Problem 19
Question
Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(\left[\begin{array}{rrr}-1 & -4 & -2 \\ -4 & -5 & -6 \\ 4 & 8 & 7\end{array}\right]\) [Hint: The eigenvalues of \(A \text { are } \lambda=-1,1 \pm 2 i .]\)
Step-by-Step Solution
Verified Answer
The general solution to the linear system \(\mathbf{x}^{\prime}=A\mathbf{x}\) is:
$$\mathbf{x}(t)=c_{1}\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}e^{-t}+c_{2}\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}e^{t}\cos(2t)+c_{3}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}e^{t}\sin(2t),$$
where \(c_{1},c_{2},\) and \(c_{3}\) are constants.
1Step 1: Write down the given eigenvalues
The hint mentioned provides us with the eigenvalues of matrix \(A\). These eigenvalues are:
$$\lambda_{1}= -1,\quad \lambda_{2} =1+2i,\quad\text{and}\quad\lambda_{3}=1-2i.$$
2Step 2: Finding Eigenvectors
To find the eigenvectors, we'll consider each eigenvalue and find the null space for the matrix \((A-\lambda I)\), where \(I\) is the identity matrix.
For \(\lambda_{1}=-1\), we have the matrix:
$$A+I=\left[\begin{array}{rrr}0 & -4 & -2 \\ -4 & -4 & -6 \\ 4 & 8 & 8\end{array}\right]$$
Row reducing, we get the matrix:
$$\begin{bmatrix}1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$
The null space is given by \(\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}\), so the eigenvector for \(\lambda_{1}=-1\) is:
$$\mathbf{v}_{1}=\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}.$$
Next, we consider the eigenvalue \(\lambda_{2}=1+2i\). In this case, we have the matrix:
$$\begin{bmatrix}-2-2i & -4 & -2 \\ -4 & -6-2i & -6 \\ 4 & 8 & 6-2i\end{bmatrix}$$
Row reducing, we get the matrix:
$$\begin{bmatrix}1 & 2 & i \\ 0 & 0 & 1-2i \\ 0 & 0 & 0\end{bmatrix}$$
The null space is given by \(\begin{bmatrix}-2-i \\ 1 \\ i\end{bmatrix}\), so the eigenvector for \(\lambda_{2}=1+2i\) is:
$$\mathbf{v}_{2}=\begin{bmatrix}-2-i \\ 1 \\ i\end{bmatrix}.$$
Since the matrix is real, the eigenvector for \(\lambda_{3}=1-2i\) is given by the complex conjugate of \(\mathbf{v}_{2}\):
$$\mathbf{v}_{3}=\begin{bmatrix}-2+i \\ 1 \\ -i\end{bmatrix}.$$
3Step 3: Write down the general solution
Now that we have the eigenvalues and their corresponding eigenvectors, we can write down the general solution to the given linear system. For complex eigenvalues, we take the real and imaginary parts of the eigenvectors to form the general solution:
$$\mathbf{x}(t)=c_{1}\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}e^{-t}+c_{2}\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}e^{t}\cos(2t)+c_{3}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}e^{t}\sin(2t),$$
where \(c_{1},c_{2},\) and \(c_{3}\) are constants.
So the general solution to the linear system \(\mathbf{x}^{\prime}=A\mathbf{x}\) is:
$$\mathbf{x}(t)=c_{1}\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}e^{-t}+c_{2}\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}e^{t}\cos(2t)+c_{3}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}e^{t}\sin(2t).$$
Key Concepts
Eigenvalues and EigenvectorsComplex EigenvaluesGeneral Solution of Differential Equations
Eigenvalues and Eigenvectors
In linear algebra, eigenvalues and eigenvectors play a crucial role in understanding linear systems. When dealing with a matrix, eigenvalues represent unique numbers that reveal a consistent direction for a given transformation. In simpler terms, if you apply the transformation represented by the matrix to an eigenvector, the vector’s direction doesn't change—only its magnitude does, scaled by its corresponding eigenvalue.
For a square matrix \(A\), an eigenvalue \(\lambda\) is such that the matrix equation \[ (A - \lambda I)\mathbf{v} = 0 \]has non-zero solutions for the vector \(\mathbf{v}\). Here, \(I\) represents the identity matrix, and \(\mathbf{v}\) is the eigenvector. The expression \((A - \lambda I)\) is often called the characteristic equation of the matrix \(A\). Solving this equation lets you determine the corresponding eigenvectors.
Let's get practical with our example matrix \(A\): \[ \left[\begin{array}{rrr}-1 & -4 & -2 \ -4 & -5 & -6 \ 4 & 8 & 7\end{array}\right] \]The given eigenvalues \(-1, 1 + 2i,\) and \(1 - 2i\) will help us determine the eigenvectors which we need for forming the general solution to the differential equation.
For a square matrix \(A\), an eigenvalue \(\lambda\) is such that the matrix equation \[ (A - \lambda I)\mathbf{v} = 0 \]has non-zero solutions for the vector \(\mathbf{v}\). Here, \(I\) represents the identity matrix, and \(\mathbf{v}\) is the eigenvector. The expression \((A - \lambda I)\) is often called the characteristic equation of the matrix \(A\). Solving this equation lets you determine the corresponding eigenvectors.
Let's get practical with our example matrix \(A\): \[ \left[\begin{array}{rrr}-1 & -4 & -2 \ -4 & -5 & -6 \ 4 & 8 & 7\end{array}\right] \]The given eigenvalues \(-1, 1 + 2i,\) and \(1 - 2i\) will help us determine the eigenvectors which we need for forming the general solution to the differential equation.
Complex Eigenvalues
When a matrix has complex eigenvalues, the eigenvectors associated with these eigenvalues tend to be complex as well. This scenario is common when dealing with matrices that have real numbers but display behaviors involving oscillation or wave patterns.
In our given problem, notice the eigenvalues \(1 + 2i\) and \(1 - 2i\). Here, the real number \(1\) denotes potential growth or decay, whereas \(2i\)—the imaginary component—introduces oscillation, similar to how waves behave.
To find complex eigenvectors, the process follows the same steps as for non-complex eigenvalues. Set up the matrix \((A - \lambda I)\), simplify it, and solve for the vector \(\mathbf{v}\) that satisfies the equation. Let's consider \(\lambda = 1 + 2i\) for our example:
- Construct the matrix \((A - \lambda I)\) using this eigenvalue.- Perform row reduction to solve for the null space.
The resulting eigenvector may include complex entries, as illustrated by \[ \mathbf{v}_{2} = \begin{bmatrix}-2-i \ 1 \ i\end{bmatrix}. \] This eigenvector is complex, and its complex conjugate will serve for the eigenvalue \(1 - 2i\). It’s crucial to remember that solutions involving complex eigenvectors often manifest as trigonometric functions when the actual differential equation is solved.
In our given problem, notice the eigenvalues \(1 + 2i\) and \(1 - 2i\). Here, the real number \(1\) denotes potential growth or decay, whereas \(2i\)—the imaginary component—introduces oscillation, similar to how waves behave.
To find complex eigenvectors, the process follows the same steps as for non-complex eigenvalues. Set up the matrix \((A - \lambda I)\), simplify it, and solve for the vector \(\mathbf{v}\) that satisfies the equation. Let's consider \(\lambda = 1 + 2i\) for our example:
- Construct the matrix \((A - \lambda I)\) using this eigenvalue.- Perform row reduction to solve for the null space.
The resulting eigenvector may include complex entries, as illustrated by \[ \mathbf{v}_{2} = \begin{bmatrix}-2-i \ 1 \ i\end{bmatrix}. \] This eigenvector is complex, and its complex conjugate will serve for the eigenvalue \(1 - 2i\). It’s crucial to remember that solutions involving complex eigenvectors often manifest as trigonometric functions when the actual differential equation is solved.
General Solution of Differential Equations
The solution to a system of linear differential equations combines both the eigenvalues and eigenvectors. Based on their nature (real or complex), different approaches lie ahead.
For our matrix \(A\), we’ve found both a real eigenvalue \(-1\) and complex eigenvalues \(1 \pm 2i\). For real eigenvalues, such as \(-1\), the solution components look straightforwardly exponential:
For the complex eigenpairs, solutions involve trigonometric functions due to Euler's formula, a remarkable link between exponents and trig functions. Each complex pair contributes two components to the solution:
For our matrix \(A\), we’ve found both a real eigenvalue \(-1\) and complex eigenvalues \(1 \pm 2i\). For real eigenvalues, such as \(-1\), the solution components look straightforwardly exponential:
- \[ \mathbf{x}_1(t) = c_1 \mathbf{v}_1 e^{-t}, \]
For the complex eigenpairs, solutions involve trigonometric functions due to Euler's formula, a remarkable link between exponents and trig functions. Each complex pair contributes two components to the solution:
- \[ \mathbf{x}_2(t) = c_2 \text{Re}(\mathbf{v}_2) e^{t} \cos(2t), \]
- \[ \mathbf{x}_3(t) = c_3 \text{Im}(\mathbf{v}_2) e^{t} \sin(2t), \]
Other exercises in this chapter
Problem 18
Let \(A\) be a \(2 \times 2\) real matrix. Prove that all solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\) satisfy $$ \lim _{t \rightarrow \infty} \mathbf{x}(t
View solution Problem 18
Convert the given linear differential equations to a first-order linear system. $$y^{\prime \prime}+a y^{\prime}+b y=F(t), \quad a, b \text { constants. }$$
View solution Problem 19
Convert the given linear differential equations to a first-order linear system. $$y^{\prime \prime \prime}+t^{2} y^{\prime}-e^{t} y=t$$
View solution Problem 19
Solve the initial-value problem \(\mathbf{x}^{\prime}=\) \(A \mathbf{x}, \mathbf{x}(0)=\mathbf{x}_{0}\). $$A=\left[\begin{array}{rrr} 2 & -1 & 3 \\ 3 & 1 & 0 \\
View solution