Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 6 & -5 & 3 \\ 2 & -1 & 3 \\ 2 & 1 & 1 \end{array}\right] \mathbf{y} $$

Short answer

To summarize, we were given a first-order linear homogeneous ODE system with constant coefficients. We found the eigenvalues and corresponding eigenvectors of the given matrix and used them to form the general solution: \(\mathbf{y}(t) = c_1 e^t \left[\begin{array}{c}\frac{2}{3} \\\ 2 \\\ 1\end{array}\right] + c_2 e^{2t} \left[\begin{array}{c}\frac{5}{2}\\\ 2 \\\ 1\end{array}\right]\) This is the general solution to the system of ODEs.

Step-by-step solution

Find the eigenvalues of the matrix

To find the eigenvalues, we need to compute the characteristic equation of the matrix. The characteristic equation is the determinant of \((\mathbf{A} - \lambda \mathbf{I})\), where \(\mathbf{A}\) is the given matrix, \(\lambda\) represents the eigenvalues, and \(\mathbf{I}\) is the identity matrix. From the given matrix: \(\mathbf{A} = \left[\begin{array}{ccc}4 & -1 & -4 \\\ 4 & -3 & -2 \\\ 1 & -1 & -1\end{array}\right]\) Now, subtract \(\lambda\) from the diagonal elements: \((\mathbf{A} - \lambda \mathbf{I}) = \left[\begin{array}{ccc}4-\lambda & -1 & -4 \\\ 4 & -3-\lambda & -2 \\\ 1 & -1 & -1-\lambda\end{array}\right]\) Now, compute the determinant: \(|\mathbf{A} - \lambda \mathbf{I}| = (4-\lambda)((-3-\lambda)(-1-\lambda)-(-2)(-1))-(-1)(4(-2)-(-1)(-1))-(4)((-3-\lambda)(-1)+(-1)(-1-\lambda))\) After simplifying and factoring: \(-\lambda^3 + 3\lambda^2 - 10\lambda + 10 = (\lambda - 2)(\lambda -1)^2\) The eigenvalues are \(\lambda_1=1\) (with algebraic multiplicity 2) and \(\lambda_2=2\) (with algebraic multiplicity 1).

Determine the eigenvectors for each eigenvalue

Now, we solve the linear system \((\mathbf{A}-\lambda\mathbf{I})\mathbf{y}=0\) to find eigenvectors for each eigenvalue. For \(\lambda_1=1\): \((\mathbf{A}-\mathbf{I})\mathbf{y}_1 = \left[\begin{array}{ccc}3 & -1 & -4 \\\ 4 & -4 & -2 \\\ 1 & -1 & -2\end{array}\right]\mathbf{y}_1=0\) Performing row operations: \(R_2=R_2-\frac{4}{3}R_1 \quad R_3=R_3-\frac{1}{3}R_1\) We have: \(\left[\begin{array}{ccc}3 & -1 & -4 \\\ 0 & 0 & 2 \\\ 0 & 0 & 0\end{array}\right]\mathbf{y}_1 = 0\) Taking k as the free variable, \(\mathbf{y}_1 = k\left[\begin{array}{c}\frac{2}{3} \\\ 2 \\\ 1\end{array}\right]\) For \(\lambda_2=2\): \((\mathbf{A}-2\mathbf{I})\mathbf{y}_2 = \left[\begin{array}{ccc}2 & -1 & -4 \\\ 4 & -5 & -2 \\\ 1 & -1 & -3\end{array}\right]\mathbf{y}_2=0\) Performing row operations: \(R_2=R_2 - 2R_1 \quad R_3=R_3 - 0.5R_1\) We have: \(\left[\begin{array}{ccc}2 & -1 & -4 \\\ 0 & -3 & 6 \\\ 0 & 0 & 0\end{array}\right]\mathbf{y}_2 = 0\) Taking k as the free variable, \(\mathbf{y}_2 = k\left[\begin{array}{c}\frac{5}{2} \\\ 2 \\\ 1\end{array}\right]\)

Form the general solution

We know that the general solution of a homogeneous linear ODE with constant coefficients is: \(\mathbf{y}(t)=c_1e^{\lambda_1 t}\mathbf{y}_1+c_2e^{\lambda_2 t}\mathbf{y}_2\) So, the general solution is: \(\mathbf{y}(t) = c_1 e^t \left[\begin{array}{c}\frac{2}{3} \\\ 2 \\\ 1\end{array}\right] + c_2 e^{2t} \left[\begin{array}{c}\frac{5}{2}\\\ 2 \\\ 1\end{array}\right]\)

Key concepts

Eigenvalues

In linear algebra, eigenvalues play a crucial role in understanding matrices. They are the special scalars \(\lambda\) associated with a square matrix \(\mathbf{A}\). They tell us about the matrix's inherent properties.
To find eigenvalues, we use the characteristic equation, derived from the matrix determinant. This equation is given by \(|\mathbf{A} - \lambda \mathbf{I}| = 0\), where \(\mathbf{I}\) is the identity matrix. Solving this equation reveals the eigenvalues.
In our case, the matrix leads to the characteristic polynomial \(-\lambda^3 + 3\lambda^2 - 10\lambda + 10\). Factoring this gives eigenvalues \lambda_1=1\ and \lambda_2=2\.

Eigenvectors

Eigenvectors are vectors linked to each eigenvalue of a matrix. They define directions in which a particular transformation stretches.
To find them, solve \( (\mathbf{A}-\lambda \mathbf{I})\mathbf{y}=0\). For each eigenvalue, this linear system must be satisfied.
When \(\lambda_1 = 1\), solving gives \{\mathbf{y}_1 = k[\frac{2}{3}, 2, 1]^T\}\. Similarly, for \(\lambda_2 = 2\), we have \{\mathbf{y}_2 = k[\frac{5}{2}, 2, 1]^T\}\. These vectors shape the matrix's behavior.

Characteristic Equation

The characteristic equation is a polynomial equation key for finding eigenvalues. Obtained from the determinant \(|\mathbf{A} - \lambda \mathbf{I}|= 0\), it represents how matrix properties resist transformations.
This equation simplifies matrix operations into algebraic ones and helps uncover matrix traits.
In the exercise, the characteristic equation \(-\lambda^3 + 3\lambda^2 - 10\lambda + 10 = 0\) was solved, yielding \(\lambda_1 = 1\) and \(\lambda_2 = 2\). These solutions describe the inherent nature of the matrix.

Homogeneous Linear Differential Equations

These are fundamental in many scientific fields, often appearing as \(\mathbf{y}’ = \mathbf{Ay}\). Solutions are vectors multiplied by exponential functions.
Their solutions are influenced by eigenvalues and eigenvectors. For example, if \(\lambda\) is an eigenvalue with \(\mathbf{y}\) as its eigenvector, the solution is \(\mathbf{y}(t) = c e^{\lambda t} \mathbf{y}\).
For this exercise, the solution combines results for each eigenvalue, \(\mathbf{y}(t) = c_1 e^t [\frac{2}{3}, 2, 1]^T + c_2 e^{2t} [\frac{5}{2}, 2, 1]^T\). This captures the general behavior of the system.