we can construct the general solution to the system of ODEs: $$ \mathbf{y} = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3 $$ Substituting our values, we get the general solution: $$ \mathbf{y} = c_1 e^{t} (-3, 2, 1) + c_2 e^{2t} (1, 0, -1) + c_3 e^{3t} (1, 1, 1) $$

Key concepts

Eigenvalues

To find the general solution of a differential equation involving matrices, we first need to find the eigenvalues of the matrix. Eigenvalues are special numbers associated with a matrix that give us valuable insights into the matrix's behavior. They are derived from the characteristic equation:

• The characteristic equation is found by solving the determinant of \( (A - \lambda I) = 0 \).
• Here, \( A \) is the given matrix, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix.

In our example, solving the determinant equation gives us three eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = 2 \), \( \lambda_3 = 3 \). These eigenvalues are crucial as they help determine the matrix's properties and the form of the general solution.
Understanding eigenvalues aids in analyzing stability, oscillations, and other dynamic behaviors in systems, making them vital in engineering and physics.

Eigenvectors

Once the eigenvalues are determined, the next step is finding the eigenvectors. Eigenvectors are vectors associated with each eigenvalue that describe the directions in which the matrix acts by simply stretching or compressing the vector.

• To find them, solve the system \( (A - \lambda I)\mathbf{v} = 0 \) for each eigenvalue.
• This results in distinct eigenvectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \) corresponding to each eigenvalue.

In our specific example:

• For \( \lambda_1 = 1 \), the eigenvector is \( \mathbf{v}_1 = (-3, 2, 1)^T \).
• For \( \lambda_2 = 2 \), the eigenvector is \( \mathbf{v}_2 = (1, 0, -1)^T \).
• For \( \lambda_3 = 3 \), the eigenvector is \( \mathbf{v}_3 = (1, 1, 1)^T \).

Eigenvectors, together with eigenvalues, provide a comprehensive picture of how linear transformations behave in vector spaces and are widely used in data analysis, quantum mechanics, and more.

Characteristic Equation

The characteristic equation is a polynomial equation derived from setting the determinant of \( (A - \lambda I) \) to zero. It is a key step in finding eigenvalues and is crucial for understanding matrix transformations.

• The equation is constructed as \( \text{det}(A - \lambda I) = 0 \).
• The degree of the polynomial corresponds to the size of the matrix.
• For a 3x3 matrix, it results in a cubic polynomial.

Solving this equation gives us the eigenvalues, which are essential for forming the general solution of differential equations.
This characteristic polynomial reflects the intrinsic properties of the matrix and helps in solving various matrix-related problems in linear algebra and applied mathematics.